TEACHING NOTES ELECTROSTATICS-1 JEE Syllabus : Coulomb’s law; Electric field and potential; Electrical potential energy of a system of point charges and of electrical dipoles in a uniform electrostatic field; Electric field lines; Flux of electric field; Gauss’s law and its application in simple cases, such as, to find field due to infinitely long straight wire, uniformly charged infinite plane sheet and uniformly charged thin spherical shell.
Complete in 12 LECTURE Introduction A number of simple experiments demonstrate the existence of electric forces and charges. For example, after running a comb through your hair on a dry day, you will find that the comb attracts bits of paper. The attractive force is often strong enough to suspend the paper. Another simple experiment is to rub an inflated balloon with wool. The balloon then adheres to a wall, often for hours. When materials behave in this way, they are said to be electrified or to have become electrically charged. You can easily electrify your body by vigorously rubbing your shoes on a wool rug. Evidence of the electric charge on your body can be detected by lightly touching (and startling) a friend. Under the right conditions, you will see a spark when you touch, and both of you will feel a slight tingle. (Experiments such as these work best on a dry day because an excessive amount of moisture in the air can cause any charge you build up to “leak” from your body to the Earth. Electric and magnetic phenomenon are generally bracketed together, since both derive from charged particles. Magnetism, arises from charges in motion. However, in the frame of reference where all charges are at rest, the forces are purely electrical. The subject of electrostatics, as the name suggest deals with the physics of charges at rest. Electro (Related to charge) + statics (stationary). Hence it deals with stationary charges Properties of Electric Charges (i) Charge comes in two varieties, which are called “plus” and “minus”, Like charges repel each other and unlike charges attract each other. Only two kinds of electric charges exist because any unknown charge that is found experimentally to be attracted to a positive charge is also repelled by a negative charge. No one has ever observed a charged object that is repelled by both a positive and a negative charge. (ii)
Charge is conserved : The charge of an isolated system is conserved. The algebraic sum of charges in any electrically isolated system does not change.
(iii)
Charge is quantized : Protons and electron are considered the only charge carriers. All desirable charges must be integral multiples of e. If an object contains n1 protons and n2 electrons, the net charge on the object is n1 (e) + n2 (– e) = (n1 – n2) e. Thus, the charge on any object is always an integral multiple of e and can be charged only in steps of e, i.e. charge is quantized. The step size e is usually, so small that we can easily neglect the quantization. Now, 1C contains n units of basic charge e where n=
1C ~ 6 × 1012 1.6 10 19 C
The step size is thus very small as compared to the charges usually found on many cases we can assume a continuous charge variation. This was verified by millikan oil drop experiment. Page-1
Meaning of a charged body A material is said to be charged if there are more of one kind of charge than the other. A negatively charged body has excess electrons over protons, while a positively charged body has excess positive charges over electrons. The protons are tightly bound in the nucleus, making them very difficult to remove. Charging a body therefore involves the removal, addition, and rearrangement of the orbital electrons. A body becomes positively charged if it loses electrons, and negatively charged if electrons are added to it. Now we study ways in which this addition, removal or rearrangement is achieved in practice. WAYS OF CHARGING 1. Charging by friction Charging by friction is the oldest form of charging. It was found that when an amber rod is rubbed with fur, the rod became negatively charged. The two bodies acquire opposite signs of electricity ; one gets positively charged, while the other becomes negatively charged. When two bodies are charged by friction, they acquire the same magnitude of charge. Furthermore, the bodies retain these excess charges even when they are separated from each other. Note : Charging involves transformation of mass. 2.
Charging by conduction In charging by contact, an uncharged body is brought into contact with a charged body. The uncharged body acquires the same sign of charge as the charged body. The total charge is distributed between the two bodies. uncharged body
A
3.
+ + + + +
+ + +
B + + +
+ + + +
bodies retain charge on separation
charging by contact
charged body
+
+ +
+
+
+
+
+
A
+ +
B + + +
+
+ +
+
A +
+ +
+
+
+ +
+
+
B +
+ +
+
Charging by induction In charging by induction, an uncharged body is brought close to, but not touching, a charged body. Charging by induction is an example of the rearrangement of charges between bodies. The sign of the induced charge is opposite to that of the inducing charge. Furthermore, the induced charges last only while the inducing charge is present. The induced charged disappear when the inducing charge is taken away. (Charge can be retained if use grounding)
Asking question Three objects are brought close to each other, two at a time. When objects A and B are brought together, they attract. When objects B and C are brought together, they repel. From this, we conclude that (a) objects A and C possess charge of the same sign. (b) objects A and C possess charges of opposite sign. (c) all three of the objects possess charges of the same sign (d) one of the objects is neutral (e) we need to perform additional experiment to determine information about the charges on the objects. Note : Neutral does not mean “chargeless”. Note : The bodies around is are almost neutral because there is microscopic balance of -ve and +ve charge.
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Asking question If a negatively charged body is attracting a body of unknown nature of charged body What are the possible signs of the unknown charged body. – –– – – – – – –
?
-ve charged body
Answer It can have any charge as a –ve charged body can definitely attract +ve, charged body, It can as well attract neutral or -ve charged body because of induction positive charge on a neutral body will be closer to the -ve charge. Although the charge on +ve body should be small so that effect of the distance overrule the slight effect of repulsion. Note : Attraction is not the proof the nature of charge on bodies.
COULOMB’S LAW From experimental observations on the electric force, Coulomb’s law can be expressed, giving the magnitude of the electric force of interaction between two point charges: Fe = ke
| q1 || q 2 | r2
where ke is a constant The Coulomb constant ke in SI units has the value ke = 8.987 5 × 109 N·m2/C2 9 × 109 N·m2/C2 This constant is also written in the form 1 ke = 4 0
Ex.1
Sol.
where the constant 0 (lowercase Greek epsilon) is known as the permittivity of free space and has the value. 0 = 8.854 2 × 10–12 C2 / N · m2 The electric force is a conservative force. The electron and proton of a hydrogen atom are separated (on the average) by a distance of approximately 5.3 × 10–11 m. Find the magnitudes of the electric force and the gravitational force between the two particles. Charge and mass of the electron and proton Particle Charge (C) Mass (kg) –19 Electron (e) – 1.602 1917 × 10 9.109 5 × 10–31 –19 Proton (P) + 1.602 191 × 10 1.672 61 × 10–27 From Coulomb’s law, (1.60 1019 C) 2 | e | | e | 9 2 2 Fe = ke = (8.99 × 10 N · m / C ) = 8.2 × 10–8 N (5.3 10 11 m) 2 r2 Using Newton’s law of universal gravitation
Fg = G
me m p
= (6.67 × 10–11
r2 The ratio Fe / Fg 2 × 1039.
N·
m2
/ kg2) ×
(9.11 10 31 kg ) (1.67 10 27 kg) = 3.6 × 10–47 N (5.3 10 11 m) 2
Gravitational force between charged atomic particles is negligible when compared with the electric force. Page-3
Coulomb’s law in vector form Remember that force is a vector quantity and must be treated accordingly. In vector form the force on charge q1 due to charge q2 is expresed as q1q 2 rˆ r2
F12 = ke
{q1 & q2 should be substituted with sign.}
ˆr is a unit vector directed from q2 toward q1 F12
r
–
+
q1 +
F21
F12
^r
q2
+
F21 (b)
(a)
When the charges are of the same sign, the force is repulsive. When the charges are of opposite signs, the force is attractive. The electric force obeys Newton’s third law, the electric force exerted by q2 and q1 is equal in magnitude to the force exerted by q1 on q2 and in the opposite direction ; that is, F21 = – F12. If q1 and q2 are of opposite sign, as shown in figure (b), the product q1q2 is negative. A negative product indicates an attractive force, so that the charges each experience a force toward the other. Coulomb forces follows principle of Superposition. i.e. the resultant force on any one of the charge is equal to the vector sum of the forces exerted by the various individual charges. For example, if four charges are present, then the resultant force exerted by charge 2, 3 and 4 on charge 1 is F1 F12 F13 F14 Charges in medium : Consider two point charges q1 and q2 kept in a medium of permittivity r (r 1 and is 1 for vacuum). The medium will be consisting of atoms which are neutral but not chargeless. If any substance is present in vicinity of a charge, the positive and negative charges of atom (nuclei and electrons) experience electric force, which in turn leads to a partial separation of these charges. These partially separated charges present on atoms apply additional electric force on charges which in combination with the force of interaction between q1 and q2 gives the resultant field. If we know the force of interaction and the additional electric force due to induced charges, we can forget about the presence of the substance itself while calculating the resultant force, since the role of the substance has already been taken into account with the help of induced charges. Thus, the resultant force in the presence of a substance is determined simply as the superposition of the external field and the field of induced charges.
r +
q•1
– q•2
(medium is )
F1 = F12 + F force on 1 due to polarisation.
1 q1 q 2 F1 represents net force on 1 and = 4 r2 r 0
q1 q 2 But force on q1 due to q2 is still F12 = 4 r 2 0 1/ 2 Page-4
Eg.
Very simple Eg Consider three point charges located at the corners of a right triangle as shown in figure, q1 = q3 = 5.0 C, q2 = = – 2.0C, and a = 0.10m. Find the resultant force exerted on q3. y q2 –
a
F13
F23
a
+ q3
2a x
q1 +
Sol.
F23 = 9.0 N in the coordinate system shown in figure, the attractive force F23 is to the left (in the negative x direction). The magnitude of the force F13 exerted by q1 on q3 is F3 = 11N the repulsive force F23 makes an angle of 45° with the x axis. Combing F13 with F23 by the rules of vector addition. To demostrate the principle of superposition.
