Electrostatics Class 12- Part 3

ELECTROSTATICS - III
- Electrostatic Potential and Gauss’s
Theorem1. Line Integral of Electric Field
2. Electric Potential and Potential Difference
3. Electric Potential due to a Single Point Charge
4. Electric Potential due to a Group of Charges
5. Electric Potential due to an Electric Dipole
6. Equipotential Surfaces and their Properties
7. Electrostatic Potential Energy
8. Area Vector, Solid Angle, Electric Flux
9. Gauss’s Theorem and its Proof
10.Coulomb’s Law from Gauss’s Theorem
11.Applications of Gauss’s Theorem:
Electric Field Intensity due to Line Charge, Plane
Sheet of Charge and Spherical Shell
Created by C. Mani, Principal, K V No.1, AFS, Jalahalli West, Bangalore

Line Integral of Electric Field (Work Done by Electric Field):
Negative Line Integral of Electric Field represents the work done by the electric
field on a unit positive charge in moving it from one point to another in the
electric field.
+q0
O
Z
Y
X
+ q
F
A
rA
B
rB
r
WAB = dW = - E . dl
A
B
Let q0 be the test charge in place of the unit
positive charge.
The force F = +q0E acts on the test charge
due to the source charge +q.
It is radially outward and tends to accelerate
the test charge. To prevent this
acceleration, equal and opposite force –q0E
has to be applied on the test charge.
Total work done by the electric field on the test charge in moving it from A to B
in the electric field is
=
11qq0
4πε0
][ -
rB rA
WAB = dW = - E . dl
A
B

1. The equation shows that the work done in moving a test charge q0 from
point A to another point B along any path AB in an electric field due to +q
charge depends only on the positions of these points and is independent
of the actual path followed between A and B.
2. That is, the line integral of electric field is path independent.
3. Therefore, electric field is ‘conservative field’.
4. Line integral of electric field over a closed path is zero. This is another
condition satisfied by conservative field.
Note:
Line integral of only static electric field is independent of the path followed.
However, line integral of the field due to a moving charge is not independent
of the path because the field varies with time.
E . dl = 0
A
B
=
11qq0
4πε0
][ -
rB rA
WAB = dW = - E . dl
A
B

Electric potential is a physical quantity which determines the flow of charges
from one body to another.
It is a physical quantity that determines the degree of electrification of a body.
Electric Potential at a point in the electric field is defined as the work done in
moving (without any acceleration) a unit positive charge from infinity to that
point against the electrostatic force irrespective of the path followed.
Electric Potential:
=
11qq0
4πε0
][ -
rB rA
WAB = - E . dl
A
B
=
11q
4πε0
][ -
rB rA
WAB
q0
According to definition, rA = ∞ and rB = r
(where r is the distance from the source charge
and the point of consideration)
=
q
4πε0 r
W∞B
q0
= V V =
W∞B
q0
SI unit of electric potential is volt (V) or J C-1
or Nm C-1
.
Electric potential at a point is one volt if one joule of work is done in
moving one coulomb charge from infinity to that point in the electric field.
or

Electric Potential Difference between any two points in the electric field is
defined as the work done in moving (without any acceleration) a unit positive
charge from one point to the other against the electrostatic force irrespective of
the path followed.
Electric Potential Difference:
=
11qq0
4πε0
][ -
rB rA
WAB = - E . dl
A
B
=
11q
4πε0
][ -
rB rA
WAB
q0
1. Electric potential and potential difference are scalar quantities.
2. Electric potential at infinity is zero.
3. Electric potential near an isolated positive charge (q > 0) is positive and that
near an isolated negative charge (q < 0) is negative.
4. cgs unit of electric potential is stat volt. 1 stat volt = 1 erg / stat coulomb
or
=
11q
4πε0
-
rB rA
WAB
q0
q
4πε0
= VB - VA
WAB
q0
VB - VA = ∆V =

Electric Potential due to a Single Point Charge:
r
+ q B
x
Q ∞
dx q0EE +q0
Let +q0 be the test charge
placed at P at a distance x
from the source charge +q.
To prevent this acceleration, equal and opposite force –q0E has to be applied
on the test charge.
The force F = +q0E is radially
outward and tends to
accelerate the test charge.
Work done to move q0 from P to Q through ‘dx’ against q0E is
dW = F . dx = q0E . dx dW = q0E dx cos 180° = - q0E dx
dW = - dx
q q0
4πε0 x2
E =
q
4πε0 x2
or
Total work done to move q0 from A to B (from ∞ to r ) is
W∞B =
∞
B
dW = - dx
q q0
4πε0 x2
∞
r
= - dx
q q0
4πε0 x2
∞
r
x2
1
=
q
4πε0 r
W∞B
q0
V =
q
4πε0 r
P

