GAUSS’S THEOREM

Introduction to Gauss Theorem

This theorem gives the relation between the electric flux

passing through closed surface and the net charge enclosed
by that surface.

This surface is called Gaussian surface. It is a closed

hypothetical surface.

Its validity is shown by experiments.

It is used to determine the electric field due to some

symmetric charge distributions.
Gauss's law is Stated as Given Below

The surface integral of the electric field intensity over any closed
hypothetical surface (called Gaussian surface) in free space is
equal to times the total charge enclosed within the surface.
Here, ε is the permittivity of free space.

According to Gauss's law,

where qinside is the total charged enclosed inside the Gaussian

surface

The surface integral of the electric field intensity over any closed
hypothetical surface is,
Electric Flux & Gauss's Law

GAUSS’S THEOREM

net
0

enclosed
net
0

net
0

Outside charges –q4 and +q5 does not play any role in net
flux passing out through close surface
When net flux is coming When net flux is coming in,
out, then positive charge then negative charge is
is enclosed by Gaussian enclosed by Gaussian
surface surface

+ –

Gaussian surface Gaussian surface

When net flux is zero, then net charge enclosed by Gaussian
surface is also zero.

+
Gaussian Surface

net
When net flux is zero, then net charge enclosed by Gaussian
surface is also zero.
When net flux is zero, then net charge enclosed by Gaussian
surface is also zero.

+Q -Q
Gaussian surface

net
When net flux is zero, then net charge enclosed by Gaussian
surface is also zero.
NOTE

Flux through Gaussian surface is independent of location

of charges within Gaussian surface.

+ +

Gaussian Surface Gaussian Surface

Gaussian Surface
Flux through Gaussian surface is independent of shape
and size of Gaussian surface.

+ –

Gaussian Surface
Gaussian Surface
Flux through Gaussian surface is independent of shape
and size of Gaussian surface.

–
+

Gaussian Surface Gaussian Surface

 In Gauss Theorem

+q1 +q2

Gaussian Surface

Here it is clear that,

 due to all charges
but φclosed  due to enclosed charges
 Important

When an imaginary closed surface placed in external

electric field, net flux passing through the close surface
is always zero

As net flux passing through close surface is zero

So, net charge enclosed by surface must be zero.
When an imaginary closed surface placed in external
electric field, net flux passing through the close surface
is always zero.
When an imaginary closed surface placed in external
electric field, net flux passing through the close surface
is always zero.
When an imaginary closed surface placed in external
electric field, net flux passing through the close surface
is always zero.
When an imaginary closed surface placed in external
electric field, net flux passing through the close surface
is always zero.
When an imaginary closed surface placed in external
electric field, net flux passing through the close surface
is always zero.
When an imaginary closed surface placed in external
electric field, net flux passing through the close surface
is always zero.
Gauss theorem is valid only when force follows inverse
square law.
Example : electrostatic force , gravitational force

Both forces follows inverse square law.

 Applications of Gauss’s theorem

Electric field due to point charge

Electric field due to conducting sphere of charge +q
Electric field due to nonconducting sphere of charge +q

Electric field due to large (infinite) sheet

Electric field due to large (infinite) conducting sheet

Electric field due to long (infinite) wire

 Type of Gaussian Surface

(a) Point charge/sphere Spherical

(b) Line charge Cylindrical

(c) Surface charge Cylindrical

 Electric field due to point charge

According to Gauss’s theorem Gaussian surface

in

Here angle between and is 00

 Electric field due to point charge

Here angle between and is 00

2
Variation of E with r

2
 Electric field due to charged conducting sphere

(a) Outside the sphere

Gaussian surface

in
According to Gauss’s theorem

Angle between and is 00

 Electric field due to charged conducting sphere

(a) Outside the sphere

Angle between and is 00

2
R

2 2

Thus for point lying outside the sphere, sphere behaves as

point charge centered at centre.
 Electric field due to charged conducting sphere

(b) On the surface of sphere

R
 Electric field due to charged conducting sphere

(b) On the surface of sphere

For the point on surface (r = R)

2
(max. value of electric field)

2
0 0
R
 Electric field due to charged conducting sphere

(c) Inside the sphere

Here it is clear that charge enclosed Gaussian surface
by gaussian surface is zero
According to Gauss’s theorem
in

R
 Electric field due to charged conducting sphere

For conducting sphere

2 2

Variation of E with r

max

out 2
in
Q. Electric field at a distance 20cm from a charged metallic
sphere of radius 10 cm is E. Find electric field at a distance
(a) r = 5cm (b) r = 10cm (c) r = 30cm, from centre

Sol.

