Electric Field and Potential

Chapter # 29

Electric Field & Potential

Solved Examples EXAMPLE 29.1

Two charges 10 C and C are placed at points A and B separated by a distance of 10 cm. Find the electric field at a point P on the perpendicular bisector of AB at a distance of 12 cm from its middle point.

Sol.

EA 

P

E



EB 12 cm 



10 c A

- 10 c B

C 10 cm

The situation is shown in figure . The distance AP = BP =

(5 cm) 2  (12 cm )2 = 13 cm.

The field at the point P due to the charge 10 C is 10 C

EA =

4 0 (13 cm )

2

=

(10  10

C )  (9  10 9 N  m 2 / C 2 )

6

169  10

4

m2

= 5.3 × 10 6 N/C. This field is along AP. The field due to – 10 C at P is E B = 5.3 × 10 6 N/C along PB. As EA and EB are equal in magnitude, the restultant will bisect the angle between the two. The geometry of the figure shows that this resulatant is parallel to the base AB. The magnitude of the resultant field is E = EA cos + EB cos = 2 × ( 5.3 ×10 6 N/C ) ×

5 13

= 4.1 × 10 6 N/C. If a charge distribution is continuous, we can use the technique of intengration to find the resultant electric  field at a point . A small element dQ is chosen in the distrubtion and the field dE due to dQ is calculated. The  resultant field is then calculated by intergrating the componenets of dE under proper limits. EXAMPLE 29.2 A ring of radius a contains a charge q distribhuted uniformly over its length. Find the electric field at a point on the axis of the ring at a distance x from the centre.

A dQ a o

P x



dEcos dE

Sol. Fig. shows the sistuation. Let us consider a small element of the ring at the point A having a charge dQ. The field at P due to this element is dQ

dE =

4 0 ( AP )2

By symmetry, the field at P will be along the axis OP. The component of dE along this direction is

manishkumarphysics.in

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Chapter # 29

Electric Field & Potential dQ

dE cos =

4 0 ( AP )

2

 OP     AP 

xdQ

=

4 0 (a 2  x 2 )3 / 2

The net field at P is xdQ  4 (a 2  x 2 )3 / 2 0

E =  dE cos = x

=

xQ  dQ = 4 (a 2  x 2 )3 / 2 0

4 0 (a  x ) 2

2 3/2

EXAMPLE 29.3

Three particles, each having a charge of 10 C, are placed at the vertices of an equilaterial triangle of side 10 cm. Find the work done by a person in pulling them apart to infinite separations.

Sol. The potential energy of the system in the initial condition is U=

3  (10 C )  (10 C ) 4 0 (10 cm )

(3  10 10 C 2 )  (9  10 9 N - m 2 /C 2 ) = 27 J. 0.1 m When the charges are infinitely separated, the potential energy is reduced to zero. If we assume that the charge do not get kinetic energy in the process, the total mechanical energy of the system decreases by 27J. Thus, the work done by the person on the system is – 27 J. =

EXAMPLE 29.4

Two charges + 10 C and + 20 C are placed due to the pair at the middle point of the line joining the two adding the two charges.

Sol.

Q Using the equation V = 4 r , the potential due to + 10 C is 0 V1 =

(10  10   C )  (9  10 9 N - m 2 /C 2 ) 1 10 -2 m

= 9 MV..

The potential due to + 20 C is V2 =

( 20  10   C )  (9  10 9 N - m 2 / C 2 ) 1 10  2 m

= 18 MV..

The net potential at the given point is 9 MV + 18 MV + 27 MV. If the charge distribtion is continuous, we may use the technique of intergration to find the electric, potential.

EXAMPLE 29.5 Figure shows two metallic plates A and B placed parallel to each other at a separation d. A uniform electric field E exists between the plates in the direction from plate B to plate A. Find the potential difference berween the plates. A d

E

B

Sol. Let us take the origin at plate A and X–axis along the direction from A to plate B. We have  rB

VB – VA = –



 rA

  E.dr = –

d

  E dx = Ed. 0

If we work in Cartesian coordinate system manishkumarphysics.in

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Chapter # 29


   E = Ex i + Ey j + Ez k     d r = dx i + dy j + dz k .

and Thus, from

dV = – Ex dx – Ey dy – Ez dz. ............(i) If we change x to x+dx keeping y and z constant, dy = dz = 0 and from (i) Ex = -

V . x

Similarly,

Ey = -

V y

and

Ez = -

V . z

 etc. are used to indicate that while differentiating with respect to one coordinate, the others x are kept constant.

The symbols

Worked out Examples 1.

Charges 5.0 × 10–7 C, –2.5 × 10–7 C and 1.0 × 10–7 C are held fixed at the three corners A, B, C of an equilateral triangle of side 5.0 cm. Find the electric force on the charge at C due to the rest two.

Sol.

The force on C due to A



1 (5  10 7 C) (1 10 7 C) 4 0 (0.05 m)2

 9  10 9

N  m2

5  10 14 C 2

 0.18 N C2 25  10  4 m 2 This force acts along AC. The force on C due to B





1 (2.5  10 7 C) (1 10 7 C)  0.09 N 4 0 (0.05 m)2

This attractive force acts along CB. As the triangle is equilateral, the angle between these two forces is 120°. The resultant electric force on C is [(0.18 N)2 + (0.09 N)2 + 2(0.18 N) (0.09 N) (cos 120°)]1/2 = 0.16 N. The angle made by this resultant with CB is tan 1

2.

Sol.

0.18 sin120  90 0.09  0.18 cos 120 

Two particles A and B having charges 8.0 × 10–6 C and – 2.0 × 10–6 C respectively are held fixed with a separation of 20 cm. Where should a third charged particle be placed so that is does not experience a net electric force ? As the net electric force on C should be equal to zero, the force due to A and B must be opposite in direction. Hence, the particle should be placed on the line AB. As A and B have charges of opposite signs, C cannot be between A and B. Also, A has larger magnitude of charge than B. Hence, C should manishkumarphysics.in

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Chapter # 29


be placed closer to B than A. The situation is shown in figure.

Suppose BC = x and the charge on C is Q. The force due to A 

The force due to B 

(8.0  10 6 C)Q 4 0 (20 cm  x )2 (2.0  10 6 C)Q 4  0 x 2

.

.

They are oppositely directed and to have a zero resultant, they should be equal in magnitude. Thus, 8 ( 20 cm  x )



2 x2

20 cm  x  2 , giving x = 20 cm. x

or, 3.

2

Three equal charges, each having a magnitude of 2.0 × 10–6 C, are placed at the three corners of a rightangled triangle of sides 3 cm, 4 cm and 5 cm. Find the force on the charge at the right-angle corner.

Sol.

The situation is shown in figure. The force on A due to B is

F1 

(2.0  10 6 C) (2.0  10 9 C) 4 0 ( 4 cm)2

= 9 × 109

N  m2 C

2

 4.0  10 12 C 2 

1 16  10  4 m 2

= 22.5 N This force acts along BA. Similarly, the force on A due to C is F 2 = 40 N in the directin of CA. Thus, the net electric force on A is

F  F12  F22  (22.5 N)2  ( 40 N)2 = 45.9 N. This resultant makes an angle  with BA where tan = 4.

Sol.

40 16  22.5 9

Two small iron particles, each of mass 280 mg, are placed at a distance 10 cm apart. If 0.01% of the electron of one particle are transferred to the other, find the electric force between them. Atomic weight of iron is 56 g/mol and there are 26 electrons in each atom of iron. The atomic weight of iron is 56 g/mol. Thus, 56 g of iron contains 6 × 1023 atoms and each atom contains 26 electrons. Hence, 280 mg of iron contains

280 mg  6  10 23  26 = 7.8 × 1022 electrons. 56 g The number of electrons transferred from one particle to another 

0.01  7.8  10 22  7.8  1018 . 100


Page # 4

Chapter # 29


The charge transferred is, therefore, 1.6 × 10–19 C × 7.8 × 1018 = 1.2 C. The electric force between the particles is 2   (1.2 C)2  9  109 N  m   C2  (10  10 2 m)2  = 1.3 × 1012 N. This equals the load of approximately 2000 million grown-up persons ?

5. Sol.

A charge Q is to be divided on two objects, What should be the values of the charges on the objects so that the force between the objects can be maximum ? Suppose one object receives a charge q and the other Q – q. The force between the objects is F

q(Q  q) , 4 0 d2

where d is the separation between them. For F to be maximum, the quantity y = q(Q – q) = Qq – q2 should be maximum. This is the case when,

dy = 0 or, Q – 2q = 0 or, q = Q/2 dq Thus, the charge should be divided equally on the two objects. 6.

Sol.

Two particles, each having a mass of 5 g and charge 1.0 × 10–7 C, stay in limiting equilibrium on a horizontal table with a separation of 10 cm between them. The coefficient of friction between each particle and the table is the same. Find the value of this coefficient. The electric force on one of the particles due to the other is

 N  m2  1 F   9  109  (1.0  10 7 C)2  2   C (0.10 m)2   The frictional force in limiting equilibrium f =  × (5 × 10-3 kg) × 9.8 m/s 2 = (0.049 ) N. As these two forces balance each other, 0.049  = 0.009 or  = 0.18. 7. Sol.

A vertical electric field of magnitude 4.00 × 10 5 N/C just prevents a water droplet of mass 1.00 × 10–4 kg from falling. Find the charge on the droplet. The forces acting on the droplet are  (i) the electric force qE and  (ii) the force of gravity m g . To just prevent from falling, these two forces should be equal and opposite. Thus, q(4.00 × 105 N/C) = (1.00 × 10–4 kg) × (9.8 m/s 2) or q = 2.45 × 10–9 C.

8.

Three charges, each equal to q, are placed at the three corners of a square of side a. Find the electric field at the fourth corner.

Sol.

Let the charges be placed at the corners A, B and C (figure). We shall calculate the electric field at the fourth corner D. The field E1 due to the charge at A will have the magnitude manishkumarphysics.in

q 4 0 a 2

and will be along Page # 5

Chapter # 29


AD. The field E2 due to the charge at C will have the same magnitude and will be along CD. q

The field E3 due to the charge at B will have the magnitude

4 0 ( 2a )2

and will be along BD. As E1

and E2 are equal in magnitude, their resultant will be along the bisector of the angle between E1, E2 and hence along E3. The magnitude of this resultant is

E12  E 22 as the angle between E1 and E2 is /2.

The resultant electric field at D is, therefore, along E3 and has magnitude

E12  E 22  E 3 2

2

 q   q  q      4 a 2   4 a 2  40 ( 2a )2 0 0    



9.

q  2 1  q  2  2   (2 2  1) . 4 0  a 2a  8 0 a 2

A charged particle of mass 1.0 g is suspended through a silk thread of length 40 cm in a horizontal electric field of 4.0 × 104 N/C. If the particle stays at a distance of 24 cm from the wall in equalibrium, find the charge on the particle.