Electric Potential due to a Group of Point Charges:
+1 C
q2
qn
q4
P
q1
r1
r2
r3r4
rnVP = V1 + V2 + V3 + V4 + …………+ Vn
│ │
1
4πε0
V = ∑
i=1
n qi
r - ri
( in terms of
position vector )
The net electrostatic potential at a point in the
electric field due to a group of charges is the
algebraic sum of their individual potentials at that
point.
1. Electric potential at a point due to a charge is not affected by the presence
of other charges.
2. Potential, V α 1 / r whereas Coulomb’s force F α 1 / r2
.
3. Potential is a scalar whereas Force is a vector.
4. Although V is called the potential at a point, it is actually equal to the
potential difference between the points r and ∞.
1
4πε0
V = ∑
i=1
n qi
ri
q3

Electric Potential due to an Electric Dipole:
ll
x
P+ q- q
p
A B +1 C
q
VP
q+
=
4πε0 (x – l)
1
VP = VP
q+
+ VP
q-
VP
q-
=
4πε0 (x + l)
1 - q
VP =
4πε0
q
(x – l)
1
[ -
(x + l)
1
]
VP =
1
4πε0
q . 2l
(x2
– l2
)
VP =
1
4πε0
p
(x2
– l2
)
i) At a point on the axial line:
O

+ q- q
p
A Bθθ
y
O
ii) At a point on the equatorial line:
q
VQ
q+
=
4πε0 BQ
1
VQ = VP
q+
+ VP
q-
VQ
q-
=
4πε0 AQ
1 - q
VQ =
4πε0
q
BQ
1
[ -
AQ
1
]
VQ = 0 BQ = AQ
The net electrostatic potential at a point in the electric field due to an electric
dipole at any point on the equatorial line is zero.
Q
ll

Equipotential Surfaces:
A surface at every point of which the potential due to charge distribution is
the same is called equipotential surface.
i) For a uniform electric field:
E
V1 V2
V3
ii) For an isolated charge:
E
Plane Equipotential Surfaces
Spherical Equipotential Surfaces
+

Properties of Equipotential Surfaces:
2. The electric field is always perpendicular to the element dl of the
equipotential surface.
1. No work is done in moving a test charge from one point to another on an
equipotential surface.
WAB
q0
VB - VA = ∆V =
If A and B are two points on the equipotential surface, then VB = VA .
WAB
q0
= 0 or WAB
= 0
WAB = - E . dl
A
B
= 0
Since no work is done on equipotential surface,
i.e. E dl cos θ = 0
As E ≠ 0 and dl ≠ 0, cos θ = 0 or θ = 90°

4. Two equipotential surfaces can not intersect.
If two equipotential surfaces intersect, then at the points of intersection,
there will be two values of the electric potential which is not possible.
(Refer to properties of electric lines of force)
3. Equipotential surfaces indicate regions of strong or weak electric fields.
dV
dr
E = -
Electric field is defined as the negative potential gradient.
or
dV
E
dr = -
Since dV is constant on equipotential surface, so
E
1
dr α
If E is strong (large), dr will be small, i.e. the separation of equipotential
surfaces will be smaller (i.e. equipotential surfaces are crowded) and vice
versa.
Note:
Electric potential is a scalar quantity whereas potential gradient is a vector
quantity.
The negative sign of potential gradient shows that the rate of change of
potential with distance is always against the electric field intensity.

Electrostatic Potential Energy:
The work done in moving a charge q from infinity to a point in the field
against the electric force is called electrostatic potential energy.
W = q V
i) Electrostatic Potential Energy
of a Two Charges System:
O
Z
Y
X
A (q1)
r1
B (q2)
r2
- r1
r2
U =
q1q2
4πε
0
│ │- r1
r2
1
or
U =
q1q2
4πε
0
r12
1

ii) Electrostatic Potential Energy
of a Three Charges System:
O
Z
Y
X
A (q1)
r1
B (q2)
r2
or
C (q3)
r3
- r1r3
- r2r3
U =
q1q2
4πε
0
│ │- r1
r2
1
+
q1q3
4πε
0
│ │- r1
r3
1
+
q2q3
4πε
0
│ │- r2
r3
1
U =
q1q2
4πε
0
r12
1
[
q1q3
r31
q2q3
r32
+ + ]
iii) Electrostatic Potential Energy of an n - Charges System:
1
4πε0
U = ∑
j=1
i≠j
n qi qj
rj - ri│ │
∑
i=1
n
[2
1
]
- r1r2