Eg.
Two identical small charged spheres, each having a mass of 3.0 × 10–2 kg, hang in equilibrium as shown in figure. The length of each string is 0.15m, and the angle is 5.0°. Find the magnitude of the charge on each sphere. //////////////////////////////// L
L
q
a
q
L=0.15m = 5.0°
Sol.
Fx = T sin – Fe = 0 Fy = T cos – mg = 0 |q| = 4.4 × 10–8C There is no way we could find the sign of the charge from the information given. In fact, the sign of the charge is not important. The situation will be exactly the same whether both spheres are positively charged or negatively charged. (1) (2)
Asking question Suppose we propose solving this problem without the assumption that the charges are of equal magnitude. We claim that the symmetry of the problem is destroyed if the charges are not equal, so that the strings would make two different angles with the vertical, and the problem would be much more complicated. How would you respond? Ans. You should argue that the symmetry is not destroyed and the angles remain the same. Newton’s third law requires that the electric forces on the two charges be the same, regardless of the equality or none quality of the charges. The solution of the example remains the same. The symmetry of the problem would be destroyed if the masses of the spheres were not the same. In this case, the strings would make different angle with the vertical and the problem would be more complicated.
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The electric field The concept of a field was developed by Michael Faraday. An electric field is said to exist in the region of space around a charged object. When another charged object, qo (the test charge) enters this space, we say the test charge experiences an electric force, Fe due to this field. We define the electric field due to the source charge at the location of the test charge to be the electric force on the test charge per unit charge. Q q0 +
+ + + +
Fe E q 0
E
+ + + + + + ++
The vector E has the SI units of newtons per coulomb (N/C). The direction of E, as shown figure, is the direction of the force a positive test charge experiences when placed in the field It is important to remember that Electric field E is produced by charge Q, which is separate from the field produced by the test charge itself. Also, note that the an electric field is a property of its source, the presence of the test charge is not necessary for the field to exist. The test charge is used to detect the electric field. Fe When using E q , we must assume that the test charge q0 is small enough that it does not disturb the 0
charge distribution responsible for the electric field. If vanishingly small test charge q0 is placed near a uniformly charged metallic sphere, as in figure (a) the charge on the metallic sphere, which produces the electric field, remains uniformly distributed. If the test charge is great enough (q0 >> q0), as in figure (b) the charge on the metallic sphere is redistributed and the ratio of the force to the test charge is different : (F’e / q’0 Fe/q0). That is, because of this redistribution of charge on the metallic sphere, the electric field it sets up is different from the field it sets up in the presence of the much smaller test charge q0.. + q0 – – –
– – – –
(a)
– –
– – – –
– – –
+ q0 >> q0
(b)
Why concept of electric field necessary? Modern understanding of electric interaction between two charges is visualized in terms of the electric field concept. A charge produces an electric field around itself ; this field then exerts force on the other charge. Thus, the interaction between two charges is a two step process.
For two charges, the measurable quantity is the force on a charge which can be directly determined using Coulomb’s law . Why then introduce this intermediate quantity called the electric field? When charges are stationary, the concept of electric field is convenient, but not really necessary. Electric field in electrostatics is an elegant way of characterising the electrical environment of a system of charges. The true physical significance of the concept of electric field, however,emerges only when we go beyond electrostatics and deal with time-dependent electromagnetic phenomena. Suppose we consider the force between two distant charges q1, q2 in accelerated motion. The greatest Page-6
speed with which a signal or information can go from one point to another is c, the speed of light. Thus, the effect of any motion of q1 on q2 cannot arise instantaneously. There will be some time delay between the effect (force on q2) and the cause (motion of q1). It is precisely here that the notion of electric field (strictly, electromagnetic field) is natural and very useful. The field picture is this: the accelerated motion of charge q1 produces electromagnetic waves, which then propagate with the speed c, reach q2 and cause a force on q2. The notion of field elegantly accounts for the time delay. Thus, even though electric and magnetic fields can be detected only by their effects (forces) on charges, they are regarded as physical entities, not merely mathematical constructs. They have an independent dynamics of their own, i.e., they evolve according to laws of their own. They can also transport energy. Thus concept of field is now among the central concepts in physics. Field due to a point charge According to Coulomb’s law, the force exerted by point charge q on the test charge q0 is : qq Fe k e 20 rˆ r q0
q +
^r
F
F
P
r
q + (a)
^r
P r (b)
where rˆ is a unit vector directed from q toward q0. By E = Fe/q0, the electric field created by q is : q E e k e 2 rˆ r The source charge sets up an electric field at point P, directed away from q. q0 E q +
^r
P F (c)
q ^r –
P
r (d)
If q is negative, as in figure (c), the force on the test charge is toward the source charge, so the electric field at P is directed toward the source charge, as in figure (d)
Superposition of electrostatic fields If we are dealing with many charges then electric field at a point p is the vector sum q E k e 2i rˆi i r1 where ri is the distance from the ith source charge q, to the point P and rˆi is a unit vector directed from qi toward P. If some more charge are added, more terms are added to the summation. However, there is no change to the terms that were already there, provided that the original charges do not move. If we know the electric fields generated by two different sets of charges separately, the electric field generated by both together is simply the vector sum of the two separate fields. The two fields, which each occupy three-dimensional space, are superimposed on one another. Because it has this property, the electric field is said to satisfy the principle of superposition.
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Illustrations Ex. Two identical positive point charges q are placed on the x-axis at x = – a and x = + a, as shown in figure. (i) Plot the variation of E along the x-axis. (ii) Plot the variation of E along the y-axis. E
+q
+q x=–a
Sol.
x
x=+a
(i)
E on point charge is undefined The variation of along the x-axis. Electric field directed along the positive x-axis is taken as positive
(ii)
The direction of electric field along the positive y-axis is taken as positive. Ey =
2kqy ( y a 2 )3 / 2
E
2
Emax =
–a /2
8 kq 27 a 2
a /2
y
occurs at
a 2 Three point charges lie along the x axis as shown in figure. The positive charge q1 = 24.0 C is at x = 3.00m, the positive charge q2 = 6.00 C is at the origin, and the resultant force acting on q3 is zero. What is the x coordinate of q3? y=±
Ex.
3.00m x q2
Sol.
ke
+
F23
3.00m-x – q3
F13
+ q1
x
| q1 | | q 3 | | q 2 | | q3 | = k e (3.00 x ) 2 x2
Noting that ke and |q3| are common to both sides and so can be dropped, we solve for x and find that (3.00 – x)2 |q2| = x2 |q1| This can be reduced to the following quadratic equation : solving this quadratic equation for x, we find that the positive root is x = 1m. There is also a second root, x = –3m. This is another location at which the magnitudes of the forces on q3 are equal, but both forces are in the same direction at this location.
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Asking question Suppose charge q3 is constrained to move only along the x axis. From its initial position at x = 1m, it is pulled a very small distance along the x axis. When released, will it return to equilibrium or be pulled further from equilibrium? That is, is the equilibrium stable or unstable? Ans. If the charge is moved to the right, F13 becomes larger and F23 becomes smaller. This results in a net force to the right, in the same direction as the displacement. Thus, the equilibrium is unstable. Note that if the charge is constrained to allowed to move up and down along y -axis in figure, the equilibrium is stable. In this case, if the charge is pulled upward (or downward) and released, will move back toward the equilibrium position and undergo oscillation Explain using graph of E vs x.} To learn stable, unstable and nuetral equilibrium.
Ex.
Four identical charges are fixed at the corners of a square of side a. Find electric field at point p which is at a distance z lying on the line perpendicular to the plane of the square passing through the centre of square. p
q
q q
Ans.
q
qz . 2 2 q 0a 3
Electric field of a continuous charge distribution : The total electric field at P due to all elements in the charge distribution is approximately. q E k e 2 i rˆi r1 i Considering the charge distribution as continuous, the total field at P in the limit qi 0 is q lim E k e qi 0 2 i rˆi = k dq rˆ e 2 r i r1 OPTIONAL
1. 2. 3.
Problem solving tactics for calculating the electric field from continuous charge distributions Identify the type of charge distribution and compute the charge density or . Divide the charge distribution into infinitesimal charges dq, each of which will act as a tiny point charge. The amount of charge dq, i.e., within a small element dl, dA or dV is dq = dl (charge distributed in length) dq = dA (charge distributed over a surface) dq = dV (charge distributed throughout a volume) Page-9
4.
5. 6. 7.
Draw at point P the dE vector produced by the charge dq. The magnitude of dE is 1 dq dE = 4 2 0 r Vector dE is along radial line joining dq to P, dE is directed away for positive charge dq while directed towards dq for negative dq. Resolve the dE vector into its components. Identify any special symmetry features to show whether any component(s) of the field that are not cancelled by other components. Write the distance r and any trigonometric factors in terms of given coordinates and parameters. The electric field is obtained by summing over all the infinitesimal contributions. E dE =
dq
4 r 2 0
8.
Perform the indicated integration over limit of integration that include all the source charges.
Ex.
A thin rod of length has a uniform positive charge per unit length . Calculate the electric field at a point P that is located along the long axis of the rod and a distance a from one end y x E
Ans.
P
dq=dx
dx
x a
l
k el a (l a )
Asking question Suppose we move to a point P very far away from the rod. What is the nature of the electric field at such a point ? Ans. If P is far from the rod (a >> ), then in the denominator of the final expression for E can be neglected, and E kel/a2. Electric field due to finite rod at perpendicular distance x from the wire. y = x tan x dy = x sec2 d
K dy cos dEx = (x sec ) 2
dEy = so
Ex =
P
y
K dy sin ( x sec ) 2
dy
K (sin 2 – sin 1) x
K (cos 1 – cos 2) x 1 and 2 are to be used with sign.