Sol.

The situation is shown in figure The forces acting on the particle are (i) the electric force F = qE horizontally, (ii) the force of gravity mg downward and (iii) the tension T along the thread. As the particle is at rest, these forces should add to zero. Taking components along horizontal and vertical, T cos = mg and T sin = F or, F = mg tan ...........(i) From the figure, sin = Thus, tan =

24 3  . 40 5

3 . From (i), 4

q(4.0 × 104 N/C) = (1.0 × 10–3 kg) (9.8 m/s 2) giving 10.

3 , 4

q = 1.8 × 10–7 C.

A particle A having a charge of 5.0 × 10–7 C is fixed in a vertical wall. A second particle B of mass 100 g and having equal charge is suspended by a silk thread of length 30 cm from the wall. The point of suspension is 30 cm above the particle A. Find the angle of the thread with the vertical when it stays in equilibrium.

Sol. manishkumarphysics.in

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Chapter # 29


The situation is shown in figure. Suppose the point of suspension is O and let  be the angle between the thread and the vertical. Forces on the particle B are (i) weight mg downward (ii) tension T along the thread and (iii) electric force of repulsion F along AB. For equilibrium, these forces should add to zero. Let X’ BX be the line perpendicular to OB. We shall take the components of the forces BX. This will give a relation between F, mg and . The various angles are shown in the figure. As

 . 2 The other angles can be written down directly. Taking components along BX, we get OBA  OAB  90 

F cos

 = mg cos(90° – ) 2 = 2 mg sin

  cos 2 2

 F  2 2 mg .

or,

sin

Now,

 N  m2 F   9  109  C2 

and

AB = 2(OA) sin

Thus,

F

........ (i)

   (5.0  10 7 C)2  1  AB 2 

 . 2

9  10 9  25  10 14 4  (30  10 2 )2  sin 2

 2

.

1 2mg

.............(ii)

 F 9  10 9  25  10 11N 1 . From (i) and (ii) sin 2  2mg =  2mg 4(30  10 2 )2  sin 2 2 or, sin3

  2

9  10 9  25  10 14 N 4  9  10  2  2  (100  10 3 kg)  9.8

m s2

= 0.0032 or, 11.

sin

 = 0.15, giving  = 17°. 2

Four particles, each having a charge q, are placed on the four vertices of a regular pentagon. The distance of each corner from the centre is x. Find the electric field at the centre of the pentagon.

Sol.

Let the charges be placed at the vertices A, B, C and D of the pentagon ABCDE. If we put a charge q at the corner E also, the field at O will be zero by symmetry. Thus, the field at the centre due to the charges at A, B, C and D is equal and opposite to the field due to the charge q at E alone. The field at O due to the charge q at E is


Page # 7

Chapter # 29

Electric Field & Potential q

along EO.

4 0 a 2

q

Thus, the field at O due to the given system of charges is 12.

4 0 a 2

along OE.

Find the electric field at a point P on the perpendicular bisector of a uniformly charged rod. The length of the rod is L, the charge on it is Q and the distance of P from the centre of the rod is a.

Sol.

Let us take an element of length dx at a distance x from the centre of the rod (figure). The charge on this element is

Q dx . L The electric field at P due to this element is dQ 

dE 

dQ 4 0 ( AP )2

.

By symmetry, the resultant field at P will be along OP (if the charge is positive). The component of dE along OP is dQ

dE cos =

4 0 ( AP )

2

.

OP a Q dx  AP 4 0L(a 2  x 2 )3 / 2

Thus, the resultant field at P is



E  dE cos  aQ  4  0L

L/2

dx



(a  x 2 )3 / 2 L / 2 2

.

....... (i)

We have x = a tan or dx = a sec 2  d Thus,

 (a 

dx 2

 x 2 )3 / 2

1 a

2





a sec 2  d a 3 sec 3  1

 cos  d  a

2

sin  

1

x

a ( x  a 2 )1 / 2 2

2

From (i), L/2

  x E 2  2 2 1/ 2  4 0La  ( x  a )  L / 2 aQ



13.

  2L  2 2 1/ 2  4 0La  (L  4a )  aQ

2



Q

.

2 0 a L2  4a 2

A uniform electric field E is created between two parallel, charged plates as shown in figure. As electron enters the field symmetrically between the plates with a speed v0. The length of each plate is l. Find the angle of deviation of the path of the electron as it comes out of the field.


Page # 8

Chapter # 29

Sol.


eE in the upward direction. The horizontal velocity remain v0 as m there is no acceleration in this direction. Thus, the time taken in crossing the field is

The acceleration of the electron is a =

t

l . v0

....... (i)

The upward component of the velocity of the electron as it emerges from the field region is

v y  at 

eEl . mv 0

The horizontal component of the velocity remains vx = v0. The angle  made by the resultant velocity with the original direction is given by

tan  

vy



vx

eEl mv 02

.

Thus, the electron deviates by an angle   tan 1

14. Sol.

eEl

.

mv 02

In a circuit, 10 C of charge is passed through a battery in a given time. The plates of the battery are maintained at a potential difference of 12 V. How much work is done by the battery ? By definition, the work done to transport a charge q through a potential difference V is qV. Thus, work done by the battery = 10 C × 12 V = 120 J.

15.

Charges 2.0 × 10–6 C and 1.0 × 10–6 C are placed at corners A and B of a square of side 5.0 cm as shown in figure. How much work will be done against the electric field in moving a charge of 1.0 × 10–6 C from C to D ?

Sol.

The electric potential at C 

q  1  q1  2   4 0  AC BC 

 9  10 9

N  m 2  2.0  10 6 C 1.0  10 6 C   0.05 m  C 2  2  0.05 m

 2 2    (9000 V)  . 2  0 . 05   The electric potential at D 

1 4 0

q   q1  2   AD BD  

 9  10 9

N  m 2  2.0  10 6 C 1.0  10 6 C   2  0.5 m  C 2  0.05 m


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Chapter # 29


 2 2 1    (9000 V)  .  2  0.05  The work done against the electric field in moving the charge 1.0 × 10-6 C from C to D is q(VD – VC)  2 2  1 2  2    = (1.0 × 10-6 C) (9000 V)   2  0 . 05   = 0.053 J. 16.

  The electric field in a region is given by E  ( A / x 3 ) i . Write a suitable SI unit for A. Write an expression for the potential in the region assuming the potential at infinity to be zero.   fdlh {ks=k esa fo|qr {ks=k E  ( A / x 3 ) i }kjk fn;k tkrk gSA A ds fy, mi;qDr SI ek=kd crkb;sA bl {ks=k es foHko dks mi;qDr

inksa esa fyf[k, ] ;g ekurs gq, fd vuUr ij foHko 'kwU; gSA Sol.

the unit of A is

N  m3 or V – m 2. C ( x, y, z )

V(x, y, z) = 



( x, y, z )

  E. d r

 



17.

Sol.

HCV_Ch-29_WOE_16

The SI unit of electric field is N/C or V/m. Thus,





A dx x

3



A 2x 2

Three point charges q, 2q and 8q are to be placed on a 9 cm long straight line. Find the positions where the charges should be placed such that the potential energy of this system is minimum. In this situation, what is the electric field at the charge q due to the other two charges ? The maximum contribution may come from the charge 8q forming pairs with others. To reduce its effect, it should be placed at a corner and the smallest charge q in the middle. This arrangement shown in figure ensures that the charges in the strongest pair 2q, 8q are at the largest separation.

The potential energy is U

q2 4 0

2  16 8    .  x 9 cm 9 cm – x 

This will be minimum if

A

For this,

2 8  is minimum. x 9 cm  x

dA 2 8   0 2 dx x (9 cm  x )2

or, 9 cm – x = 2x or, x = 3 cm. The electric field at the position of charge q is

q 4 0

 2  8     x 2 (9 cm  x )2   

=0 18.

Sol.

An HCl molecule has a dipole moment of 3.4 × 10–30 C–m. Assuming that equal and opposite charges lie on the two atoms to form a dipole, what is the magnitude of this charge ? The separation between the two atoms of HCl is 1.0 × 10–10 m. If the charges on the two atoms are q, –q, q(1.0 × 10–10 m) = 3.4 × 10–30 C-m. or, q = 3.4 × 10–20 C. Note that this is less than the charge of a proton. Can you explain, how such a charge can appear on an atom ?


Page # 10

Chapter # 29


19.

Figure shows an electric dipole formed by two particles fixed at the ends of a light rod of length l. The mass of each particle is m and the charges are – q and + q. The system is placed in such a way that the dipole axis is parallel to a uniform electric field E that exists in the region. The dipole is slightly rotated about its centre and released. Show that for small angular displacement, the motion is angular simple harmonic and find its time period.

Sol.

Suppose, the dipole axis makes an angle  with the electric field at an instant. The magnitude of the torque on it is   |  |  | pE | = qlE sin. This torque will tend to rotate the dipole back towards the electric field. Also, for small angular displacement sin   so that  = qlE. The moment of inertia of the system about the axis of rotation is 2

ml 2 . l I  2  m   2 2

Thus, the angular acceleration is = where 2 =

 2qE    2  I ml 2qE . ml

Thus, the motion is angular simple harmonic and the time period is T = 2

ml . 2qE

QUESTIONS FOR SHORT ANSWER 1.

The charge on a proton is + 1.6 × 10–19 C and that on an electron is –1.6 × 10–19 C. Does it mean that the electron has a charge 3.2 × 10–19 C less than the charge of a proton ?

2.

Is there any lower limit to the electric force between two particles placed at a separation of 1 cm ?

3.

Consider two particles A and B having equal charges and placed at some distance. The particle A is slightly displaced towards B. Does the force on B increase as soon as the particle A is displaced ? Does the force on the particle A increase as soon as it is displaced ?

4.

Can a gravitational field be added vectorially to an electric field to get a total field ?

5.

Why does a phonograph-record attract dust particles just after it is cleaned ?

6.

Does the force on a charge due to another charge depend on the charges present nearby ?

7.

In some old texts it mentioned that 4 lines of force originate from each unit positive charge. Comment on the statement in view of the fact that 4 is not an integer.

8.

Can two equipotential surface cut each other ?

9.

If a charge is placed at rest in an electric field, will its path be along a line of force ? Discuss the situation when the lines of force are straight and when they are curved.

10.

Consider the situation shown in figure. What are the signs of q1 and q2 ? If the lines are drawn in proportion to the charge, what is the ratio q1/q2 ? manishkumarphysics.in

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Chapter # 29


11.

A point charge is taken from a point A to a point B in an electric field. Does the work done by the electric field depend on the path of the charge ?

12.

