Chapter – 1…….. ELECTRIC CHARGES AND FIELDS
Coulomb’s law:
The force of interaction between two stationary point charges in vacuum is directly proportional to product of charges and inversely proportional to the square of separation between them .
here F k
1
qq k 1 22 r 9
, here k is is proportionality constant.The value of
2
9 10 Nm / C
40
2
.
Here permittivity of free space , its value is 0
12
0
8.85 8.8510
2
1
2
C N m
.
1 Coulomb is defined as that charge which when placed at a distance of 1m from another charge charge of same magnitude in vacuum , experiences an electric force of 9 10 N. In practice
we generally use mC or micro coulomb If q1
then, F
q2
C
9
.
1C , and r 1m
9 10
9
11 2
1
9
9 10 N
Dielectric Constant or Relative permittivity:
The force between two charges q 1 and q2 located at distance ‘r’ apart in medium may be expressed as : F vacuum
1
q1q2
40
r 2
, Here is absolute permittivity of medium.
0
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F medium 1
Fvacuum
1 4
r 2
F medium
q1q2
4
r 2
, Here is absolute permittivity of medium.
q1q 2 r 2 q1q 2
40
1
0
r
, here is relative permittivity of medium with r
respect to vacuum. It is also denoted by ‘K’ also known an dielectric constant of medium. ( )
K r
0
Fvacuum
Fmedium
K 0
or F medium
F vacuum K
Coulomb’s law in Vector form:
Let two like charges q1 and q2 are present at separation ‘r’.
Therefore , According to Coulomb’s law force between them is F
kq1q2 r 2
Let F21
kq1q2 r
2
r 12
F12
kq1q2 r2
r 21
12
is unit vector pointing from q1 to q2.
kq1q2 r12 r2
Let
r
.
r r
21
is unit vector pointing from q2 to q1 .
kq1q2 r21 . r2 r
But since ,
r
12
r
21
or ,
r
21
r
12
, therefore
F12
F 21
.
Limitations of Coulomb’s Law :
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(i) Electric charges should must be at rest. (ii) The charges should must be point charges .it is not directly applicable for extended charged body.
Coulomb’s force is very much stronger than gravitational 10 F force . F . 30
electric
gravitational
Superposition principle:
The coulomb's law obeys the principle of superposition, which means that the force between two particles is not affected by the presence of other charges. This principle is used to find the net force exerted on a given particle by other charged particles. Ftotal
F12
F13
F14
........
F 1n
q 3
q4
F13
F12
q 2
q1
F14 Fig.(3) Forces acting on q1 due to q2, q3 and q4 are shown
Continuous Charge distribution:
(i) Linear charge distribution: In this distribution charge distributed on a line.Forexample : charge on a wire, charge on a ring etc. Relevant parameter parameter is which is called linear charge density.
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charge length
(ii) Surface charge density : In this distribution charge distributed on the surface. For example : Charge on a conducting sphere, charge on a sheet etc. Relevant parameter is
which is called surface charge
density .
charge Area
(iii) Volumetric Charge density :In this distribution charge distributed in the whole volume of the body. For example : Non conducting charged sphere. Relevant parameter parameter is which is called volume charge density i.e.,
charge Volume
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Electric Field:
Eelctric field at a point is defined as the force experienced experienced by a unit positive charge placed at that point. The charge which produces electric field is called Source charge (Q) and charge (q) which experiences experiences force is called test charge . Electric Field Intensity:
The electric field intensity is defined as force experienced per unit positive test charge placed at that point without disturbing the source charge . Mathematically Mathematically , E
Lim q0 0
F q0
.
Electric field is a vector quantity whose direction is same as the force experienced by a unit positive test charge . The S.I. unit of electric field is N/C or Volt per metre(V/m) . The dimensions of electric field are [ MLT MLT
3
1
A
]
.