Ey =
For eg in this figure
1 =
,2 = 3 4
/3 /4
P
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(i)
Results: For semi infinite wire E x
K = x
E y
=
O
K x
2 K at 45° with OP. x For infinite wire
E= (ii)
O
K x Ey = 0
Ex = 2
Electric field due to arc = charge distribution / unit length. Find field at the centre due to arc. So only the horizontal components are cancelled and only vertical components are added. /2
dE = 0
°r
K ( R d) cos R2
dr
R
d
2K sin 2 E net = R
Rd
d
Electric field due to the ring (at a point P lying a distance x from its center along the central axis perpendicular to the plane of the ring) dq + + +
+ + + +
a
r
+
+
+
+
x
P dEx
+
+ +
+
dE dE
The magnitude of the electric field at P due to the segment of charge dq is dq r2 This field has an x component dEx = dE cos along the x-axis and a component dE perpendicular to the x-axis. The resultant field at P must lie along the x-axis because the perpendicular components of the field created by any charge element is canceled by the perpendicular component created by an element on the opposite side of the ring.
dE = ke
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1 +
+ + + +
+ +
+
+
dE2
+
+
+
+ +
dE1
+
2
kex dq x dq dEx = dE cos = k e 2 = 2 ( x a 2 )3 / 2 r r All elements of the ring make the same contribution to the field at P because they are all equidistant from this point. Thus, we can integrate to obtain the total field at P k x
( x 2 ea 2 )3 / 2 dq
Ex =
=
kex Q ( x a 2 )3 / 2 2
1
dq + + +
+ +
a
+
r
+
x
+
+
+
+
P dEx
+
+
+ +
+ +
+
+
Ex.
+ + +
+ +
+
+
dE dE
+
+ +
A thread carrying a uniform charge per unit length has the configurations shown in Fig. a and b. Assuming a curvature radius R to be considerably less than the length of the thread, find the magnitude of the electric field strength at the point O.
O R
R (a) [Ans. (a) E =
Ans.
dE1
+
2
O
Ex.
dE2
(b)
2 ; (b) E = 0 ] 40 R
Suppose a negative charge is placed at the center of the ring in and displaced slightly by a distance x << a along x-axis. When released, what type of motion does it exhibit ? In the expression for the field due to a ring of charge, we let x << a, which results in
k eQ x a3 Thus, from Fe = qE , the force on a charge –q placed near the center of the ring is Ex =
k e qQ x a3 Because this force has the form SHM, the motion will be simple harmonic. Fx = –
Always look for the condition of SHM as F –x Page-12
Asking question What will be the changes in the result if the ring is non uniformly charged.
Ex.
A system consists of a thin charged wire ring of radius R and a very long uniformly charged thread oriented along the axis of the ring, with one of its ends coinciding with the centre of the ring. The total charge of the ring is equal to q. The charge of the thread (per unit length) is equal to . Find the interaction force between the ring and the thread.
q [Ans. F = 4 R ] 0
Electric field due to disk (at a point P that lies along the central perpendicular axis of the disk and a distance x from the center of the disk. Let disk has radius R has a uniform surface charge density . dq R r x
P
dr
The ring of radius r and width dr shown in has a surface area equal to 2rdr. The charge dq on this ring is 2r dr. Using this result of field due to the ring kex dEx = 2 2 3 / 2 (2r dr) (x r ) To obtain the total field at P, we integrate this expression over the limits r = 0 to r = R. x is a constant R
Ex = kex
2r dr
(x 2 r 2 )3 / 2 0
x = 2k e 1 2 2 1/ 2 ( x R ) This result is valid for all values of x > 0 and x<0. but for x = 0 the answer is not valid as thereis discontinuty at x = 0. We can calculate the field close to the disk along the axis by assuming that R >> x ; thus, the expression in bracket reduces to unity to give us the near-field approximation. Ex = 2ke = 2 0 * Explain that the field created by a uniformly charged infinite sheet is same..
E
x
O
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ELECTRIC FIELD LINES and FLUX Concept of field lines We have already studied electric field in the few lectures. It is a vector quantity and can be represented as we represent vectors. Let us try to represent E due to a point charge pictorially. Let the point charge be placed at the origin. Draw vectors pointing along the direction of the electric field with their lengths proportional to the strength of the field at each point. Since the magnitude of electric field at a point decreases inversely as the square of the distance of that point from the charge, the vector gets shorter as one goes away from the origin, always pointing radially outward. Figure. shows such a picture.
In this figure, each arrow indicates the electric field, i.e., the force acting on a unit positive charge, placed at the tail of that arrow. Connect the arrows pointing in one direction and the resulting figure represents a field line. We thus get many field lines, all pointing outwards from the point charge. Relative density of field lines represent magnitude of electric field Have we lost the information about the strength or magnitude of the field now, because it was contained in the length of the arrow? No. Now the magnitude of the field is represented by the density of field lines. E is strong near the charge, so the density of field lines is more near the charge and the lines are closer. Away from the charge, the field gets weaker and the density of field lines is less, resulting in wellseparated lines. Another person may draw more lines. But the number of lines is not important. In fact, an infinite number of lines can be drawn in any region. It is the relative density of lines in different regions which is important. We draw the figure on the plane of paper, i.e., in two dimensions but we live in three-dimensions. So if one wishes to estimate the density of field lines, one has to consider the number of lines per unit cross-sectional area, perpendicular to the lines. Since the electric field decreases as the square of the distance from a point charge and the area enclosing the charge increases as the square of the distance, the number of field lines crossing the enclosing area remains constant, whatever may be the distance of the area from the charge. We started by saying that the field lines carry information about the direction of electric field at different points in space. Having drawn a certain set of field lines, the relative density (i.e., closeness) of the field lines at different points indicates the relative strength of electric field at those points. The field lines crowd where the field is strong and are spaced apart where it is weak. Figure shows a set of field lines.
Relative density of field lines is inversely propotional to square of distance We can imagine two equal and small elements of area placed at points R and S normal to the field lines there. The number of field lines in our picture cutting the area elements is proportional to the magnitude of field at these points. The picture shows that the field at R is stronger than at S. To understand the dependence of the field lines on the area, or rather the solid angle subtended by an area element, let us Page-14
try to relate the area with the solid angle, a generalization of angle to three dimensions. Recall how a (plane) angle is defined in two-dimensions. Let a small transverse line element l be placed at a distance r from a point O. Then the angle subtended by l at O can be approximated as = l/r. Likewise, in three-dimensions the solid angle subtended by a small perpendicular plane area S, at a distance r, can be written as = S/r2. We know that in a given solid angle the number of radial field lines is the same. In Fig., for two points P1 and P2 at distances r1 and r2 from the charge, the element of area subtending the solid angle is r12 at P1 and an element of area r22 at P2, respectively. The number of lines (say n) cutting these area elements are the same. The number of field lines, cutting unit area element is therefore n / (r12 ) at P1 and n / (r22 ) at P2, respectively. Since n and are common, the strength of the field clearly has a 1 / r2 dependence. Drawing field lines The picture of field lines was invented by Faraday to develop an intuitive non- mathematical way of visualizing electric fields around charged configurations. Faraday called them lines of force. This term is somewhat misleading, especially in case of magnetic fields. The more appropriate term is field lines (electric or magnetic) that we will use. Electric field lines are thus a way of pictorially mapping the electric field around a configuration of charges. An electric field line is, in general, a curve drawn in such a way that the tangent to it at each point is in the direction of the net field at that point.An arrow on the curve is obviously necessary to specify the direction of electric field from the two possible directions indicated by a tangent to the curve. A field line is a space curve, i.e., a curve in three dimensions. Figure shows the field lines around some simple charge configurations.
As mentioned earlier, the field lines are in 3-dimensional space, though the figure shows them only in a plane. The field lines of a single positive charge are radially outward while those of a single negative charge are radially inward. The field lines around a system of two positive charges (q, q) give a vivid pictorial description of their mutual repulsion, while those around the configuration of two equal and opposite charges (q, –q), a dipole, show clearly the mutual attraction between the charges. Properties of field lines: The field lines follow some important general properties: (i) Field lines start from positive charges and end at negative charges. If there is a single charge, they may start or end at infinity. (ii) In a charge-free region, electric field lines can be taken to be continuous curves without any breaks. (iii) Two field lines can never cross each other. (If they did, the field at the point of intersection will not have a unique direction, which is absurd.) Page-15
(iv)
Electrostatic field lines do not form any closed loops. This follows from the conservative nature of electric field.
Illustration 1. Figure shows the sketch of field lines for two point charges 2Q and – Q.