It is said that the separation between the two charges forming an electric dipole should be small. Small compared to what ?

13.

The number of electrons in an insulator is of the same order as the number of electrons in a conductor. What is then the basic difference between a conductor and an insulator ?

14.

When a charged comb is brought near a small piece of paper, it attracts the piece. Does the paper become charged when the comb is brought near it ?

OBJECTIVE – I 1.

Fig. shown some of the electronic field lines corresponding to an electric to an electric field. The figure suggests that fuEu iznf'kZr fp=k esa ,d fo|qr {ks=k ls lEc) fo|qr cy js[kk,sa n'kkZ;h xbZ gS -

(A) EA > EB > EC (C*) EA = EC > EB

(B) EA = EB = EC (D) EA = EC < EB

2.

When the seperation between two charges is incresased, the electric potential energy of the charges (A) increases (B) decresaes (C) remains the same (D*) may increase or decrease tc nks vkos'kksa ds e/; nwjh c<+kbZ tkrh gS rks vkos'kksa dh fo|qr fLFkfrt ÅtkZ (A) c<+rh gS (B) ?kVrh gS (C) iwoo Z r~ jgrh gSA (D*) c<+ ;k ?kV ldrh gS

3.

If a positive charge is shifted from a low-potential region to a high-potential region, the electric potential energy (A*) increases (B) decresaes (C) remains the same (D) may increase or decrease

vxj ,d /kukRed vkos'k dks U;wu foHko {ks=k ls mPp foHko {ks=k dh vksj ys tkrs gS rks fo|qr ÅtkZ (A*) c<+rh gSA (B) ?kVrh gSA (C) ,d leku jgrh gSA (D) ?kV ldrh gS ;k c<+ ldrh gSA 4.

HCV_29_Obj I_3

Two equal positive charges are kept at points A and B. The electric potential at the points between A and B (excluding these points) is studied while moving from A to B. The potential [HCV_Chp.29_Obj_4] (A) continuously increases (B) continuosly decreases (C) increases then decreases (D*) decreases than increases nks leku /kukRed vkos'k fcUnq A o B ij j[ks gSAa A ls B dh vksj pyrs gq, fcUnq A o B ds e/; fo|qr foHko dk v/;;u ¼bu fcUnqvksa ds vfrfjDr½ fd;k tkrk gSA fo|qr foHko manishkumarphysics.in

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Chapter # 29 (A) yxkrkj (C) yxkrkj


c<+rk gS ?kVrk gS

(B) igys c<+rk gS fQj ?kVrk gS (D*) igys ?kVrk gS fQj c<+rk gS

5.

The electric field at the origin is along the positive X-axis. A small circle is drawn with the centre at the origin cutting theaxes at points A, B, C and D having coordinates (a,0), (0,a), (-a,0), (0,-a) respectively. Out of the points on the periphery of the circle, the potential is minimum at ewy fcUnq ij fo|qr {ks=k /kukRed X-v{k ds vuqfn'k gS ewy fcUnq dks dsUnz ekurs gq, ,d NksVk o`Ùk [khapk tkrk gS tks v{kksa dks A, B, C o D fcUnqvksa ij dkVrk gSA bu fcUnqvksa ds funs±'kkad Øe'k% (a,0), (0,a), (-a,0), (0,-a) gSA o`Ùk dh ifjf/k ij fLFkr fcUnqvksa esa ls U;wure foHko gksxk (A*) A (B) B (C) C (D) D

6.

If a body is charged by rubbing it, is weight (A) remains precisely constant (B) increases slightly (C) decreases slightly (D*) may increase slightly or may decrease slightly ;fn ,d fudk; dks jxM+dj vkosf'kr fd;k tkrk gS rks bldk nzO;eku (A) fu;r jgrk gSA (B) FkksM+k lk c<+rk gSA (C) FkksM+k lk ?kVrk gSA (D*) FkksMk+ lk c<+ vFkok FkksMk+ lk ?kV ldrk gS

7.

An electric dipole is placed in a uniform electric field. The net electric force on the dipole (A*) is always zero (B) depends on the orientation of the dipole (C) can never be zero (D) dependw on the strength of the dipole ,d fo|qr f}/kzoq dks ,d leku fo|qr {ks=k esa j[kk tkrk gSA f}/kzoq ij dqy fo|qr cy (A*) ges'kk 'kwU; gksrk gS (B) f}/kzo q ds vfHkfoU;kl ij fuHkZj djrk gS (C) dHkh Hkh 'kwU; ugha gks ldrk gSA (D) f}/kzqo ds lkeF;Z ij fuHkZj djrk gS

8.

Consider the situtaion fig. The work done in taking a point charge from P to A is W A, from P to B is W B and from P to C is W C fuEu iznf'kZr fp=k ij fopkj fdft,A fcUnq vkos'k dks P ls A rd ys tkus esa W A P ls B rd ys tkus esa W B rFkk P ls C rd ys tkus esa W C dk;Z djuk iM+rk gS &

(A) W A < W B < W C (C*) W A = W B = W C 9.

(B) W A > W B > W C (D) none of these

A point charge q is rotated along a circle in the electric field generated by another point charge Q. The work done by the electric field on the roatating charge in one complete revolution is (A*) zero (B) positive (C) negative (D) zero if the charge Q is at the centre and nonzero otherwise ,d fcUnq vkos'k q dks ,d vU; fcUnq vkos'k Q }kjk mRiUu fo|qr {ks=k esa ,d o`Ùk ds vuqfn'k ?kqek;k tkrk gSA fo|qr {ks=k esa ,d o`Ùk ds vuqfn'k ?kqek;k tkrk gSA fo|qr {ks=k }kjk ,d iw.kZ pØ esa ?kwf.kZr vkos'k ij fd;k x;k dk;Z gksxk (A*) 'kwU; (B) /kukRed (C) _.kkRed (D) 'kwU; ;fn vkos'k Q dsUnz ij gS vFkok v'kwU;A

OBJECTIVE – II 1.

Mark out the correct options. (A*) The total charge of the universe is constant. (B) The total positive charge of the universe is constant. (C) The total negative charge of the universe is constant (D) The total number of charged particles in the universe is constant fuEu esa ls lgh fodYi gS manishkumarphysics.in

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Chapter # 29


(A*) fo'o dk dqy vkos'k fu;r gSA (C) fo'o dk dqy _.kkRed vkos'k fu;r 2.

gSA

(B) fo'o (D) fo'o

dk dqy vkos'k fu;r gSA esa vkosf'kr d.kksa dh dqy la[;k fu;r gSA

A point charge is brought in an electric field. The electric field at a nearby point

HCV II _Ch-29_Obj.2_2

fo|qr {ks=k esa ,d fcUnq vkos'k dks yk;k tkrk gSA utnhd ds fcUnq ij fo|qr {ks=k (A) will increase if the charge is positive (C*) may increase if the charge is positive (A) c<+x s k vxj vkos'k /kukRed gSA (C*) c<+ ldrk gS vxj vkos'k /kukRed gSA

(B) will decrease if the charge is negative (D*) may decrease if the charge is negative (B) ?kVsxk vxj (D*) ?kV ldrk

vkos'k _.kkRed gSA gS vxj vkos'k _.kkRed gSA

3.

The electric field and the electric potential at a point are E and V respectively (A) If E = 0, V must be zero (B) If V = 0, E must be zero (C) If E  0, V cannot be zero (D) If V  0, E cannot be zero fdlh fcUnq ij fo|qr {ks=k ,oa fo|qr foHko Øe'k% E o V gS (A) ;fn E = 0, V Hkh 'kwU; gksxkA (B) ;fn V = 0, E Hkh 'kwU; gksxkA (C) ;fn E  0, V 'kwU; ugha gks ldrkA (D) ;fn V  0, E 'kwU; ugha gks ldrkA Ans. None of these buesa ls dksbZ ugha A

4.

The electric potential decreases uniformly from 120 V to 80 V as one moves on the X-axis from x = – 1 cm to x = + 1 cm. The electric field at the origin (A) must be equal to 20V/cm (B*)may be equal to 20V/cm (C*) may be greater than 20V/cm (D) may be less than 20V/cm x-v{k ds vuqfn'kx = –1 lseh ls x = +1 lseh rd xfr djus ij fo|qr foHko ,d leku :i ls 120 oksYV ls 80 oksYV rd ?kVrk gSA dsUnz fcUnq ij fo|qr {ks=k (A) 20 oksYV@lseh ds cjkcj gksuk pkfg, (B*) 20 oksYV@lseh ds cjkcj gks ldrk gSA (C*) 20 oksYV@lseh ls T;knk gks ldrk gSA (D) 20 oksYV@lseh ls de gks ldrk gSA

5.

Which of the following quantites do not depend on the choice of zero potential or zero potential energy (A) potential at a point (B*) potential difference between two points (C) potential energy of a two - charge system (D*) change in potential energy of a two-charge system fuEu esa ls dkSulh jkf'k;k¡ 'kwU; foHko ;k 'kwU; fLFkfrt ÅtkZ ds p;u ij fuHkZj ugha djrh gS (A) ,d fcUnq ij foHko (B*) nks fcUnqvksa ds e/; foHkokUrj (C) nks vkos'kksa ds fudk; dh fLFkfrt ÅtkZ (D*) nks vkos'kksa ds fudk; dh fLFkfrt ÅtkZ esa ifjorZu

6.

An electric dipole is placed in an electric field generated by a point charge (A) The net electric force on the dipole must be zero (B) The net electric force on the dipole may be zero (C) The torque on the dipole due to the field must be zero (D*) The torque on the dipole due to the field may be zero ,d fo|qr f}/kzoq ij fcUnq vkos'k }kjk mRiUu fo|qr {ks=k esa j[kk gS (A) f}/kzo q ij dqy oS|rq cy 'kwU; gksuk pkfg,A (B) f}/kzo q ij dqy oS|rq cy 'kwU; gks ldrk gSA (C) {ks=k ds dkj.k f}/kzo q ij cy vk?kw.kZ 'kwU; gksuk pkfg,A (D*) {ks=k ds dkj.k f}/kzo q ij cy vk?kw.kZ 'kwU; gks ldrk gSA

7.

A proton and an electron are placed in a unifrorm electric field. [HCV_CHP_29_OBJ-2_7] (A) The electric forces acting on them will be equal (B*) The magnitudes of the forces will be equal (C) Their acelerations will be equal (D) The magnitudes of the accelerations will be equal. ,d izkVs kWu o ,d bysDVªkWu ,d le:i fo|qr {ks=k esa j[ks x, gSa (A) mu ij yxus okyk oS|qr cy cjkcj gksaxsA (B*) cyksa ds ifjek.k cjkcj gksx a As (C) muds Roj.k cjkcj gksaxsA (D) muds Roj.k ds ifjek.k cjkcj gksx a sA

8.