+q
-q
Electric Field Due to a point charge : N avi avi n P hysics hysi cs C lasse lasses ,W ,What hatss A pp no. 9211 92119992 999279 795| P a g e
The direction of E is that of the force on a positive charge. In other words, positive charges experience forces parallel to the field, and negative charges experience forces opposite to the field Once the field strength is known, the force on any charge q can be found from F
qE
From Coulomb's law, the electric field created by a point charges q is given by kq where the unit vector has it's origin at the E r r
2
source charge q. Proof : O
P
r
q0 test charge charge
sourcecharge source charge
F
Let us consider a point charge q placed at origin O . Let us place another test charge q placed at point P. According to Coulomb’s Law , the force on charge q is , 1 qq is a unit vector in direction from qtoq . F r , here r 4 0
0
0
r
2
0
0
Electric field at point P is E
F q0
1
qq0
40
r 2
q0
r
1
q
40
r 2
r
Therefore , the magnitude of electric field is here
E
1 2
r
E
1
q
40
r
2
,
.
Electric Field Due to a System of Charges
Since the principle of superposition is valid for Coulomb's law, law, it is also valid for the electric field . To calculate the field strength
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at point due to a system of charges, we first find the individual field intensity E intensity E 1 due to q1, E 2 due to q2 and so on. Charge particles :
For N For N point point charges, the resultant field intensity is the vector sum E
E1
E2
E3 ........ E n
q 2
E1
E2 P
E3 q 3 q1 Fig.(6) Electric field at a point is the superposition of individual contribution contribution of e ach charge.
Electric field at a point is the superposition of individual contribution of each charge.
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Electric Field Intensity at any Point on the Axis of a Uniformly Charged Ring
Consider a circular loop of wire of negligible thickness, radius a andcentreO andcentreO held perpendicular to the plane of the paper. Let the loop carry a total charge q distributed uniformly over its circumference. We have to determine electric field intensity at any point P point P on on the axis of the loop, where OP = = r , figure below. N avi avi n P hysics hysi cs C lasses lasses ,W ,What hatss A pp no. 9211 92119992 999279 7910| P a g e
Consider a small element AB element AB of the loop. Let length of element AB = dl and and C be be the centre of the element. Charge on the element AB element AB is is dq
q 2a
dl
... (1)
Electric field intensity at P due to the charge element AB is 1 dq along PC along PC ´ at with the axis. | dE | 4 CP
2
0
| dE |
dE
dq
1 40 (r
2
a2 )
can be resolved into two rectangular components.
sin along PY dE along PX along PX , the axis of the loop, and dE sin along PY , perpendicular to the axis. cos cos
For a pair of diametrically opposite elements of the loop, components of electric field intensity perpendicular to the axis will cancel, whereas the components along the axis of the loop will add. As the loop can be considered to be made up of a large number of pairs of diametrically opposite elements, therefore, dE sin 0
Hence the resultant resultant electric field intensity intensity at P P is is | E |
dE cos
In DOPC DOPC ,
cos
OP CP
r
(r 2
a
2 1/ 2
)
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| E |
dq
1
4
0
r
2 1/ 2 (r a ) ( r a ) 2
2
2
Using (1), we get | E | 41 qr (2 a) 40 2a(r
2
r q dl 2 2 3/ 2 0 2a ( r a )
| E |
2 3/ 2
a )
=
qr 40 2a(r 2 a2 )3/ 2
qr
40 (r
2
2 3/ 2
a )
dl =
………(3)
The direction of is along PX along PX , the axis of the loop. Special Cases
(i) When P lies at the centre of the loop. r = = 0, therefore from (3), (3),
E
0
(ii) When r >> a (i.e. a (i.e. P P lies lies far off from the loop), neglecting a2 in comparison to r 2 in (3), we get | E |
qr
40 r 3
q
40 r 2
,
along PX
This is expression for | E | at a distance r from from a point charge q. Hence a circular loop of charge behaves as a point charge when the observation point ( P P ) is at very large distance from the loop, compared to the radius of the loop.
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