E=0 2Q
–Q
•
The pattern of field lines can be deduced by considering the following points: (a) Symmetry : For every point above the line joining the two charges there is an equivalent point below it. Therefore, the pattern must be symmetrical about the line joining the two charges. (b) Near field : Very close to a charge, its own field predominates. Therefore, the lines are radial and spherically symmetric. (c) Far field : Far from the system of charges, the pattern should look like that of a single point charge of value (2Q – Q) = + Q, i.e., the lines should be radially outward. (d) Null point : There is one point at which E = 0. No lines should pass through this point. (e) Number of lines : Twice as many lines leave + 2Q as entre – Q. Figure (a) shows the incorrectly and (b)shows the correctly drawn field lines for a collection of four charges – Q, 4Q, – 2Q and +Q. -2Q
-2Q
-Q
+4Q
P
(a)
+4Q
-Q
-Q
(b)
-Q
ELECTRIC FLUX Analogy with flow of water and concept of flux Consider flow of a liquid with velocity v, through a small flat surface dS, in a direction normal to the surface. The rate of flow of liquid is given by the volume crossing the area per unit time v dS and represents the flux of liquid flowing across the plane. If the normal to the surface is not parallel to the direction of flow of liquid, i.e., to v, but makes an angle with it, the projected area in a plane perpendicular to v is v dS cos . Therefore the flux going out of the surface dS is v . nˆ dS. For the case of the electric field, we define an analogous quantity and call it electric flux. We should however note that there is no flow of a physically observable quantity unlike the case of liquid flow. In the picture of electric field lines described above, we saw that the number of field lines crossing a unit area, placed normal to the field at a point is a measure of the strength of electric field at that point. This means that if we place a small planar element of area S normal to E at a point, the number of field lines crossing it is proportional to ES. Now suppose we tilt the area element by angle . Clearly, the number of field lines crossing the area element will be smaller. The projection of the area element normal to E is S cos. Thus, the number of field lines crossing S is proportional to ES cos . When = 90°, field lines will be parallel to S and will not cross it at all (Figure). Page-16
Area as a vector The orientation of area element and not merely its magnitude is important in many contexts. For example, in a stream, the amount of water flowing through a ring will naturally depend on how you hold the ring. If you hold it normal to the flow, maximum water will flow through it than if you hold it with some other orientation. This shows that an area element should be treated as a vector. It has a magnitude and also a direction. How to specify the direction of a planar area? Clearly, the normal to the plane specifies the orientation of the plane. Thus the direction of a planar area vector is along its normal. How to associate a vector to the area of a curved surface? We imagine dividing the surface into a large number of very small area elements. Each small area element may be treated as planar and a vector associated with it, as explained before. Notice one ambiguity here. The direction of an area element is along its normal. But a normal can point in two directions. Which direction do we choose as the direction of the vector associated with the area element? This problem is resolved by some convention appropriate to the given context. For the case of a closed surface, this convention is very simple. The vector associated with every area element of a closed surface is taken to be in the direction of the outward normal. This is the convention used in Fig.
Thus, the area element vector S at a point on a closed surface equals S nˆ where S is the magnitude of the area element and nˆ is a unit vector in the direction of outward normal at that point. Vectorial definition of flux We now come to the definition of electric flux. Electric flux through an area element S is defined by = E.S = E S cos which, as seen before, is proportional to the number of field lines cutting the area element. The angle here is the angle between E and S. For a closed surface, with the convention stated already, is the angle between E and the outward normal to the area element. Notice we could look at the expression E S cos in two ways: E (S cos ) i.e., E times the projection of area normal to E, or ES i.e., component of E along the normal to the area element times the magnitude of the area element. The unit of electric flux is N C–1 m2. Page-17
The basic definition of electric flux given by Eq. = E.S = E S cos can be used, in principle, to calculate the total flux through any given surface. All we have to do is to divide the surface into small area elements, calculate the flux at each element and add them up. Thus, the total flux through a surface S is ~ E ·S The approximation sign is put because the electric field E is taken to be constant over the small area element. This is mathematically exact only when you take the limit S 0 and the sum in Eq. ~ E ·S is written as an integral.
Gauss’s Law Let us again consider a positive point cahrge q located at the centre of a sphere of radius r, as shown in figure Gaussian surface r +
q
dA E
q ˆ we know that the magnitude of the electric field everywhere on the surface r2 r of the sphere is E = keq/r2. The field lines are directed radially outward and hence are perpendicular to the surface at every point on the surface. That is, at each surface point, E is parallel to the vector Ai representing a local element of area Ai srrounding the surface point. Therefore, E ·Ai = EAi
from equation E = ke
and from equation E = E · dA = E n dA we find that net flux through are gaussian surface is E = E · dA = E dA = E dA where we have moved E outside of the integral because, by symmetry, E is constant over the surface and given by E = keq / r2. Furthermore, because the surface is spherical dA = A = 4r2. Hence, the net flux through the gaussian surface is k eq (4r2) = 4keq r2 we know that ke = 1/4 0, we can write this equation in the form
E =
q E = 0 q Note from equation E = that the net flux through the spherical surface is proportional to the charge 0
inside. The flux is indepdent of the radius r because the area of the spherical surface is proportional to r2, whereas the electric field is proportional to 1/r2. Thus, in the product of area and electric field, the dependence on r cancels.
Page-18
Asking question What if the charge in figure were not located at the center of the spherical gaussian surface ? Gaussian surface r +
q
Ans.
dA E
In this case, the situation does not possess enough symmetry to evaluate the electric field. Because the charge is not at the center, the magnitude of E would vary over the surface of the sphere and the vector E would not be everywhere perpendicular to the surface. But still the flux through the spherical surface is q / 0 Now consider several closed surfaces surrounding a charge q, as shown in. S3 S2
S1
q Surface S1 is spherical, but surface S2 and S3 are not. From equation E = , the flux that passes 0
through S1 has the value q / 0. As we discussed in the preceding section, flux is proportional to the number f electric field lines passing through a surface. The construction shown in figure shows that the number of lines through S1 is equal to the number of lines through the nonspherical surfaces S2 and S3. Thefore, we conclude that the net flux through any closed surface surrounding a point charge q is given by q / 0 and is independent of the shape of that surface. Flux through closed surface for a charge kept outside the surface Now consider a point charge located outside a closed surface of arbitary shape, as shown.
q
As you can see from this construction, any electric field line that enters the surface leaves the surface at another point. The number of electric field lines entering the surface equals the number leaving the surface. Therefore, we conclude that the net electric flux through a closed surface that surrounds no charge is zero. Page-19
Gauss’ Law for multiple charges Let us calculate flux for many charges. We once again use the superposition principle, which states that the electric field due to many charges is the vector sum of the electric fields produced by the individual charges. Therefore, we can express the flux through any closed surface as : E . d A ( E E .....). d A 1 2 where E is the total electric field at any point on the surface produced by the vector addition of the electric fields at that point due to the individual charges. Consider the system of charges shown in figure. S q•1
q•4
•q2
• q3 S S
The surface S surrounds only one charge, q1 ; hence, the net flux through S is q1 / 0. The flux through S due to charges q2, q3 and q4 outside it is zero because each electric field line that enters S at one point leaves it at another. The surface S’ surrounds charges q2 and q3 ; hence, the net flux through it is (q2 + q3)/ 0. Finally, the net flux through surface S” is zero because there is no charge inside this surface. That is, all the electric field lines that enter S” at one point leave at another. Notice that charge q4 does not contribute to the net flux through any of the surfaces because it is outside all of the surface. Gauss’s law, which is a generalization of what we have just described, states that the net flux through any closed surface is : q E . d A in E = 0
where qin represents the net charge inside the surface and E represents the electric field at any point on the surface. Some points to be emphasized about the gauss law :
(i)
It is true for any closed surface no matter what its shape is size.
(ii)
The q includes sum of all charges enclosed by the surface.
(iii)
(iv)
In situations when the surface is so chosen that there are some charges inside and some outside, the E (whose flux appear in the equation) is due to all charges, just term ‘q’ in the law represents only total charge inside.
The gaussian surface should not pass through any discrete charge however, it can pass through a continuous charge distribution.
Application of gauss law Definition of a Gaussian surface : While applying Gauss’s law we are interested in evaluating the integral E =
E · dA
The closed surface for which the flux is calculated is generally an imaginary or hypothetical surface, called a Gaussian surface. Whenever we apply Gauss’s law we may choose a surface of any size and shape as our Gaussian surface. But selecting a proper size and shape for a Gaussian surface is a key factor for determining flux. Here are the list of different types of the Gaussian surfaces to be chosen for a given charge distribution. Page-20
Charge distribution Point charge Spherical charge distribution Line of charge Planar charge Ex
Gaussian surface Spherical Spherical Cylindrical Cylindrical
Electric field Radial Radial Radial Normal to surface
Consider a cube of edge a, kept in a uniform electric field of magnitude E, directed along x-axis as shown in Figure. dA3
3
D C A
E
B
dA1
l G
1
dA2 x
z
E l 4
2
F dA
Sol.
The net flux is algebraic sum of the flux through all the faces of the cube. Note that flux through faces ABFE, BCGE, ADHE, CDHG is zero because E is normal to area vector on these faces. Flux through EFGH, (E)EFGH = E(a2) cos 180° = – Ea2 Area vector and electric field vector are opposite to each other. Flux through ABCD, (E)ABCD = E(a2) cos 0° = + Ea2 Area vector and electric field vector are parallel to each other. Net flux over all the six faces is (E) = (E)ABFE + (E)BCGF + (E)ADHE + (E)CDHG + (E)ABCD + (E)EFHG = 0 + 0 + 0 + 0 + ( – Ea2) + Ea2 =0
Ex.
Find electric flux through square of side a, due to charge placed at distance a/2 from centre of a square. a/2
Sol.
Let us enclose the charge q by a cubical gaussian surface with q at its centre.
q•
a/2
• a
By symmetry all face have equal flux Page-21
q = 6 0
Objective : Flux from a point charge is distributed symmetrically
Ex.