The electric field in a region is directed outward and is propertional to the distance r form the origin. Taking the electric potential at the origin to be zero manishkumarphysics.in

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Chapter # 29


(A) it is uniform in the region (C*) it is proportional to r 2

(B) it is proportional to r (D) it increases as one goes away from the origin ls nwjh r lekuqikrh gSA ewy fcUnq ij fo|qr foHko dk eku 'kwU;

,d {ks=k esa fo|qr {ks=k ckgj dh vksj gS rFkk ewy fcUnq ij (A) foHko {ks=k esa ,d leku gSA (B) foHko r ds lekuqikrh gSA 2 (C*) foHko r ds lekuqikrh gSA (D) foHko ewy fcUnq ls nwj tkus ij c<+rk gSA

ysus

EXERCISES 1. Ans. 2.

Find the dimensional formula of  0.  0 dk foeh; lw=k Kkr dhft,A I2M–1 L–3T4 A charge of 1.0 C is placed at the top of the your college building and another equal charge at the top of your house. Take the separation between the two charges to be 2.0 km. Find the force exerted by the charges on each other. How many times of your weight is this force ? 1.0 C dk ,d vkos'k vkids egkfo|ky; Hkou ds f'k[kj ij j[kk gS] tcfd nwljk leku vkos'k vkids ?kj ds f'k[kj ij j[kk gSA nksuksa vkos'kksa ds e/; nwjh 2.0 fdeh eku fyft,A vkos'kksa }kjk ,d nwljs ij yxk;k x;k cy fudkfy,A ;g cy vkids

Hkkj dk fdruk xquk gS\

Ans. 3.

Ans. 4.

Ans. 5.

9 × 103 N = 2.25 × 103 N 4 At what separations should two equal charges, 1.0 C each, be placed so that the force between them equals the weight of a 50 kg person ? (g = 9.8 m/s 2) 1.0 C ds nks leku vkos'kksa dks ,d nwljs ls fdruh nwjh ij j[kuk pkfg,] rkfd muds e/; cy 50 fdyks ds O;fDr ds Hkkj ds cjkcj gksA (g = 9.8 m/s2)

30 × 103 m 7 Two equal charges are placed at a separation of 1.0 m. What should be the magnitude of the charges so that the force between them equals the weight of a 50 kg person ? (g = 9.8 m/s 2) nks leku vkos'k ,d nwljs ls 1.0 eh- nwjh ij j[ks gSAa vkos'kksa dk ifjek.k fdruk gksuk pkfg, rkfd muds e/; cy 50 fdyks ds O;fDr ds Hkkj ds cjkcj gksA (g = 9.8 m/s2) 7 × 10–4 C 3

Find the electric force between two protons separated by a distance of 1 fermi (1 fermi = 10 –15 m). The protons in a nucleus remain at a separation of this order.

nks izkVs kWu tks ,d nwljs ls 1 QehZ (1 QehZ = 10–15 eh) nwjh ij gS] ds e/; fo|qr cy fudkfy;sA ukHkd esa izksVkWuksa ds e/; nwjh bl dksfV dh gksrh gSA Ans.

230.4 N

6.

Two charges 2.0 × 10–6 C and 1.0 × 10–6 C are placed at a separation of 10 cm. Where should a third charge be placed such that it experiences no net force due to these charges ? nks vkos'k 2.0 × 10–6 dwyke rFkk 1.0 × 10–6 dwykWe ,d nwljs ls 10 lseh dh nwjh ij j[ks x;s gSAa ,d rhljk vkos'k dgk¡

j[kk tkuk pkfg,] ftlls ;g bu vkos'kksa ds dkj.k dksbZ ifj.kkeh cy vuqHko uk djsAa

Ans. 7.

10 2 2 1

 20  10 2 = 5.9 cm from the larger charge in between the two charges

Suppose the second charge in the previous problem is – 1.0 × 10–6 C. Locate the position where a third charge will not experience a net force. eku fyft, iwoZ iz'u esa nwljk vkos'k – 1.0 × 10–6 C gSA og fLFkfr fu/kkZfjr fdft, tgk¡ ,d rhljk vkos'k dksbZ ifj.kkeh

cy vuqHko u djsAa


Page # 15

Chapter # 29


10

Ans.

8.

Ans. 9.

Ans. 10.

 10( 2  1) = 24.1 cm from the charge of smaller magnitude on the line joining the charge in 2 1 the side of the smaller charge. Two charged particles are placed at a distance 1.0 cm apart. What is the minimum possible magnitude of the electric force acting on each charge ? nks vkosf'kr d.k ,d nwljs ls 1.0 lseh nwj j[ks gSAa izR;sd vkos'k ij yxus okys oS|rq cy dk U;wure laHko eku D;k gS\ 2.3 × 10–24 N Estimate the number of electrons in 100 g of water. How much is the total negative charge on these electrons ? 100 xzke ikuh esa bysDVªkWuksa dh la[;k dk vkdyu dhft,A bu bysDVªkWuksa ij dqy _.k vkos'k fdruk gS\ 3.35 × 1025, 5.35 × 106 C Suppose all the electrons of 100 g water are lumped together to form a negatively charged particle and all the nuclei are lumped together to form a positivively charged particle. If these two particles are placed 10.0 cm away from each other, find the force of attraction between them. Compare it with your weight. eku fyft, dh 100 xzke ikuh ds lHkh bysDVªkuW ksa dks bdB~Bk dj ,d _.k vkosf'kr d.k cuk fn;k tkrk gS rFkk lHkh ukfHkdksa dks bdB~Bk dj /ku vkosf'kr d.k cuk fn;k tkrk gSA ;fn ;s nks d.k ,d nwljs ls 10.0 lseh nwj j[ks tkrs gSa rks muds e/

; vkd"kZ.k cy fudkfy,A bl cy dh vius Hkkj ls rqyuk dhft,A

Ans.

2.56 × 1025 N

11.

Consider a gold nucleus to be a sphere of radius 6.9 fermi in which protons and neutrons are distributed. Find the force of repulsion between two protons situated at largest separation. Why do these protons not fly apart under this repulsion ? lksus ds ,d ukfHkd dks 6.9 QehZ f=kT;k ds xksys ds :i esa ekfu;s ftlesa izkVs kWu o U;wVkª uW forfjr gS] ,d nwljs ls vf/kdre

nwjh ij fLFkr nks izkVs kWuksa ds e/; izfrd"kZ.k cy Kkr dhft,A ;s izksVkWu bl izfrd"kZ.k ds dkj.k ,d nwljs ls ijs D;ksa ugha gks tkrs\

Ans.

1.2 N

12.

Two insulating small spheres are rubbed aganist each other and placed 1 cm apart. If they attract each other with a force of 0.1 N, how many electrons were transferred from one sphere to the other during rubbing ? [HCV_CHP-29_EX.12] nks NksVs vpkyd xksys ,d nwljs ls jxM+ dj 1 lseh dh nwjh ij j[k fn;s tkrs gSAa ;fn os ,d nwljs dks 0.1 N ds cy ls

Ans.

2 × 1011

13.

NaCl molecule is bound due to the electric force between the sodium and the chlorine ions when one electron of sodium is transferred to chlorine. Taking the separation between the ions to be 2.75 × 10–8 cm, find the force of attraction between them. State the assumptions (if any) that you have made. NaCl v.kq lksfM;e ,oa Dyksfju vk;uksa ds e/; oS|rq cy ds dkj.k c) voLFkk esa gS] tcfd lksfM;e dk ,d bysDVªkuW Dyksjhu dks LFkkukUrfjr gksrk gSA vk;uksa ds e/; nwjh 2.75 × 10–8 lseh ysrs gq,] muds e/; vkd"kZ.k cy Kkr dhft,A vkids }kjk

Ans.

3.05 × 10–9 N

14.

Find the ratio of the electric and gravitational forces between two protons.

Ans.

1.23 × 1036

15.

Suppose an attractive nuclear force acts between two protons which may be written as F = Ce–kr/r2. (a) Write down the dimensional formulae and appropriate SI units of C and k. (b) Suppose that k = 1 f erm i –1 and that the r epuls ive elec tric f orc e between the pr otons is j us t balanced by the attractive nuclear when the separation is 5 fermi. Find the value of C. eku fyft, dh nks izkVs kWu ds e/; ,d ukfHkdh; vkd"kZ.k cy dk;Z djrk gS] ftls F = Ce–kr/r2 }kjk fy[kk tk ldrk gS (a) C ,oa K ds foeh; lw=k ,oa mi;qDr SI ek=kd fyf[k,A (b) eku fyft, fd k = 1 QehZ–1 rFkk izkVs kWuksa ds e/; oS|r q izfrd"kZ.k cy ukfHkdh; vkd"kZ.k cy ls Bhd larfq yr gksrk gS tc muds e/; nwjh 5 QehZ gS rks C dk eku Kkr dhft,A (a) ML3T–2, L–1, N–m 2, (b) 3.4 × 10–26 N–m 2

vkdf"kZr djrs gSa rks jxM+us ds nkSjku ,d xksys ls nwljs ij fdrus bysDVªkWu LFkkukarfjr gq,\

dh xbZ dYiukvksa ¼;fn dksbZ gS½ dks O;Dr dhft,A

nks izksVkWuksa ds e/; oS|rq ,oa xq:Roh; cyksa dk vuqikr Kkr dhft,A

Ans. 16.

Three equal charges, 2.0 × 10–6 C each, are held fixed at the three corners of an equilateral triangle of side 5 cm. Find the coulomb force experienced by one of the charges due to the rest two. manishkumarphysics.in

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Chapter # 29


2.0 × 10–6 dwyke

ds rhu leku vkos'k 5 lseh Hkqtk ds ,d leckgq f=kHkqt ds 'kh"kks± ij j[ks x, gSAa fdlh ,d vkos'k ij vU; nksuksa vkos'kksa ds dkj.k yxk;k x;k dwykWe cy Kkr dhft,A Ans.

24.9 N at 30° with the extended sides from the charge under consideration.

17.

Four equal charges 2.0 × 10–6 C each are fixed at the four corners of a square of side 5 cm. Find the coulomb force experienced by one of the charges due to the rest three. 2.0 × 10–6 dwykWe ds pkj leku vkos'k 5 lseh Hkqtk ds ,d oxZ ds dksuksa ij j[ks x, gSAa fdlh ,d vkos'k ij vU; rhuksa

Ans.

27.5 N at 45° with the extended sides of the square from the charge under consideration.

18.