Area Vector:
Small area of a surface can be represented by a vector.
dS
S
dS
n
Electric Flux:
Electric flux linked with any surface is defined as the total number of electric
lines of force that normally pass through that surface.
dS
E
dS
θ
90°
θ
dS
θ
dS
S
Electric flux dΦ through a small area
element dS due to an electric field E at an
angle θ with dS is
= E dS cos θE . dSdΦ =
Total electric flux Φ over the whole
surface S due to an electric field E is
Φ = E . dS
S
= E S cos θ = E . S
dS = dS n
Electric flux is a scalar quantity. But it is a
property of vector field.
SI unit of electric flux is N m2
C-1
or J m C -1
.
dS

Solid Angle:
Solid angle is the three-dimensional equivalent of an ordinary two-
dimensional plane angle.
SI unit of solid angle is steradian.
Solid angle subtended by area element dS at the
centre O of a sphere of radius r is
dΩ =
dS cos θ
r2
= 4π steradianΩ = dΩ =
dS cos θ
r2
S S
1. For 0° < θ < 90°, Φ is positive.
2. For θ = 90°, Φ is zero.
3. For 90° < θ < 180°, Φ is negative.
Special Cases:
r
θ
dS
n
r
dΩ

Gauss’s Theorem:
The surface integral of the electric field intensity over any closed hypothetical
surface (called Gaussian surface) in free space is equal to 1 / ε0 times the net
charge enclosed within the surface.
E . dS =
S
ΦE =
1
ε0
∑
i=1
n
qi
Proof of Gauss’s Theorem for Spherically Symmetric Surfaces:
E . dSdΦ =
r2
1
4πε0
=
q
r . dS n
dΦ =
r2
1
4πε0
q dS
r n.
Here, = 1 x 1 cos 0° = 1r n.
dΦ =
r2
1
4πε0
q dS
S
ΦE = dΦ
r2
1
4πε0
q
= 4π r2
ε0
q
=dS
S
r2
1
4πε0
q
=
O •
r
r
dS
E
+q

Proof of Gauss’s Theorem for a Closed Surface of any Shape:
EE . dSdΦ =
r2
1
4πε0
=
q
r . dS n
dΦ =
r2
1
4πε0
q dS
r n.
Here, = 1 x 1 cos θ
= cos θ
r n.
dΦ =
r2
q
4πε0
dS cos θ
S
ΦE = dΦ
q
4πε0
= 4π
ε0
q
=dΩ
q
4πε0
S
=
dΩ
r
θ
dS
n
r
+q •

O •
r
r
dS
E
+q
Deduction of Coulomb’s Law from Gauss’s Theorem:
From Gauss’s law,
E . dS =
S
ΦE =
q
ε0
E dS =
S
ΦE =
q
ε0
or dS =
S
ΦE =
q
ε0
E
E =
q
4πε0 r2
E x 4π r2
q
ε0
=
If a charge q0 is placed at a point where E
is calculated, then
Since E and dS are in the same direction,
which is Coulomb’s Law.
or
F =
qq0
4πε0 r2

Applications of Gauss’s Theorem:
1. Electric Field Intensity due to an Infinitely Long Straight Charged
Wire:
Gaussian surface is a
closed surface,
around a charge
distribution, such that
the electric field
intensity has a single
fixed value at every
point on the surface.
From Gauss’s law,
E . dS =
S
ΦE =
q
ε0
E . dS =
S
E . dS +
A
E . dS +
B
E . dS
C
E . dS =
S
E dS cos 90° +
A B
E dS cos 90° +
C
E dS cos 0° =
C
E dS = E x 2 π r l
- ∞ + ∞B A
C
E
E E
dSdS
dS
l
r

q
ε0
=
λ l
ε0
(where λ is the liner charge density)
E x 2 π r l =
λ l
ε0
or E =
2 πε0
1 λ
r
or E =
4 πε0
1 2λ
r
In vector form, E (r) =
4 πε0
1 2λ
r
r
The direction of the electric field intensity is radially outward from the positive
line charge. For negative line charge, it will be radially inward.
Note:
The electric field intensity is independent of the size of the Gaussian surface
constructed. It depends only on the distance of point of consideration. i.e. the
Gaussian surface should contain the point of consideration.