A point charge q is placed on the apex of a cone of semi-vertex angle . Show that the electric flux through the base of the cone is
Sol.
q (1 cos ) . 2 0
Consider a Gaussian sphere with its centre at the apex and radius the slant length of the cone. The flux through the whole sphere is q / 0. Therefore, the flux through the base of the cone,
A q E = A · 0 0 Here, A0 = area of whole sphere = 4R2 and A = area of sphere below the base of the cone. Consider a differential ring of radius r and thickness dr. dA = (2r) R d r Rd x
R
C
(b)
= (2R sin ) R d as r = R sin = (2R2) sin d
A = ( 2R 2 ) sin d 0
A = 2R2 (1 – acos )
A q The desired flux is E = A · 0 0 q (1 cos ) (2R 2 ) (1 cos ) q = · = 2 2 0 (4 R ) 0
Objective : Introduction to application of solid angle
Page-22
OPTIONAL Ex. A charge is placed at the centre of a cylindrical surface. Find total flux passing through lateral curved surface. 2R R
Sol.
45°
q
Total flux linked can be divided into parts (i) Flux through lateral surface L (ii) Flux through end caps A If end cap substends solid angle at centre. q L + 2A = 0 q A = × 4 0 Note : Solid angle is given by = 2 (1 – cos )
=
q 0
1 2 1 2 4 2
1 q 1 2 = 0 2 q
L =
2 0
]
Calculation of electric field using Gauss Law : I. Shell A charged shell with total charge Q distributed uniformly on the surface of shell find field at x. (i) x > R Student will attemp it like this Possible field Q Let gaussian surface of r = x
Q dA = E 0
Q E 4x2 = 0
Q E = 4 x 2 0
directions
R x
{Emphasize this point}
Note : But this soln is incomplete because, this law only gives E along dA, and how do we know that E is along dA .
Page-23
Proper Logic Field can’t be along tangent no tangential component can exist because of symmetry of charge distribution. Only direction possible is radial.
Q 0
Q 0
E · dA =
E · dA =
Also by “symmetry” field at each point an gaussian surface has same magnitude is same, so E can be taken out of integral
(ii)
Q E dA = 0
Q E 4x2 = 0
Q = E 40 x 2 ]
x
x Gaussian surface
Emphasize that
E · dA
= 0 may also means E to dA or E is 0 or E · dA = 0
Emphasize proper method E · dA = 0 By symmetry E is along dA Also E · dA 0 as all will be either +ve or –ve by symmetry E dA = 0 Now
dA
0
E 0
Page-24
II.
Uniformly charged sphere with const volumetric charge density. (i) field at outside point x > R By symmetry we see that tangentical components of E is cancelled and net field is radial only.
4 R 3 E · dA = 3 0
Also there is uniform charge distribution E at every point has same magnitude and directed radially.
4 R 3 E dA = 3 0
x
R
(iii)
4R 3 2 E 4x = 3 0
x > R ; Hence proved.
4 R 3 E= 3 2 40 x
E
Gaussian surface
Field at inside points x
Gaussian surface
4 x 3 E · dA = 3 0
x
Again by symmetry E is only along radial direction.
4 x 3 E dA = 3 0
4 x 3 E 4x2 = 3 3 0
x E = 3 0
x
Hence proved. OPTIONAL Ex. Charge is distributed throughout a spherical region of space in such a manner that its volume charge density is given by = ar2, 0 r R where a s is a constant. Find the field at distance r from the center.
Page-25
Y R
dr r
Z
Sol.
dv = 4r2dr X
Since charge distribution is volumetric we choose a small volume element dV within the sphere. Because the charge distribution is spherically symmetric, we choose a thin spherical shell of radius r and thickness dr. We can think of entire spherical charge to be made up of concentric shells. Area of sphere is 4r2. Therefore, dV = 4r2dr R
q = dV = (ar 2 ) (4r 2 dr ) 0
=
4 ar 5 5
1 4 E = 4 . . ar3 5 0
Ex.
Sol.
Field inside a cavity : Consider a spherically symmetric charge distribution of charge density , with a cavity inside it. Field inside cavity can be obtained by super position of electric field due to complete sphere and electric field due to cavity part i.e. at the given point subtract the contribution of electric field due to cavity part from field of complete sphere. r Contribution of complete sphere E sphere = 3 0 x Ecavity = – 3 0 r x Resultant field E = E sphere + E cavity l
E = 3 ( r x ) 0 E inside cavity = 3 l 0
So field inside cavity is uniform. Objective : (i) Cavity can be considered having equal amount of positive and negative charge (ii) Cavity of uniform charged non conducting sphere has uniform E. III.
Infinite sheet, constant charge density We choose a small closed cylinder whose axis is perpendicular to plane as our gaussian surface. From symmetry we expect E to be directed perpendicular to the plane on both sides as shown, and to be uniform over the end caps of the cylinder. Page-26
Gaussion surface
A E dA
•
E
x
A = E · d A 0 E . dA +
curved
E . dA
left cap
+
A . = E dA 0 right cap
Now if the consider a plane through x, charge above and below is same, by symmetry again E is only along dA . through lateral surface is zero as E is along x-axis and here dA is perpendicular to it. E . dA = 0 curved Thus, E = 2 0
Asking question : Why we should consider the cylindrical gaussian surface in front as well as behind the sheet? IV.
Hollow cylinder (infinite length) What type of charge distribution cylinder can have? (i) Field outside cylinder x > R Linear charge distribution and surface charge distribution Let us consider a coaxial cylindrical gaussian surface.
l = E · d A ε0
x
By previous proved reason
l
l E x 2 x l = ε 0
E = 2x 0
(ii) Field inside cylinder x
2 3 1
3 Page-27
ELECTRIC POTENTIAL The electrostatic Potential Energy The concepts of work and potential energy were discussed in class XI. The work done by a force was defined as Work done = force × distance moved by the point of application of force. Examples considered in class XI included the work done against the gravitational force in lifting a mass, and against the restoring force of a spring when it is stretched. In both cases energy must be expended to do the work, but this energy does not disappear. It is stored as potential energy, which may later be released : gravitational potential energy may, for example, be released by allowing an object to fall. Work done by electric charges The concepts of work and potential energy also apply when the forces are electrical. Consider the two positive charges q and q1 as shown in figure. A q1
ri
q
B
C
rf
Lets say the charge q1 is fixed, but q may be moved. There is a repulsive force between the two charges and when they are separated by a distance r the magnitude of the force is given by equation
qq1 F = 4 r 2 0 The force on q acts in the direction AB. If q moves a distance dr along the line BC, the work done by the force is Fdr. The amount of work done when q moves from B to C, rf
rf
r
qq1 1 f qq1 qq1 W = Fdr = r = 4 2 dr = 4 ri 0 0 ri 4 0 r ri
1 1 ri rf
This work represents the difference in the electrical potential energy of the system of the two charges when q moves from B to C. It is natural to choose the potential energy to be zero at rf = , and with this choice the total potential energy U of the two charges when they are at A and B, separated by a distance ri, is qq1 U = W = 4 r 0 i qq1 Equation U = W = 4 r is still valid, but if q and q1 have different signs the right-hand side is 0 i
negative as the work must be done to pull the charges apart. The potential energy of q depends only on its distance from q1 and not on the direction. Note : Electric force is a Central force. Central force : Any force satisfying | F | f (r ) i.e. function of distance from a fixed point Fˆ rˆ i.e. directed along line joining All Central forces are conservative force. Page-28
The Electrostatic Potential The potential energy per unit charge U/q0 is independent of the value of q0 and has a value at every point in an electric field. This quantity U/q0 is called the electric potential (or simply the potential) V. Thus, the electric potential at any point in an electric field is : U V= q 0
Since potential energy is a scalar quantity so the electric potential also is a scalar quantity. If the test charge is moved between two positions A and B in an electric field, the charge-field system experiences a change in potential energy. Potential is defined as the change in potential energy of the system when a test charge is moved between the points divided by the test charge q0. U V = q = – 0
1.
E . dl
B
A
Units :
Joule / coulomb = volt
2.
Dimension :
w M1L2T 2 = = M1L2T–3A–1 q AT
3.
It is a scalar quantity. As work can be +ve or –ve, the electric potential can also be +ve or –ve.
4.
Just as with potential energy, only differences in electric potential are meaningful. We often take the value of the electric potential to be zero at some convenient point in an electric field.The most general reference point is infinite & potential at is assumed to be zero. (This fails for line charge & sheet).
5.
Electric potential is a scalar characteristic of an electric field, independent of any charges that may be placed in the field.
Electric fields lines always point in the direction of heighest decreasing electric potential A
A d
d
q
m
B
B
can be explained using analogy with gravity.. E
g
Now suppose that a test charge q0 moves from A to B. We can calculate the change in the potential energy of the charge-field system from Equations : U = q0V = –q0Ed From this result, we see that if q0 is positive, then U is negative. Page-29
Ex.
Calculate Potential difference between points A and B in front of Infinite charge sheet. B B VB – VA = V = – E . d s = – (E cos 0)ds = – E . d s B
A
A
A
Because E is constant, we can remove it from the integral sign ; this gives B
V = – E ds Ed A
The negative sign indicates that the electric potential at point B is lower than at point A ; that is, VB < VA. Ex.
If E 3ˆi 4ˆj N/C and potential at origin is zero find potential at (2, 4) and find work required to take a particle of charge –10C from (1,1) to (2,2).
Asking Questions Q.1 In figure, two points A and B are located within a region in which there is an electric field. The potential difference V = VB – VA is (a) positive (b) negative (c) zero. Q.2
In figure. a negative charge is placed at A and then moved to B. The change in potential energy of the charge-field system for this process is (a) positive (b) negative (c) zero.