A hydrogen atom contains one proton and one electron. It may be assumed that the electron revolves in a circle of radius 0.53 angstrom (1 angstrom = 10–10 m and is abbreviated as Å) with the proton at the centre. The hydrogen atom is said to be in the ground state in this case. Find the magnitude of the electric force between the proton and the electron of a hydrogen atom in its ground state.

vkos'kksa ds dkj.k yxk;k x;k dwykWe cy Kkr dhft,A

,d gkbMªkt s u ijek.kq esa ,d bysDVªkuW o ,d izkVs kWu gksrk gSA izkVs kWu dks dsUnz ij ysrs gq, ;g ekuk tk ldrk gS fd bysDVªkuW 0.53 Å f=kT;k ds o`Ùk ls pDdj yxkrk gSA (1 Å angstrom = 10–10 m) A bl fLFkfr esa gkbMªkt s u ijek.kq dh ewy voLFkk esa dgk tkrk gSA gkbMªkt s u ijek.kq dh ewy voLFkk esa dgk tkrk gSA gkbMªkt s u ijek.kq dh ewy voLFkk esa izkVs kWu o bysDVªkuW ds e/; oS|rq cy dk eku Kkr dhft,A Ans.

8.2 × 10–8 N

19.

Find the speed of the electron in the ground state of a hydrogen atom. The description of ground state is given in the previous problem.

Ans.

2.18 × 106 m/s

20.

Ten positively charged particle are kept fixed on the X-axis at points x = 10 cm, 20 cm, 30 cm, ....., 100 cm. The first particle has a charge 1.0 × 1.0–8 C, the second 8 × 10–8 C, the third 27 × 10–8 C and so on. The tenth particle has a charge 1000 × 10–8 C. Find the magnitude of the electric force acting on a 1 C charge placed at the origin. nl /ku vkosf'kr d.k X-v{k ij x = 10 lseh, 20 lseh, 30 lseh, ....., 100 lseh. fcUnqvksa ij j[ks x, gSAa igys d.k ij vkos'k 1.0 × 1.0–8 C, nwljs ij 8 × 10–8 C, rhljs ij 27 × 10–8 C vkSj blh izdkj nlosa d.k ij vkos'k 1000 × 10–8 C gSA ewy fcUnq ij j[ks 1 C ds vkos'k ij dk;Zjr oS|qr cy dk eku Kkr dhft,A 4.95 × 105 N

,d gkbMªkt s u ijek.kq dh ewy voLFkk esa bysDVªkWu dh pky Kkr dhft,A ewy voLFkk ds ckjs esa iwoZ iz'u esa crk;k x;k gSA

Ans. 21.

Two charged particles having charge 2.0 × 10–8 C each are joined by an insulating string of length 1 m and the system is kept on a smooth horizontal table. Find the tension in the string. nks vkosf'kr d.k] ftuesa izR;sd ij vkos'k 2.0 × 10–8 dwykWe gS] 1 eh- dh dqpkyd Mksjh ls tksMs+ x, gSAa fudk; dks ,d fpduh

{ksfrt Vsfcy ij j[kk x;k gSA Mksjh esa ruko Kkr dhft,A

Ans.

3.6 × 10–6 N

22.

Two identical balls, each having a charge of 2.00 × 10–7 C and a mass of 100 g, are suspended from a common point by two insulating strings each 50 cm long. The balls are held at a separation 5.0 cm apart and then released. Find (a) the electric force on one of the charged balls (b) the components of the resultant force on it along and perpendicular to the string (c) the tension in the string (d) the acceleration of one of the balls. Answers are to be obtained only for the instant just after the release. 2.00 × 10–7 C vkos'k o 100 xzke nzO;eku izR;sd dh nks leku xsna ksa dks ,d mHk;fu"B fcUnq ls 50 lseh izR;sd yEckbZ dh dqpkyd Mksfj;ksa }kjk yVdk;k x;k gSA nksuksa xsna ksa dks ,d nwljs ls 5.0 lseh nwj ys tkdj NksM+ fn;k tkrk gSA Kkr dhft, : (a) fdlh ,d vkosf'kr xsna ij oS|rq cy (b) xsna ij Mksjh ds vuqfn'k ,oa yEcor~ ifj.kkeh cy ds ?kVd (c) Mksjh esa ruko (d) fdlh ,d xsan dk Roj.kA xsna ksa dks NksM+us ds rqjUr i'pkr~ ds d.k ds fy, gh mÙkj Kkr djus gSaA (a) 0.144 N (b) zero, 0.095 N away from the other charge (c) 0.986 N and (d) 0.95 m/s 2 perpendicular to the string and going away from the other charge

Ans.

23.

Two identical pith balls are charged by rubbing against each other. They are suspended from a horizontal rod through two strings of length 20 cm each, the separation between the suspension points being 5 cm. In equilibrium, the separation between the balls is 3 cm. Find the mass of each ball and the tension in the strings. The charge on each ball has a magnitude 2.0 × 10–8 C. nks leku xsna ksa dks ,d nwljs ls jxM+dj vkosf'kr fd;k tkrk gSA mUgsa ,d {kSfrt NM+ ls 20 lseh yEckbZ dh nks Mksfj;ksa }kjk yVdk;k x;k gSA] fuyEcu fcUnqvksa ds e/; nwjh 5 lseh gSA lkE;koLFkk esa xsna ksa ds e/; nwjh 3 lseh gSA izR;sd xsna dk nzO;eku ,oa Mksfj;ksa esa ruko Kkr dhft,A izR;sd xsna ij vkos'k dk ifjek.k 2.0 × 10–8 C gSA manishkumarphysics.in

Page # 17

Chapter # 29

Electric Field & Potential –2

Ans.

8.2 g, 8.2 × 10 N

24.

Two small spheres, each having a mass of 20 g, are suspended from a common point by two insulating strings of length 40 cm each. The spheres are identically charged and the separation between the balls at equilibrium is found to be 4 cm. Find the charge on each sphere. 20 xzke nzO;eku ds nks NksVs xksyksa dks ,d mHk;fu"B fcUnq ls 40 lseh yEckbZ dh nks dqpkyd Mksfj;ksa }kjk yVdk;k x;k gSA nksuksa xksys leku :i ls vkosf'kr gS ,oa muds e/; lkE;koLFkk esa nwjh 4 lseh gSA izR;sd xksys ij vkos'k Kkr dhft,A 4.17 × 10–8 C

Ans. 25.

Two identical pith balls, each carrying a charge q, are suspended from a common point by two strings of equal length l. Find the mass of each ball if the angle between the strings is 2 in equilibrium. [HCV_CHP-29_EX.25] q vkos'k dh nks leku xsna ksa dks ,d mHk;fu"B fcUnq ls leku yEckbZ dh Mksfj;ksa }kjk yVdk;k x;k gSA ;fn lkE;koLFkk esa Mksfj;ksa ds e/; dks.k 2 gks rks izR;sd xsna dk nzO;eku Kkr dhft,A

q2 cot  Ans.

16  0 gl 2 sin 2 

26.

A particle having a charge of 2.0 × 10–4 C is placed directly below and at a separation of 10 cm from the bob of a simple pendulum at rest. The mass of the bob is 100 g. What charge should the bob be given so that the string becomes loose. 2.0 × 10–4 C vkos'k dk ,d d.k ,d fLFkj ljy yksyd ds Bhd uhps xksyd ls 10 lseh nwjh ij j[kk x;k gSA xksyd dk nzO;eku 100 xzke gSA xksyd dks fdruk vkos'k fn;k tk, rkfd Mksjh <+yh gks tk,\ 5.4 × 10–9 C

Ans. 27.

Two particles A and B having charges q and 2q respectively are placed on a smooth table with a separation d. A third particle C is to be clamped on the table in such a way that the particles A and B remain at rest on the table under electrical forces. What should be the charge on C and where should it be clamped ? q o 2q vkos'k ds nks d.k A o B ,d fpduh Vsfcy ij ,d nwljs ls d nwjh ij j[ks gq, gSa ,d rhljk d.k C Vsfcy ij bl rjg ls tdM+ dj (clamped) j[kk tkuk gS fd oS|rq cyksa ds izHkko esa A o B Vsfcy ij fLFkj jgsA C ij vkos'k fdruk gksxk

,oa bls dgk¡ tdM+ dj j[kk tkuk pkfg,\

Ans.

– (6 – 4 2 ) q, between q and 2q at a distance of ( 2 – 1) d from q.

28.

Two identically charged particles are fastened to the two ends of a spring of spring constant 100 N/m and natural length 10 cm. The system rests on a smooth horizontal table. If the charge on each particle is 2.0 × 10–8 C, find the extension in the length of the spring. Assume that the extension is small as compared to the natural length. Justify this assmuption after you solve the problem. 100 U;wVu@eh- fLizx a fu;rkad ,oa 10 eh- dqy yEckbZ dh ,d fLizxa ds nksuksa fljksa ls nks leku vkosf'kr d.k ca/ks gSA fudk; ,d ?k"kZ.k jfgr {kSfrt est ij fLFkjkoLFkk esa gSA ;fn izR;sd d.k ij vkos'k 2.0 × 10–8 C gS rks fLizxa dh yEckbZ esa fdruk

izlkj gksxk\ eku fyft, fd izlkj ewy yEckbZ dh rqyuk esa vYi gSA iz'u dks gy djus ds i'pkr~ bl ekU;rk dk vkWfpR; crkb;sA

Ans.

3.6 × 10–6 m

29.

A particle A having a charge of 2.0 × 10–6 C is held fixed on a horizontal table. A second charged particle of mass 80 g stays in equilibrium on the table at a distance of 10 cm from the first charge. The coefficient of friction between the table and this second particle is  = 0.2. Find the range within which the charge of this second particle may lie. 2.0 × 10–6 C vkos'k dk ,d d.k ,d {kSfrt est ij LFkkfir gSA 80 xzke dk ,d nwljk vkosf'kr d.k igys d.k ls 10 lseh nwjh ij est lkE;koLFkk esa gSA bl nwljs d.k o est ds e/; xq.kkad  = 0.2 gSA nwljs d.k ds vkos'k dh ijkl Kkr dhft,A between ± 8.71 × 10–8 C

Ans. 30.

A particle A having a charge of 2.0 × 20–6 C and a mass of 100 g is placed at the bottom of a smooth inclined plane of inclination 30° Where should another particle B, having same charge and mass, be placed on the incline so that it may remain in equilibrium ? 100 xzke nzO;eku o 2.0 × 20–6 C vkos'k dk ,d d.k 30° >qdko okys ,d fpdus vkur ry ds fuEure fcUnq ij j[kk gSA leku vkos'k ,oa nzO;eku dk ,d vU; d.k B vkur ry ij dgk¡ j[kk tk, rkfd ;g lkE;koLFkk esa jgsA

Ans.

27 cm from the bottom


Page # 18

Chapter # 29 31.