E
dS
C
B
A
E E
dS
dS r
l
2. Electric Field Intensity due to an Infinitely Long, Thin Plane Sheet of
Charge:
From Gauss’s law,
σ
E . dS =
S
ΦE =
q
ε0
E . dS =
S
E . dS +
A
E . dS +
B
E . dS
C
E . dS =
S
E dS cos 0° +
A B
E dS cos 0° +
C
E dS cos 90° = 2E dS = 2E x π r2
TIP:
The field lines remain
straight, parallel and
uniformly spaced.

(where σ is the surface charge density)
or E =
2 ε0
σ
In vector form, E (l) =
2 ε0
σ
l
The direction of the electric field intensity is normal to the plane and away
from the positive charge distribution. For negative charge distribution, it will
be towards the plane.
Note:
The electric field intensity is independent of the size of the Gaussian surface
constructed. It neither depends on the distance of point of consideration nor
the radius of the cylindrical surface.
q
ε0
=
σ π r2
ε0
2 E x π r2
=
σ π r2
ε0
If the plane sheet is thick, then the charge distribution will be available on
both the sides. So, the charge enclosed within the Gaussian surface will be
twice as before. Therefore, the field will be twice.
E =
ε0
σ

3. Electric Field Intensity due to Two Parallel, Infinitely Long, Thin
Plane Sheet of Charge:
σ1
σ2
E1
E1E1
E2
E2E2
E E E
Region I Region II Region III
E = E1 + E2
E =
2 ε0
σ1 + σ2
E = E1 - E2
E =
2 ε0
σ1 - σ2
E = E1 + E2
E =
2 ε0
σ1 + σ2
σ1 > σ2
( )
Case 1: σ1 > σ2

Case 2:
σ1
σ2
E1
E1E1
E2
E2E2
E E E
E = E1 - E2
E =
2 ε0
σ1 - σ2
E = E1 + E2
E =
2 ε0
σ1 + σ2
E = E1 - E2
E =
2 ε0
σ1 - σ2
σ1 > σ2
( )
σ1 > σ2
( )
+ σ1 & - σ2

Case 3:
σ1
σ2
E1
E1E1
E2
E2E2
E = 0 E ≠ 0
E = 0
E = E1 - E2
E =
2 ε0
σ1 - σ2
= 0
E = E1 - E2
E =
2 ε0
σ1 - σ2
= 0
E = E1 + E2
E =
2 ε0
σ1 + σ2
=
ε0
σ
+ σ & - σ

4. Electric Field Intensity due to a Uniformed Charged This Spherical
Shell:
dS
E
q
r
R
•P
From Gauss’s law,
E . dS =
S
ΦE =
q
ε0
E dS =
S
ΦE =
q
ε0
or dS =
S
ΦE =
q
ε0
E
E x 4π r2
q
ε0
=
or E =
q
4πε0 r2
i) At a point P outside the shell:
Since q = σ x 4π R2
,
E =
ε0 r2
σ R2
Electric field due to a uniformly
charged thin spherical shell at
a point outside the shell is such
as if the whole charge were
concentrated at the centre of
the shell.
HOLLOW
……… Gaussian Surface
O •

dS
EFrom Gauss’s law,
E . dS =
S
ΦE =
q
ε0
E dS =
S
ΦE =
q
ε0
or dS =
S
ΦE =
q
ε0
E
E x 4π R2
q
ε0
=
or E =
q
4πε0 R2
ii) At a point A on the surface of the shell:
Electric field due to a uniformly
charged thin spherical shell at
a point on the surface of the
shell is maximum.
Since q = σ x 4π R2
,
E =
ε0
σ
q R
HOLLOW
O •
•A

dS
E
r’
Oq R
HOLLOW
•
B
•
From Gauss’s law,
E . dS =
S
ΦE =
q
ε0
E dS =
S
ΦE =
q
ε0
or dS =
S
ΦE =
q
ε0
E
E x 4π r’2
q
ε0
=
or E =
0
4πε0 r’2
iii) At a point B inside the shell:
This property E = 0 inside a cavity is
used for electrostatic shielding.
(since q = 0 inside the Gaussian surface)
E = 0
r
E
R
O
Emax
END

Electrostatics Class 12- Part 3

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Electrostatics Class 12- Part 3