E
B A
Equipotential surfaces : For an isolated charge q1 the electrostatic potential at a distance r from q1 is q1 / (40r). All points on the surface of a sphere of radius r are at the same potential. Spherical surfaces centred on q1 are equipotential surfaces.
q1
Electric field is perpendicular to the equipotential surface. This is obvious for a single charge, for which the field lines are radial and the equipotential are spherical. Electric field is always perpendicular to equipotential surfaces, no matter what the distribution of charge. This is easily proved by considering a small movement of a test charge on an equipotential surface. No work is done, and it follows that the electric field does not have a component lying in the surface, that is, the field is perpendicular to the surface. Properties of equipotential surfaces. 1.
These are imaginary surfaces where potential of all points are equal.
2.
W.D. to take a charge from one point on equipotential surface to other is zero.
3.
Two equipotential surface do not intersect each other.
4.
Direction of electric field at any point on equipotential surface is to the surface and in direction in which v is decreasing. (if v = 0 at ).
Page-30
Asking question Q.1 Arrange ER, EP, ER in order of magnitude. Equi Potential surfaces 10V
20V •
•
P
A
30V •
Q
B
40V
R
C
D
ER > EQ > EP
Q.2 Find the direction of field at P. A
B
P C
D
20V 10V 5V
C is direction of E field
Q.3
l •
•
Calculate shape of equipotential surfaces.
Potential due to combination of charges : (superposition principle) If a point is located in field of more than one charge then potential of that point is summation of individual potential with sign, where all potentials are calculated with same reference point. Methods of calculation of potential (1) Find E in space and calculate potential at a point using V = - E . d s (2) Calculate potential due to elements and add them up. dq
Potential on the axis of ring + + +
+ + + +
a
+
+
+
+
x
P
+
+ +
(1)
r
+
Potential at P using relation between E & V q x Eeff = 4 2 2 3/ 2 = 0 (r x ) Page-31
V = E eff .d s = x
(2)
q 4 0
x
x
(r 2 x 2 )3 / 2
q
dx =
4 0
r
2
x2
Potential at P using superposition q
1 dV = 40
dq
2
r x
0
2
q
= 4 0
r
2
x2
Asking question Q.1
What happens if the ring is non uniformly charged?
Q.2
Draw graph of V versus r and relate it with the graph of E (i) Centre at x = 0
E=0 ; V=
q (maximum) 40 r
(ii) Infinite E = 0 ; V = 0
Q 2 R & Emax = 2 3 3 40 R 2
(iii) E is maximum at x = ±
V
E –R /2 R /2
x r
Potential difference due to Infinite charged wire
+ve r1 r2
Potential difference in moving from r1 r2. r2
So V2 – V1 =
20r (–dr) r1
V2 – V1 = 2 (nr1 – nr2) 0 So here it is clear that V1 0 as r1 Also here as V 0 kdq kdq V = dV = will not work as when we write potential due to an element as we x x assume potential at infinite to be zero.
* *
Page-32
Potential due to solid uniformly charged non conducting sphere
R r
1 Q V = 4 0 r
r>R
4 r3 3 V= + 40 r V= 0
R
4 x 2 dx
40 x
r
r
1 2 1 2 r2 q(3R 2 r 2 ) r R = 2 2 80 R 3 3
Results : (i) Centre r = 0 E=0
;
V=
(ii) Surface r = R q E = 4 R 2 0
3 1 Q 2 40 R
q ; V = 4 R 0
(Graph of field and potential due to shell) E
E Inverse square ~ I/r2
Linear ~r
O
r
a
Inverse square ~ I/r2
Linear ~r
O
r
a
parabola
1 V r
(iii) V r=Rr
Ex.
1 V r
V r=Rr
A fixed solid sphere of radius R is made of a material dielectric constant K = 1. Uniform charge density . A bullet of mass m and having charge q is fired towards its centre. From a distance x (x > R) from centre, calculate min. velocity so it can cross the sphere. Assume no other force acting on the bullet except the electrostatic force.
Finding E from V E V = – grad (V)
Page-33
ˆ ˆ ˆ i j k (x, y, z co-ordinate system) grad = y z x Ex.
V = x + y + z find E
V =1 ; x E = – ( ˆi
Ex.
V V =1 ; =1 y z ˆj kˆ )
V = xyz find E at (1, 1, 1)
V V V = yz ; = xz ; = xy y x z E = – ( yzˆi xzˆj xykˆ ) Q.
= – ( ˆi ˆj kˆ ) The electric potential varies in space according to the relation V = 3x + 4y. A particle of mass 10Kg starts from rest from point (2, 3.2) under the influence of this field. Find the velocity of the particle when it crosses the x-axis. The charge on the particle is +1C. Assume V and (x, y) are in S.I. units. [Ans. 2mm/s]
III.
Concept of Self energy / potential energy of system energy required to create a system is called self energy.
1.
Electric potential energy of a system Definition : It is the work done to assemble the charges from infinite separation to present configuration w/o change in K.E. of any particle. Methods of calculation
(i) Keep all charges at separation from each other and then bring them one by one in present configuration and calculate the work done. PEsys = Wi
(ii) Find PE of each charge dues to field other charges.
PEsys =
PE1 PE 2 PE 3 ......... 2
Where PE1 = PE12 + PE13 + ........... PE2 = P21 + PE23 useful for symmetric arrangements. A
B
q1
q2 q1 C
Page-34
Method I
Work done to bring A w1 = 0 Work done to bring B w2 =
kq1q 2 r
Work done to bring C w3 =
kq1q 3 kq 2 q 3 + r r
Self energy = w1 + w2 + w3 1 q (V VA / C + qB (VB / A VB / C ) + qC (VC/A + VC/B)] 2 A A/B
Method II : U =
1 U= 2
n
n q i Vi / r i 1 r 1
Ex.
Find energy to the break. the system
Sol.
1 U= 2
7
i 1
7 q Vi / r r 1
7 3k (q) 3kq k (q ) Vi / r = a a 2 a 3 r 1 as because of (–q) is also same.
–q
q
q
–q
–q
q –q
a
q
kq 3ka 3ka 1 Work done to distroy = – U = - 8q a 2 a 3 2 a Calculation of self energy due to continuous charge distributions 1. Shell Method I : let say that x charge has been brought and if further dx charge from inifinity. Q
U=
U=
kx dx R 0
R Q
KQ 2 2R
Method II : Here V due to other charges is
Q dq
Q
1 kQ U = R dq 2 0
multiplied by
U=
1 because every interaction counted two times 2
kQ 2 2R
Page-35
2. Solid sphere Only by Method I : Let us consider a shell of radius x has ben created and further dx them created. R
4 / 3x 3 U = 4 x (4 x2 dx)) 0 0 3kQ 2 U= 5R
Ex.
Find self energy of system r
Q1R1
Sol.
Q2R2
Because of three parts to create both sphere and then their interacles. 3kQ12 3kQ 22 kQ1Q 2 U= 5R 1 5R 2 r
Energy density of electric field * Energy / unit volume where field is
1 0E 2
2
This is the energy required to create electric field per unit volume.
Ex. Ans.
Self energy of shell Now let us consider at x, energy required to make field at x. 2
1 1 Q 2 U = 0 2 4x dx 2 4 x 0 R
1 k 2Q2 2 = 0 2 4x dx 2 x R
1 k 2Q 2 4 = 0 2 2 R
kQ 2 2R Note : We don’t write self energy of point charge.
=
Page-36
Ex.
Solid sphere :Here since E in sphere varies in different ways for x > R and x < R 2
R
1 x 4x 2dx + U = 0 2 3 0 0 U=
2
1 1 Q 2 2 0 40 x 2 4x dx R
x
3kQ 2 5R
{Optional} Irodov 3.140. A point charge q is located at the Fig. centre O of a spherical uncharged conducting layer provided with a small orifice (Fig. ) The inside and outside radii of the layer are equal to a and b respectively. What amount of work has to be performed to slowly transfer the charge q from the point O through the orifice and into infinity?
Ans.
Electric Dipole Definition : When two charges of equal magnitude and opposite sign are separated by a very small distance, then the arrangement is called electric dipole. d• • –q
• q d
(i)
Dipole moment : P = q d ; d is always taken reason -ve to +ve
(ii)
Axis of dipole : is line joining -q to +q
(iii)
Mid point of axis of dipole is centre of dipole
(iv)
bisector of line joining +q to -q is equatorial line of dipole.
(v)
All the distances are measured from centre of dipole.
Electric potential due to dipole P r A –q
O
for d << r
i.e.
BPe = r –
d cos 2
AP = r +
qB
short dipole
d cos 2 Page-37
kq kq k q d cos ~ – = kq (d 2cos ) d d r2 d 2 2 r cos r cos r cos 2 2 4 k p . ˆr kp = 2 = 2 cos r r
VP =
Electric field due to dipole : E ( V )
E
ˆ kp cos = – rˆ r r 2 r
=–
r • P
–
kP = 3 ( 2 cos rˆ sin ˆ ) r
P 1 3 cos 2
(i)
+
where (r >> d)
r3
tan where is angle of Enet with r 2
direction tan =
2P cos Er = 4 r 3 0
Er
•
2kP cos kP sin ˆ rˆ + 3 r r3
1 E = 4 0
Enet
P sin E = 4 r 3 0
;
On the axis : Maximum potential and electric field at = 0 i.e. on axis for given distance r E
E
E • P =0 V +ve
+
= V –ve
Formula notapplicable
1 2P E = 4 3 0 r 1 P V = 4 2 0 r E is in direction of P
(ii)
Equatorial line : Minimum potential and field at = 90° i.e. on equitorial plane for given distance r. E
1 P E = 4 3 0 r
Ex.