Two particles A and B, each having a charge Q, are placed a distance d apart. Where should a particle of charge q be placed on the perpendicular bisector of AB so that it experiences maximum force ? What is the magnitude of this maximum force ? Q vkos'k ds nks d.k A o B ,d nwljs ls d nwjh ij j[ks x, gSAa q vkos'k ds ,d d.k dks AB ds yEcor~ f}Hkktd ij dgk¡

j[kk tkuk pkfg,] ftlus ;g vfHkyEc cy vuqHko djs\a bl vf/kdre cy dk ifjek.k fdruk gksxk\

Ans. 32.

Qq d/2 2 , 3.08 4 0 d2

Two particles A and B, each carrying a charge Q, are held fixed with a separation d between them. A particle C having mass m and charge q is kept at the middle point of the line AB. (a) If it is displaced through a distance x perpendicular to AB, what would be the electric force experiencd by it. (b) Assuming x << d, show that this force is proportional to x. (c) Under what conditions will the particle C execute simple harmonic motion if it is released after such a small displacement ? Find the time period of the oscillations if these conditions are satisfied. Q vkos'k ds nks d.k A o B ,d nwljs ls d nwjh ij LFkkfir gSA m nzO;eku o q vkos'k dk ,d d.k C js[kk AB ds e/; fcUnq ij j[kk tkrk gSA (a) ;fn bls AB ds yEcor~ x nwjh ij foLFkkfir fd;k tkrk gS] rks blds }kjk vuqHko fd;k x;k cy fdruk gksxkA (b) x << d, ekurs gq,] n'kkZb;s fd ;g cy x ds lekuqikrh gSA (c) fdu ifjfLFkfr;ksa esa d.k C ljy vkorZ xfr

djsxk] ;fn bls blh izdkj vYi foLFkkfir dj NksM+k tkrk gS\ ;fn ;s fLFkfr;k¡ lar"q V gksrh gks rks nksyu dk vkorZdky Kkr dhft,A 1

Ans.

33.

(a)

Qqx 3/2

 m  3  0 d3  2 (c)    Qq 

 d2  2 0  x 2  4   Repeat the previous problem if the particle C is displaced through a distance x along the line AB. iwoZ iz'u dks nksgjkb;s ;fn d.k C dks nks d.k AB ds vuqfn'k x nwjh rd foLFkkfir fd;k tkrk gS\ 1

Ans.

 3  0md3  2 time period     2Qq 

34.

The electric force experienced by a charge of 1.0 × 10–6 C is 1.5 × 10–3 N. Find the magnitude of the electric field at the position of the charge. 1.0 × 10–6 C ds vkos'k }kjk vuqHko fd;k x;k oS|rq cy 1.5 × 10–3 N gSA vkos'k dh fLFkfr ij fo|qr {ks=k dk ifjek.k Kkr

Ans.

1.5 × 103 N/C

35.

Two particles A and B having charges of + 2.00 × 10–6 C and of – 4.00 × 10–6 C respectively are held fixed at a separation of 20.0 cm. Locate the point(s) on the line AB where (a) the electric field is zero (b) the electric potential is zero. 2.00 × 10–6 C o – 4.00 × 10–6 C ds nks d.k A o B ,d nwljs ls 20.0 lseh- nwj j[ks x, gSaA js[kk AB ij fcUnq@fcUnqvksa dh fLFkfr Kkr dhft,] tgk¡ : (a) fo|qr {ks=k 'kwU; gSA (b) fo|qr foHko 'kwU; gSA (a) 48.3 cm from A along BA

dhft,A

Ans.

(b) 20 cm from A along BA and 36.

Ans.

20 cm from A along AB 3

A point charge produces an electric field of magnitude 5.0 N/C at a distance of 40 cm from it. What is the magnitude of the charge ? ,d fcUnq vkos'k blesa 40 lseh nwjh ij 5.0 N/C ifjek.k dk fo|qr {ks=k mRiUu djrk gSA vkos'k dk ifjek.k fdruk gS\ 8.9 × 10–11 C

37.

A water particle of mass 10.0 mg and having a charge of 1.50 × 10–6 C stays suspended in a room. What is the magnitude of electric field in the room ? What is its direction ? 10.0 mg nzO;eku ,oa 1.50 × 10–6 C vkos'k dk ,d ikuh dk ,d d.k ,d dejs ls fuyafcr gSA dejs esa fo|qr {ks=k dk

Ans.

65.3 N/C, upward

38.

Three identical charges, each having a value 1.0 × 10–8 C, are placed at the corners of an equilateral triangle of side 20 cm. Find the electric field and potential at the centre of the triangle.

ifjek.k fdruk gS\ bldh fn'kk D;k gS\


Page # 19

Chapter # 29

Ans. 39.


rhu leku vkos'k] ftuesa izR;sd ij vkos'k 1.0 × 10–8 C gS] 20 lseh Hkqtk ds ,d leckgq f=kHkqt ds dksuksa ij j[ks x, gSAa f=kHkqt d s dsUnz ij fo|qr {ks=k ,oa foHko Kkr dhft,A zero, 'kwU; 2.3 × 103 V Positive charge Q is distributed uniformly over a circular ring of radius R. A particle having a mass m and a negative charge q, is placed on its axis at a distance x from the centre. Find the force on the particle. Assuming x << R, find the time period of oscillation of the particle if it is released from there. R f=kT;k dh ,d o`Ùkkdkj oy; ij /kukRed vkos'k Q leku :i ls forfjr gSA m nzO;eku o _.k vkos'k q dk ,d d.k blds v{k ij dsUnz ls x nwjh ij j[kk x;k gSA d.k ij cy Kkr dhft,A x << R ekurs gq,] d.k ds nksyu dk vkorZdky

Kkr dhft,A ;fn bls ogk¡ ls NksM+k tk,A

Ans. 40.

163  0mR 3    Qq  

1/ 2

A rod of length L has a total charge Q distributed uniformly along its length. It is bent in the shape of a semicircle. Find the magnitude of the electric field at the centre of curvature of the semicircle. [Q.40/HCV-2/CH-29/Exercise] [4] L yEckbZ dh ,d NM+ ij bldh yEckbZ ds vuqfn'k dqy Q vkos'k leku :i ls forfjr gSA bls ,d v)ZoÙ` k ds :i esa eksMk+

x;k gSA v)ZoÙ` k ds oØrk dsUnz ij fo|qr {ks=k dk ifjek.k Kkr dhft,A

[Q.40/HCV-2/CH-29/Exercise] Sol.

[4]

Q . Let there be a small element of length Rd at an angle  from OA. L Charge on element = Rd. Linear charge density  =

Electric field due to element = dE =

kdq R2

kRd R2

dE = – dEsin ˆi – dEcos ˆj /2

Ex = dE sin =

 k   d  sin = k  cos  / 2 = 0  / 2 R  R  / 2



/2

Ey =dEcos =

ENet = –

 k   d  cos = k sin  / 2 = 2k – / 2 R  R R  / 2



2 2k ˆj = 4 0 R

Q/L ˆ L j 

Q

ENet = –

2 0 L2 Q

ENet =

2 0 L2

, along angular bisector or ring and away from ring.

Ans.

41.

A 10 cm long rod carries a charge of + 50 C distributed uniformly along its length, Find the magnitude of the electric field at a point 10 cm from both the ends of the rod. 10 lseh yEch ,d NM+ ij + 50 C vkos'k bldh yEckbZ ds vuqfn'k leku :i ls forfjr gSA NM+ ds nksuksa fljksa ls 10

Ans.

5.2 × 107 N/C

42.

Consider a uniformly charged ring of radius R. Find the point on the axis where the electric field is maximum. R f=kT;k ds ,d leku :i ls vkosf'kr oy; ij fopkj dhft,A v{k ij og fcUnq Kkr dhft,A tgk¡ oS|r q {ks=k vf/kdre

lseh nwj fLFkr fcUnq ij oS|rq {ks=k dk ifjek.k Kkr dhft,A

gksA


Page # 20

Chapter # 29


Ans.

R/ 2

43.

A wire is bent in the form of a regular hexagon and a total charge q is distributed uniformly on it. What is the electric field at the centre ? You may answer this part without making any numerical calculations. ,d rkj dks ,d fu;fer "kV~Hkqt ds :i esa eksMk+ x;k gS rFkk bl ij dqy vkos'k q leku :i ls forfjr gSA dsUnz ij fo|qr

Ans.

zero

44.

A circular wire-loop of radius a carries a total charge Q distributed uniformly over its length. A small length dL of the wire is cut off. Find the electric field at the centre due to the remaining wire. a f=kT;k ds ,d o`Ùkkdkj rkj ds ywi ij dqy vkos'k Q bldh yEckbZ ij forfjr gSA rkj ls ,d vYi yEckbZ dL dk VqdM+k

{ks=k D;k gS\ vki bl iz'u dk mÙkj fcuk dksbZ vkafdd x.kuk,a fd;s gq, ns ldrs gSAa 'kwU;

dkVk tkrk gSA dsUnz ij 'ks"k rkj ds dkj.k fo|qr {ks=k Kkr dhft,A QdL

Ans.

8 2  0 a 2

45.

A positive charge q is placed in front of a conducting solid cube at a distance d from its centre. Find the electric field at the centre of the cube due to the charges appearing on its surface. [2] [HCV-II_Chap. 29_45] ,d /kkukRed vkos'k q ,d pkyd Bksl /ku ds lkeus blds dsUnz ls d nwjh ij j[kk x;k gSA /ku ds dsUnz ij blds i`"B ij mifLFkr vkos'kksa ds dkj.k] fo|qr {ks=k Kkr dhft,A [2]

Sol. Due to presence of +q charge, charge is redistributed on conducting solid sphere in such a way that net electric field inside conducting sphere is zero.    E = 0 = Epo int ch arg e + E induced ch arg e

 kq Epo int ch arg e = 2 rˆ d  kq so that Einduced ch arg e = – 2 rˆ d  kq  Ans.  2  ] d 

46.

A pendulum bob of mass 80 mg and carrying a charge of 2 × 10–8 C is at a distance d from its centre. Find the electric field at the centre of the cube due to the charges appearing on its surface. 80 mg nzO;eku dk ,d yksyd dk xksyd] ftl ij 2 × 10–8 C vkos'k gS] 20 kv/m ds leku ,oa {ksfrt fo|qr {ks=k esa fLFkj

gSAa Mksjh esa ruko Kkr dhft,A

Ans.

8.8 × 10–4 N

47.

A particle of mass m and charge q is thrown at a speed u against a uniform electric field E. How much distance will it travel before coming to momentary rest ? m nzO;eku o q vkos'k dk ,d d.k leku fo|qr {ks=k E ds fo:) u osx ls QSd a k tkrk gSa vkaf'kd :i ls fojkekoLFkk esa vkus

ls iwoZ ;g fdruh nwjh r; dj ysxk\

Ans. 48.

mu2 2qE

A particle of mass 1 g and charge 2.5 × 10 –4 C is released from rest in an electric field of 1.2 × 104 N/C. (a) Find the electric force and the force of gravity acting on this particle. Can one of these forces be neglected in comparison with the other for approximate analysis ? (b) How long will it take for the particle to travel a distance of 40 cm ? (c) What will be the speed of the particle after travelling this distance ? (d) How much is the work done by the electric force on the particle during this pariod ? 1 xzke nzO;eku o 2.5 × 10–4 C vkos'k dk ,d d.k 1.2 × 104 N/C ds ,d fo|qr {ks=k esa fojkekoLFkk ls NksMk+ tkrk gSA (a)

d.k ij dk;Zjr oS|rq cy ,oa xq:Ro cy Kkr dhft,A D;k buesl a s ,d cy lfUudV fo'ys"k.k ds fy, nwljs dh rqyuk esa manishkumarphysics.in

Page # 21

Chapter # 29


ux.; ekuk tk ldrk gS\ (b) d.k dks 40 lseh nwjh r; djus esa fdruk le; yxsxk\ (c) bl nwjh dks r; djus ds ckn d.k dh pky D;k gksxh\ (d) bl varjky esa oS|rq cy }kjk d.k ij fdruk dk;Z fd;k tkrk gS\ Ans.

(a) 3.0 N, 9.8 × 10–3 N, (c) 49.0 m/s

(b) 1.63 × 10–2 s (d) 1.20 J

49.

A ball of mass 100 g and having a charge of 4.9 × 10–5 C is releasd from rest in a region where a horizontal electric field of 2.0 × 104 N/C exists. (a) Find the resultant force acting on the ball. (b) What will be the path of the ball ? (c) Where will the ball be at the end of 2s ? 100 xzke nzO;eku o 4.9 × 10–5 C vkos'k dh ,d xsna dks fojkekoLFkk ls ,d {ks=k esa NksM+k tkrk gS] tgk¡ 2.0 × 104 N/C dk ,d {kSfrt oS|rq {ks=k mifLFkr gSA (a) xsna ij dk;Zjr ifj.kkeh cy Kkr dhft,A (b) xsna dk iFk D;k gksxk\ (c) 2s i'pkr~

xsan dgk¡ ij gksxh\





Ans.

(a) 1.4 N making an angle of 45° with g and E (b) straight line along the resultant force (c) 28 m from the starting point on the line of motion

50.

The bob of a simple pendulum has a mass of 40 g and a positive charge of 4.0 × 10–6 C. It makes 20 oscillations in 45 s. A vertical electric field pointing upward and of magnitude 2.5 × 104 N/C is switched on. How much time will it now take to complete 20 oscillations ? ,d ljy yksyd ds xksyd dk nzO;eku 40 g o vkos'k 4.0 × 10–6 C gSA ;g 45 s esa 20 nksyu djrk gSA 2.5 × 104 N/C ifjek.k dk ,d fo|qr {ks=k Å/okZ/kj Åij dh vksj yxk;k tkrk gSA vc bls 20 nksyu iw.kZ djus esa fdruk

Ans.

52 s

51.

A block of mass m and having a charge q is placed on a smooth horizontal table and is connected to a wall through an unstressed spring of spring constant k as shown in figure. A horizontal electric field E parallel to the spring is switched on. Find the amplitude of the resulting SHM of the block. m nzO;eku dk ,d fiaM ftl ij q vkos'k gS] ,d fpduh {kSfrt est ij j[kk gS rFkk fp=kkuqlkj nhokj ls k fLizx a fu;rkad okyh ,d vrkZur (unstressed) fLizxa }kjk tqMk+ gSA fLizxa ds lekUrj ,d {kSfrt fo|qr {ks=k E yxk;k tkrk gSA fiaM dh

le; yxsxk\

ifj.kkeh ljy vkorZ xfr dk vk;ke Kkr dhft,A

Ans.

qE/k

52.

A block of mass m containing a net positive charge q is placed on a smooth horizontal table which terminates in a vertical wall as shown in figure. The distance of the block from the wall is d. A horizontal electric field E towards right it switched on. Assuming elastic collisions (if any) find the time period of the resulting oscillatory motion. Is it a simple harmonic motion ? m nzO;eku dk ,d fiaM] ftl ij dqy /kukRed vkos'k q gS] ,d fpduh {kSfrt est ij j[kk gSA est fp=kkuqlkj ,d m/okZ/ kj nhokj ds lkFk yxh gSA fiaM dh nhokj ls nwjh d gSA nka;h vksj ,d {kSfrt fo|qr {ks=k E yxk;k tkrk gSA VDdjksa dks izR;kLFk

ekurs gq, ¼;fn dksbZ gksrh gS½] ifj.kkeh nksyuh xfr dk vkorZdky Kkr dhft,A D;k ;g ljy vkorZ xfr gS\

Ans.

8 md qE

53.

A uniform electric field of 10 N/C exists in the vertically downward direction. Find the increase in the electric potential as one goes up through a height of 50 cm. 10 N/C dk ,d leku fo|qr {ks=k Å/okZ/kj uhps dh vksj fo|eku gSA ;fn 50 lseh Å¡pkbZ rd tk;k tk, rks fo|qr {ks=k esa

Ans.

5V

o`f) Kkr dhft,A


Page # 22

Chapter # 29


54.

12 J of work has to be done against an existing electric field to take a charge of 0.01 C from A to B. How much is the potential difference VB – VA ? 0.01 C ds vkos'k dks A ls B rd ys tkus esa fo|eku fo|qr {ks=k ds fo:) 12 J dk;Z djuk iM+rk gSA foHkokUrj VB – VA

Ans.

1200 volts

55.

Two equal charges, 2.0 × 10–7 C each, are held fixed at a separation of 20 cm. A third charge of equal magnitude is placed midway between the two charges. It is now moved to a point 20 cm from both the charges. How much work is done by the electric field during the process ? 2.0 × 10–7 C ds nks leku vkos'k ,d nwljs ls 20 lseh nwjh ij fLFkr gSA leku ifjek.k dk ,d rhljk vkos'k nksuksa vkos'kksa

fdruk gS\

ds e/; fcUnq ij j[kk tkrk gsA vc bls nksuksa ds e/; fcUnq ij j[kk tkrk gSA vc bls nksuksa vkos'kksa ds e/; fcUnq ij j[kk tkrk gSA vc bls nksuksa vkos'kksa ls 20 lseh nwj fLFkr fcUnq rd ys tk;k tkrk gSA bl izfØ;k esa fo|qr {ks=k }kjk fdruk dk;Z fd;k tkrk gS\

Ans.

3.6 × 10–3 J

56.

An electric field of 20 N/C exists along the X-axis in space. Calculate the potential difference VB – VA where the points A and B are given by, (a) A = (0, 0) ; B = (4m, 2m) (b) A = (4m, 2m) ; B = (6m, 5m) (c) A = (0, 0) ; B = (6m, 5m). Do you find any relation between the answers of parts (a), (b) and (c) ? 20 N/C dk ,d fo|qr {ks=k vkdk'k esa X-v{k ds vuqfn'k fo|eku gSA foHkokUrj VB – VA dh x.kuk dhft, tgk¡ fcUnq A o B fn;s tkrs gSa (a) A = (0, 0) ; B = (4m, 2m) (b) A = (4m, 2m) ; B = (6m, 5m) (c) A = (0, 0) ; B = (6m, 5m) D;k vki Hkkx (a), (b) o (c) ds mÙkjksa ds e/; dksbZ lac/a k ikrs gS\a (a) – 80 V (b) – 40 V (c) – 120 V

Ans. 57.

Ans. 58.

Consider the situation of the previous problem. A charge of –2.0 × 10–4 C is moved from the point A to the point B. Find the change in electrical potential energy VB – VA for the cases (a), (b) and (c). iwoZ iz'u dh fLFkfr ij fopkj dhft,A –2.0 × 10–4 C dk ,d vkos'k fcUnq A ls fcUnq B rd ys tk;k tkrk gSA (a) , (b) o (c) fLFkfr;ksa ds fy, oS|qr fLFkfrt ÅtkZ esa ifjorZu VB – VA Kkr dhft,A 0.016 J, 0.008 J, 0.024 J    An electric field E  ( i 20  j 30 ) N/C exists in the space. If the potential at the origin is taken to be zeror, find the potential at (2m, 2m).    ,d fo|qr {ks=k E  ( i 20  j 30) N/C vkdk'k

esa fo|eku gSA ;fn ewy fcUnq ij foHko 'kwU; fy;k tkrk gS rks (2eh, 2eh)

ij foHko Kkr dhft,A Ans. 59.

–100 V

  An electric field E  i Ax exists in the space, where A = 10 V/m 2. Take the potential at (10 m, 20 m) to be zero. Find the potential at the origin.   ,d fo|qr {ks=k E  i Ax vkdk'k esa fo|eku gS] tgk¡ A = 10 oksYV/eh2. (10 eh, 20 eh) ij foHko 'kwU; eku yhft,A ewy fcUnq

ij foHko Kkr dhft,A

Ans.

500 V

60.

The electric potential existing in space is V(x, y, z) = A(xy + yz + zx). (a) Write the dimensional formula of A. (b) Find the expression for the electric field. (c) If A is 10 SI units, find the magnitude of the electric field at (1 m, 1 m, 1 m). [HCV_CHP-29_Ex_Q.60] vkdk'k esa fo|eku fo|qr foHko V(x, y, z) = A(xy + yz + zx) gSA (a) A dk foeh; lw=k fyf[k,A (b) fo|qr {ks=k dk O;atd fyf[k,A (c) ;fn A 10 SI ek=kd gS rks (1 eh, 1 eh, 1 eh) ij fo|qr {ks=k dk ifjek.k Kkr dhft,A

Ans.

(a) MT–3 I–1

61.

Two charged particles, having equal charges of 2.0 × 10–5 C each, are brought from infinity to within a separation of 10 cm. Find the increase in the electric potential energy during the process. 2.0 × 10–5 C leku vkos'k ds nks vkosf'kr d.k vuUr ls 10 lseh nwjh ds Hkhrj yk, tkrs gSAa bl izfØ;k esa oS|r q fLFkfrt

(b) – A {ˆi ( y  z)  ˆj ( z  x )  kˆ ( x  y )}

(c) 35 N/C

ÅtkZ esa o`f) Kkr dhft,A


Page # 23

Chapter # 29


Ans.

36 J

62.

Some equipotential surfaces are shown in figure. What can you say about the magnitude and the direction of the electric field ?

fp=k esa dqN lefoHko i`"B n'kkZ;s x, gSAa fo|qr {ks=k dh fn'kk ,oa ifjek.k ds ckjs esa vki D;k dg ldrs gS\a

Ans.