•P
–
+
V=0 Find equivalent dipole moment of a ring having linear charge densities + and – on its two halves.
. Page-38
P
Electric dipole in uniform field : (A)
(B)
(C)
Force F1 F2 = 0 Fnet 0 Torque
+q
E qE
•
E
• • –q –qE
l | | qE sin 2 2 In the vector form p E Potential Energy The work done by an external agent in rotating the dipole without change in its kinetic energy is stored an potential energy in the field and dipole system. It is given by U = –p.E. or U = –p E cos
U
O
/2
where is the angle between the dipole and the electric field vector, as shown in figure. There potential energy U as a function of the angle is plotted in figure. Note : • • •
the potential energy is minimum at = 0 the potential energy is maximum at = the dipole is in stable equilibrium at =
. 2
E
+•
Potential energy in this posn. is zero dU = – d
90°
= – PE cos d
–•
0
= – PE sin = – PE cos = – P·E (D)
It will undergo SHM if turned by small angle PE sin = –
I .d 2 dt 2
•+ –•
• –
2
PE d 2 =– I dt
=
E
stable
for small
PE I
+ •
t
• unstable +•
•– •
Page-39
Electrostatic Pressure {Optional} If a small piece of radius b is removed from a charged spherical shell of radius a (>>b), calculate electric intensity at the midpoint of the aperture, assuming the density of charge to be . ER ER
P ED
ED
a
[Sol.
Consider the shell to be made up of a disc of radius b and the remainder. If ED and ER are the intensities due to disc and the remainder respectively at P, then for a charged spherical shell (or conductor) Eout = and Ein = 0 0
Now as for outside the shell both ED and ER will be directed outwards while inside ER will be outwards while ED inwards so that : Eout = ER + ED and Ein = ER – ED ........ (2) And hence equating Eqs. (1) and (2), ER + ED = and ER – ED = 0 0
Solving these for ER and ED : ER = ED = 2 0
i.e., field at the aperture will be (/20) directed outwards. Note : As intensity on the disc (element) the to remainder is (/20), electric force on it will be,
2 dF = dq E = ( ds) 2 = 2 ds 0 0 So force per unit area on a charged conductor due to its own charge 2 dF 1 = = 0 E 2 2 0 ds 2
[as for a conductor E = ] 0
This force is called ‘mechanical force’ or electrostatic pressure.
].
Page-40
CONDUCTORS A conductor means infinite no. of free charges which move in random direction so the lattice becomes positively charged. Conductors contain charge carriers, these charge carriers are electrons. In a metal, the outer (valence) electrons part away from their atoms and are free to move. These electrons are free within the metal but not free to leave the metal. The free electrons form a kind of ‘gas’; they collide with each other and with the ions, and move randomly in different directions. The positive ions made up of the nuclei and the bound electrons remain held in their fixed positions. 1.
Inside a conductor, electrostatic field is zero Consider a conductor, neutral or charged. There may also be an external electrostatic field. In the static situation, when there is no current inside or on the surface of the conductor, the electric field is zero everywhere inside the conductor. This fact can be taken as the defining property of a conductor. A conductor has free electrons. As long as electric field is not zero, the free charge carriers would experience force and drift. In the static situation, the free charges have so distributed themselves that the electric field is zero everywhere inside. Electrostatic field is zero inside a conductor.
Further Explanation What happens if conductor is placed in an exterrnal field. Lets keep a positive charge q charge near a neutral or acharged conductor then the electrons goes close to q. The redistribution of electrons inside conductor takes place which generates an internal electric field E int . + +
E net E ext E int
¯
+
¯
¯ ¯ ¯
Eint
+
Eext
+ ¯
+
¯ neutral conductor
So an e¯ experience E net
If Eint Eext , then the e¯ move such that they will create a stronger E int which will tend to cancel Eext. This constitutes current and therefore energy conservation is not valid. E net has to be O instantaneously.. E net E ext E int 0 2.
The interior of a conductor can have no excess charge in the static situation A neutral conductor has equal amounts of positive and negative charges in every small volume or surface element. When the conductor is charged, the excess charge can reside only on the surface in the static situation.
Page-41
Explanation This follows from the Gauss’s law. Consider any arbitrary volume element v inside a conductor. If we consider any small gaussian surface inside Q
E · dA = 0
[as E = 0]
On the closed surface S bounding the volume element v, electrostatic field is zero. Thus the total electric flux through S is zero. Hence, by Gauss’s law, there is no net charge enclosed by S. qenclosed = 0 Since the surface S can be made as small as you like, i.e., the volume v can be made vanishingly small. This means there is no net charge at any point inside the conductor, and any excess charge must reside at the surface.
Note : Thus Solid “ conducting” sphere is same as a shell. Note : You may emphasise again but qin = 0 does not imply that E = 0 from gauss law. 3.
At the surface of a charged conductor, electrostatic field must be normal to the surface at every point If E were not normal to the surface, it would have some non-zero component along the surface. Free charges on the surface of the conductor would then experience force and move. In the static situation, therefore, E should have no tangential component. Thus electrostatic field at the surface of a charged conductor must be normal to the surface at every point. (For a conductor without any surface charge density, field is zero even at the surface.)
Asking question Consider a neutral conducting sphere placed near a infinite non conducting uniform sheet of charge. Draw the field near the infinite sheet. + + +
Sol.
+ +
Emphasize following facts from figure (i) Field is normal to surface conductor (ii) Field inside conductor is zero Further discussion When we place an ideal conductor in an electric field E, the free electrons experience a force in the opposite direction of the field and migrate to one side of the conductor as shown in figure (a). Page-42
– –
+
–
+ + + +
–
E
(a)
(a) The accumulation of electrons leaves one side positively charged and the other negative. This charged distribution creates an electric field in a direction opposite to the applied field. The redistribution of charge takes place till net field inside the conductor is zero. Therefore, in electrostatic equilibrium, the electric field inside an ideal conductor is zero. If we place a Gaussian surface, an infinitesimal distance below the surface, the electric field is zero at every point on this Gaussian surface because it is inside the conductor [figure (b)]. Gauss’s law then implies that the net charge contained within the Gaussian surface is zero. In electrostatic equilibrium, excess charge on an ideal isolated conductor must reside on the conductor’s surface. No free charge can exist anywhere within the electrostatic conductor. Note: Also if some external field is present then charge distribution on a Solid conducting sphere and a conducting shell will be non uniform. OPTIONAL : Electric field at the surface of a charged conductor E = nˆ 0
where is the surface charge density and nˆ is a unit vector normal to the surface in the outward direction. To derive the result, choose a pill box (a short cylinder) as the Gaussian surface about any point P on the surface, as shown in figure.
The pill box is partly inside and partly outside the surface of the conductor. It has a small area of cross section S and negligible height. Just inside the surface, the electrostatic field is zero; just outside, the field is normal to the surface with magnitude E. Thus, the contribution to the total flux through the pill box comes only from the outside (circular) cross-section of the pill box. This equals ± ES (positive for > 0, negative for < 0), since over the small area S, E may be considered constant and E and S are parallel or antiparallel. The charge enclosed by the pill box is S. By Gauss’s law Page-43
ES =
| | S 0
|| E= 0 Including the fact that electric field is normal to the surface, we get the vector relation, Eq. E = nˆ , 0
which is true for both signs of . For > 0, electric field is normal to the surface outward; for < 0, electric field is normal to the surface inward.
4.
Electrostatic potential is constant throughout the volume of the conductor and has the same value (as inside) on its surface Since E = 0 inside the conductor and has no tangential component on the surface, no work is done in moving a small test charge within the conductor and on its surface. That is, there is no potential difference between any two points inside or on the surface of the conductor. Hence, the result. If the conductor is charged, electric field normal to the surface exists; this means potential will be different for the surface and a point just outside the surface. In a system of conductors of arbitrary size, shape and charge configuration, each conductor is characterised by a constant value of potential, but this constant may differ from one conductor to the other.
Charge density on a conductor’s surface in abscence of an external field
+ ++r + +++ ++ +
+
+
+
+
+ + + 1 r
+ R + + + +
Now consider a single conductor with a nonspherical shape. If a charge is given to this conductor figure, the charge density will not be uniform on the entire surface. A portion where the surface is more “flat” may be considered as part of a sphere of larger radius. The charge density at such a portion will be smaller from equation (1). At portions where the surface is more curved, the charge density will be larger. More precisely, the charge density will be larger where the radius of curvature is small. Thus the distribution of charge Q on surface of conductor is “Non-uniform 1
and
R curvature
Charged conductor in external field : Figure shows a charge Q placed infront of a neutral conductor. Here the charge density also depends on external field and E out E ind . int = 0
¯
• Q
¯ ¯ ¯ ¯ ¯
+ • + + + +
Q1
Note : So here charge Q1 must be distributed non uniformly unrelated to Radius of curvature.
Page-44
Ex.
A point charge q is kept at a distance l from neutral sphere of radius R. Find electric potential V at P, solid conducting sphere. Q P x – q
+ + +
r C
– – –
R –
+
l
Sol.
Potential due to induced charges is zero as centre C is equidistant from all induced charges Volume of solid conductor is equipotential volume
Vc = Vp =
Kq KQ l R
Also at P VP = Vdue to q + Vdue to induced charges + Vdue to Q Kq KQ Kq KQ + = + + Vdue to induced charges l R x R Kq Kq – l x E external + E conductor + Einduced = 0 Resultant electric field inside material is zero
Vinduced charge =
Ex.