(a) 200 V/m making an angle 120° with the X-axis (b) radially outward, decreasing with distance as E

63.

6V m r2

Consider a circular ring of radius r, uniformly charged with linear charge density . Find the electric potential at a point on the axis at a distance x from the centre of the ring. Using this experssion for the potential, find the electric field at this point. r f=kT;k dh ,d o`Ùkkdkj oy; jsf[kd vkos'k ?kuRo ls leku :i ls vkosf'kr gSA v{k ij dqM a yh ds dsUnz ls x nwjh ij

fLFkr fcUnq ij fo|qr foHko Kkr dhft,A foHko ds bl O;atd dk iz;ksx djrs gq, bl fcUnq ij fo|qr {ks=k Kkr dhft,A

Ans.

64.

r 2 0 (r  x ) 2

2 1/ 2

,

r x 2 0 (r  x 2 )3 / 2 2

An electric field of magnitude 1000 N/C is produced between two parallel plates having a separation of 2.0 cm as shown in figure. fp=kkuqlkj] nks lekUrj IysVksa ds e/;] tks ,d nwljs ls 2 lseh nwj gS] 1000 N/C ifjek.k dk ,d fo|qr {ks=k yxk;k tkrk

gSA

(a) (b) (c)

(a) (b) (c)

What is the potential difference between the plates ? With what minimum speed should an electron be projected from the lower plate in the direction of the field so that it may reach the upper plate ? Suppose the electron is projected from the lower plate with the speed calculated in part. (b) and the direction of projection makes an angle of 60º with the field. Find the maximum height reached by the electron. (Neglect gravity) [HCV_II _Chap.-29_64] [4]

IysVksa ds e/; foHkokUrj fdruk gS\ uhps okyh IysV ls ,d bysDVªkWu dks {ks=k dh fn'kk esa fdl U;wure osx ls iz{ksfir fd;k tk;s rkfd ;g Åijh IysV rd igq¡p tk,A eku yhft;s fupyh IysV ls bysDVªkuW dks (b) Hkkx esa laxf.kr osx ls iz{ksfir fd;k tkrk gSA iz{ksi.k dh fn'kk esa {ks=k ls 60º dk dks.k cukrh gSA bysDVªkWu ds }kjk izkIr vf/kdre Å¡pkbZ Kkr dhft,A (Neglect gravity)

Sol.

E = 1000 N/C 2 = 20 volt 100 (b) by work energy theorem manishkumarphysics.in

(a) V = E.d. = 1000 ×

Page # 24

Chapter # 29

Electric Field & Potential KE = work done

–

1 mv2 = – eEd 2



v=

2eEd = m

2  1.6  10 19  1000  2  10 2 9.1 10 31

= 2.65 × 106 m/s

u2 sin2  (c) H = 2geff geff =

eE m

( 2.65  10 6 ) 2  sin 2 60  9.1 10 31 u 2 sin 2 m = = 0.5 × 10–2 m 2eE 2  1.6  10 19  10 3 H = 0.5 cm Ans. (a) 20 V (b) 2.65 × 106 m/s (c) 0.50 cm

H=

65.

A uniform field of 2.0 N/C exists in space in x-direction. (a) Taking the potential at the origin to be zero, write an expression for the potential at a general point (x, y, z). (b) At which points, the potential is 25 V ? (c) If the potential at the origin is taken to be 100 V, what will be the expression for the potential at a general point ? (d) What will be the potential at the origin if the potential at infinity is taken to be zero ? Is it practical to choose the potential at infinity to be zero ? 2.0 N/C dk ,d leku {ks=k vkdk'k esa x-fn'kk esa fo|eku gSA (a) ewy fcUnq ij foHko 'kwU; ekurs gq, ,d O;kid fcUnq (x, y, z) ij foHko dk O;atd fyf[k,A (b) fdu fcUnqvksa ij foHko 25 V gS ? (c) ;fn ewy fcUnq ij foHko 100 V fy;k tk, rks ,d O;kid fcUnq ij foHko dk O;atd D;k gksxk\ (d) ;fn vuUr ij foHko 'kwU; eku fy[k tk, rks ewy fcUnq foHko fdruk gksxk\ D;k vuUr ij foHko 'kwU; ysuk O;ogkfjd

Ans.

(a) –(2.0 V/m) x (c) 100 V – (2.0 V/m) x

66.

How much work has to be done in assembling three charged particles at the vertices of an equilateral triangle as shown in figure ?

gS\

(b) points on the plane x = –12.5 m (d) infinity

fp=k esa n'kkZ, vuqlkj rhu vkosf'kr d.kksa dks ,d leckgq f=kHkqt ds 'kh"kks± ij LFkkfir djus ds fy, fdruk dk;Z djuk iM+xs k\

Ans.

234 J

67.

The kinetic energy of a charged particle decreases by 10 J as it moves from a point at potential 100 V to a point at potential 200 V. Find the charge on the particle. ,d vkosf'kr d.k 100 oksYV ds ,d fcUnqls 200 oksYV foHko ds ,d fcUnq rd xfr djrk gS rks mldh xfrt ÅtkZ 10 twy

Ans.

0.1 C

68.

Two identical particles, each having a charge of 2.0 × 10–4 C and mass of 10 g, are kept at a separation of 10 cm and then released. What would be the speeds of the particles when the separation becomes large ? 10 xzke nzO;eku o 2.0 × 10–4 C vkos'k ds nks leku d.k ,d nwljs ls 10 cm nwj j[ks tkrs gSa vkSj fQj NksM+ fn;s tkrs

Ans.

600 m/s

?kV tkrh gSA d.k ij vkos'k dk eku Kkr dhft,A

gSAa tc nksuksa ds e/; nwjh vf/kd gks tkrh gS rks d.kksa dk osx D;k gksxk\


Page # 25

Chapter # 29 69.

Ans. 70.

Ans. 71.

Ans. 72.


Two particles have equal masses of 5.0 g each and opposite charges of +4.0 × 10 –5 C and –4.0 × 10–5 C. They are released from rest with a separation of 1.0 m between them. Find the speeds of the particles when the separation is reduced to 50 cm. nks d.kksa dk leku nzO;eku 5.0 xzke gS rFkk vkos'k foifjr +4.0 × 10–5 C o –4.0 × 10–5 C gSA tc muds e/; nwjh 1.0 eh- gS rks mUgsa fojkekoLFkk ls NksM+k tkrk gSA d.kksa dk osx Kkr dhft,A tcfd muds e/; dh nwjh 50 lseh jgh xbZ gksA 54 m/s for each particle A sample of HCl gas is placed in an electric field of 2.5 × 104 N/C. The dipole moment of each HCl molecule is 3.4 × 10–30 C–m. Find the maximum torque that can act on a molecule. HCl xSl dk ,d izfrn'kZ 2.5 × 104 N/C ds ,d fo|qr {ks=k esa j[kk tkrk gSA izR;sd HCl v.kq dk f}/kzo q vk?kw.kZ 3.4 × 10– 30 C–m gSA v.kq ij yx ldus okyk vf/kdre cy vk?kw.kZ Kkr dhft,A 8.5 × 10–26 N–m Two particles A and B, having opposite charges 2.0 × 10–6 C and –2.0 × 10–6 C, are placed at a separation of 1.0 cm. (a) Write down the electric dipole moment of this pair. (b) Calculate the electric field at a point on the axis of the dipole 1.0 cm away from the centre . (c) Calculate the electric field at a point on the perpendicular bisector of the dipole and 1.0 m away from the centre. foifjr vkos'k 2.0 × 10–6 C ,oa –2.0 × 10–6 C, ds nks d.k A o B ,d nwljs ls 1.0 lseh nwjh ij j[ks x, gSAa (a) bl ;qXe dk oS|rq f}/kzoq vk?kw.kZ fyf[k,A (b) f}/kzoq ds v{k ds dsUnz ls 1.0 lseh nwj fLFkr fcUnq ij fo|qr {ks=k dh x.kuk dhft,A (c) f}/kzo q ds yEcor~ f}Hkktd ij dsUnz ls 1.0 m nwj fLFkr fcUnq ij fo|qr {ks=k dh x.kuk dhft,A (a) 2.0 × 10–8 C–m (b) 360 N/C (c) 180 N/C Three charges are arranged on the vertices of an equilateral triangle as shown in figure. Find the dipole moment of the combination.

rhu vkos'k ,d leckgq f=kHkqt ds 'kh"kksZ ij fp=kkuqlkj O;ofLFkr fd;s x, gSAa la;kstu dk f}/kzoq vk?kw.kZ Kkr dhft,A

Ans.

qd 3 , along the bisector of the angle at 2q, away from the triangle

73.

Find the magnitude of the electric field at the point P in the configuration shown in figure for d >> a. Take 2qa = p. fp=k esa n'kkZ;s foU;kl esa d >> a ds fy, fcUnq P ij fo|qr {ks=k dk ifjek.k Kkr dhft,A eku yhft, fd 2qa = p.

Ans.

(a)

74.

Two particles, carrying charges – q and +q and having equal masses m each, are fixed at the ends of a light rod of length a to form a dipole. The rod is clamped at an end and is placed in a uniform electric field E with the axis of the dipole along the electric field. The rod is slightly tilted and then released. Neglecting gravity find the time period of small oscillations.

q 4 0 d2

p

(b)

4 0 d3

1

(c)

4 0 d3

q2 d2  p 2


Page # 26

Chapter # 29


– q o +q vkos'k

ds nks d.k] ftuesa izR;sd dk nzO;eku m gS] a yEckbZ dh ,d gYdh NM+ ls tqMs+ gSa ,oa ,d f}/kzoq dk fuekZ.k djrs gSAa NM+ dks ,d fljs ls dl fn;k tkrk gS rFkk bls ,d leku fo|qr {ks=k esa f}/kzoq ds v{k dks fo|qr {ks=k ds vuqfn'k j[kk tkrk gSA NM+ dks FkksM+k lk >qdkdj NksM+ fn;k tkrk gSA xq:Ro dks ux.; ekurs gq, vYi vk;ke ds nksyuksa dk vkorZ dky Kkr dhft,A ma qE

Ans.

2

75.

Assume that each atom in a copper wire contributes one free electron. Estimate the number of free electrons in a copper wire having a mass of 6.4 g (take the atomic weight of copper to be 64 g/mol). eku yhft, fd ,d rk¡cas ds rkj esa izR;sd ijek.kq ,d eqDr bysDVªkuW nsrk gSA 6.4 xzke nzO;eku ds ,d rk¡cs ds rkj esa eqDr bysDVªkWuksa dh la[;k dk vkdyu dhft,A (rkacs dk ijek.kq Hkkj 64 xzke/eksy) ys yhft,A 6 × 1022

Ans.


Page # 27

Electric Field and Potential

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