Find electric field due to induced charges at P Kq | E p | by induced charges = 2 x
Cavity inside a conductor : Cavity is aplace surrounded from all sides by the conductor such that without touching the body we can’t reach cavity. Electrostatic shielding Consider a conductor with a cavity, with no charges inside the cavity. A remarkable result is that the electric field inside the cavity is zero, whatever be the size and shape of the cavity and whatever be the charge on the conductor and the external fields in which it might be placed. We have proved a simple case of this result already: the electric field inside a charged spherical shell is zero. The proof of the result for the shell makes use ofthe spherical symmetry of the shell. But the vanishing of electric field in the (charge-free) cavity of a conductor is, as mentioned above, is a general result. A related result is that even if the conductor is charged or charges are induced on a neutralconductor by an external field, all charges reside only on the outer surface of a conductor with cavity. Whatever be the charge and field configuration outside, any cavity in a conductor remains shielded from outside electric influence: thefield inside the cavity is always zero. This is known as electrostatic shielding. The effect can be made use of in protecting sensitive instruments from outside electrical influence. Figure gives a summary of the important electrostatic propertiesof a conductor. Page-45
1.
If there is no charge present inside the cavity than field inside it is zero. (This does not mean that we are implying “no net charge”. If there is some charge present here and there in cavity such that there sum total is zero then the following disscussion will not be valid)
Proof wrong sol. Let us consider a gauesion surface just near to cavity. E · dA = 0 q in = 0 Sol.
From this we don’t have proved that no charge resides on cavity but have proved that net charge on cavity surface is O. Now suppose a conductor of arbitrary shape contains a cavity as shown in figure.
B•
•A
Let us assume that no charges are inside the cavity. In this case, the electric field inside the cavity must be zero regardless of the charge distribution on the outside surface of the conductor. Furthermore, the field in the cavity is zero even if an electric field exists outside the conductor. To prove this point, we use the fact that every point on the conductor is at the same electric potential, and therefore any two points A and B on the surface of the cavity must be at the same potential. Now imagine that a field E exists in the cavity and evaluate the potential difference VB – VA defined by equation. U V q = – 0
B
E · ds A
B
VB – VA = – E · ds A
Because VB – VA = 0, the integral of E · ds must be zero for all paths between any two points A and B on the conductor. The only way that this can be true for all paths is if E is zero everywhere in the cavity. Thus, we conclude that a cavity surrounded by conducting walls is a field-free region as along no charges are inside the cavity. Page-46
2.
Equal and opposite charge is induced on the inner surface of cavity. Figure shows a conductor with a cavity inside it. A charge q is placed inside cavity. In electrostatic equilibrium charge distribution will be as shown in figure
+q –q q
charge on inner surface of cavity is – q since material of conductor is initially neutral, equal and opposite charge appears on outer surface. +
+
+ + B
+ + +
A¯ ¯ ¯ ¯ •q ¯ ¯ ¯ •p +
+
C Gaussian surface
+
Proof: Let us consider a gaussian surface just outside cavity inside material of conductor. As E in material of conductor is zero.
E · dA = 0
qenclosed = 0 Thus equal and opposite charge is induced on the inner surface of cavity.
3.
The electric field due to charges on the outer surface of conductor is zero for all the points inside the conductor seperately Consider a charged conductor having charge +q1 and Q is kept inside the cavity. Lets call charge Q inside cavity as A, the induced charge -Q on the surface of the cavity as B and the charge on the surface of the conductor Q + q1 as C. C B A
Now field inside the conductor is, E net and E A , EB andEC are fields due to charge A,B and C inside the conductor. and, E net = E A + E B E C = 0 Now the electric field due to charges on the outer surface of conductor is zero for all the points inside the conductor seperately and the E B E A is zero seperately.. Page-47
Special case : When spherical cavity in present inside spherical conductor and charge is at the centre. Then field inside the cavity is just due to charge A only as new field lines are already to B and E due to B = 0. [Symmetrical distribution]
C
¯ ¯¯ B ¯ q•A¯ ¯ ¯
Note : When if we displace the charge q inside the cavity, it will only affect the charge distribution at B the charge distribution at C will remain unaffected. Ex.
Find field at P, the conductor is neutral. and cavityis having a charge q inside the cavity. •P + x + r •q¯A + + ¯ ¯¯ • R B + C + + +
Sol.
as
Ex.
RQ r2 E A + EB = O
Ep =
A hollow, uncharged spherical conductor has inner radius a and outer radius b. A positive point charge +q is in the cavity at the centre of the sphere. Make the graph E and potential V(r) everywhere, assuming that V = 0 at r = .
Er
V
kq r2
a
b
kq r2 Z
kq + kq – kq a b r kq b kq r a
b
r
+ – –+ a –+ +q –+ – + – b +
Page-48
Ex. Sol.
Find charge on all surfaces : Applying Gauss’ law on opposite faces. E · ds = 0 4Q P
A • B
– 3Q C
5Q D
E
F
2Q+x –x 4Q–x
3Q-x
x
–3Q+x
as through two lids E = 0 as part of conductor and through lateral surface E · ds = 0 qenclosed = 0 Important :
(i) Thus facing surfaces have equal and opposite charges. (ii) Using this also prove that outer faces of the two last plates have equal charges. (4Q x ) / Area (2Q x ) / Area – =0 2 0 2 0
Ex.
[as no field because of equal is cancelled]
4Q – x – 2Q – x = 0 2x = 2Q x = Q.
A solid conducting sphere having a charge Q is surrounded by an uncharged concentric conducting hollow spherical shell. Let the potential difference between the surface of the solid sphere and that of the outer surface of hollow shell be V. What will be the new potential difference between the same two surfaces if the shell given a charge –3Q ? Shell Sphere ++ ++++ + + +Q b ++ + ++ ++
[Sol.
In case of a charged conducting sphere, 1 q Vin = VC = VS = 4 R 0
and
1 q Vout = 4 r 0
So if a and b are the radii of sphere and spherical shell respectively, potential at their surfaces will be, 1 Q 1 Q Vsphere = 4 a and Vshell = 4 b 0 0
And so according to given problem
Q 1 1 V = Vsphere – Vshell = 4 a b ........... (i) 0 Now when the shell is given a charge (–3Q) the potential at its surface and also inside will change by V0 Page-49
1 3Q = 4 b 0 So that now 1 Q V’sphere = 4 + V0 0 a
and
1 V’shell = 4 0
Q b V0
And hence Q 1 1 V’sphere – V’shell = 4 = V [from Eq.(1)] b 0 a
i.e., if any charge is given to external shell, the potential difference between sphere and shell will not change. This is because by presence of charge on outer shell, potential everywhere inside and on the surface of shell will change by same amount and hence potential difference between sphere and shell will remain unchanged.
(I)
When two conductors are connected : When two conductors are connected charge redistributes on the connected conductors till their potential becomes equal
Ex.
A metal sphere A of radius a is charged to potential V. What will be its potential if it is enclosed by a spherical conducting shell B of radius b and the two are connected by a wire ?
Sol.
If the charge on sphere of radius a is q, 1 q V = 4 a 0 a i.e. q = (40a)V Now, when sphere A is enclosed by spherical conductor B and the two are connected by a wire, charge will reside on outer surface of B and so the potential of B will be ,
B b
A a
1 q 1 40a a VB = 4 b = 4 V= V b b 0 0 Now as sphere A is inside B so its potential, a VA = VB = (V) [< V as a < b] b
Page-50
Ex.
Find charge on all four platess assume length ang breadth of plates as very large. d
d
2d
D
A
P •
Q
Sol.
3Q
–5Q
2Q
Writing field at P we get let charge on 2nd and 4th be Q1 QA + QB = 5Q A
B
C
QA
D
+ve
QB
Now since both surfaces have same potential
Work done is zero from B to D
Writing fields in region BC and CD and calculating the work. We get, 3QA – 3QB = 2Q so
6QA = 17Q
QB = 5Q –
QA =
17Q 30 Q 17Q = 6 6
=
II.
17Q 6
13Q 6
Earthing : Earthing means connecting it with Earth.
Earthing
We assume that (i) (ii)
Earth is infinite resource and sink of charge so potential of earth will not change. The potential of earth is assumed to be “zero”. So after earthing the charge on conductors vary so that potential of conductor becomes zero. Page-51
Ex.
Consider concentric spherical shells of negaligible thickness. Initial charges on these shells are Q1 & Q2. Now inner shell is earthed. Find charge on the inner shell.
Sol.
Let the charge be q1
Q2
r2
kq1 kQ 2 + r1 r2 = 0
r1 q1 = – Q2 r 2
Q1 r1
[new charge after earthing]
This means that the inner shell is at zero potential and that electric field lines leave the outer shell and go to infinity but other electric field lines leave the outer shell and end on the inner shell. r1 [This implies that a charge + Q2 r has been transferred to earth leaving negative charge on A.] 2
Ex.
Sol.
Figure shows three thin concentric spherical shells with initial charges as shown in the figure shell A & C are connected by wire and such that it does not touch B and shell B is earthed.. Let finally charges be qA, qB and qC. qA + qC = 2Q ... (i) kq C kq B kq A + + =0 2k 2k 3k
3Q 3R
qB qC q + + A = 0 ... (ii) 2 2 3
2Q 2R R
Also
kq C kq 2 kq A kq A kq B kq C + + = + + R 2R 3R 3R 3R 3R
aC +
qB qA qA qB qC + = + + 2 3 3 3 3
B
A
–Q
... (iii)
Page-52