5cm
2 2

R=10cm

(a) 5cm (r < R), so E = 0 (for conductor E = 0 at inside points)

Sol.

(b) r = 10cm

(b) 10cm (r = R), 2

r=R

2 R=10cm
Sol.

(c) r = 30cm, from center

(c) 30cm (r = 3R),

R=10cm
Q. A charge is placed at the centre of conducting spherical
shell of radius R, find out electric field at points 1, 2.

1
Sol. Electric field at point 2 due to induce charges will be zero
At point 2 electric field is only due to charge placed at
centre. (i.e. +Q1)

1 1
2
1

1
2
2
Sol. For the points lies outside the shell, all charges supposed
to be centered at centre of sphere.
i.e. for outside points, whole system behave like a point
charge +Q1 placed at centre of sphere.
So electric field at point 1 will be
1
1
1
1

1
Q. Two concentric conducting spherical shells are shown in
figure. Find electric fields at points 1,2,3.

2 2
1

1
2
3
 Sol.
GS1
For GS1 charge enclosed
is +Q1 and +Q2 i.e. B
R2 +Q2
1
1 2 r1 A
1 +Q1

For GS2 charge enclosed 2 R1

is +Q1, i.e.
r2 −Q1 +Q1
r3
GS3
1
2 3
GS2
For GS3 charge enclosed is
zero, i.e.

3
Q. Figure shows thick conducting shell of inner radius R1 and
outer radius R2 then find out
(a)
(b) 2

1
1
Sol. Charges induced on inner and outer surfaces of shell due to
charge at centre.

2 (a)

1
2
1
(a)
1
1
Electric field due to uniformly charged nonconducting
sphere

(a) EF at the point outside the sphere Gaussian surface

According to Gauss’s theorem
qin
ϕ = E. ds =
ε + + +
+ +
Here angle between E and ds + + + +
+
is 00 + ++
+ + + +
E
q + + ds
E. ds = Eds = + + R +
ε + +
+ +
q
E ds =
ε
q q kq kq
E(4πr2) = or E = 2 = E =
ε 4πε (r ) r 2
r2
Thus for point lying outside the sphere, sphere behaves as
point charge centered at centre.
Electric field due to uniformly charged nonconducting
sphere

(b) EF at the point on surface of the sphere

+ + +
+ +
+ + + + +
+ ++
+ + +
+ + +
+ + R +
+ +
+ +
Electric field due to uniformly charged nonconducting
sphere

(b) EF at the point on surface of the sphere

2
(max. value of electric field)

+ + +
2 + +
+ + +
0 3 0
0
+ +
+ ++
+ + +
+ + +
+ +R +
+ +
+ +
 Electric field due to uniformly charged nonconducting
sphere

(c) EF at the point inside the surface of the sphere Gaussian

surface
According to Gauss’s theorem +
+ +
qin + +
ϕ = E. ds = + +
ε + +
+
+ + +
Here angle between E and ds is 00 +
+
+ + +
qin +
E. ds = Eds = + + R
ε +
+
+
qin + +
E ds =
ε
q q kqin
E 4πr = or E = = 2 ………………(i)
ε 4πε (r ) r
 Electric field due to uniformly charged nonconducting
sphere

(c) EF at the point inside the surface of the sphere Gaussian

Charge inside the Gaussian surface surface
+
+ +
q = ρdV + +
+ +
q 4 + +
q = +
4 πr + +
3 + +
πR +
3 +
+ +
qr +
+ + R
q = +
R +
+
+ +
From equation (i)

kqin k qr kq
E= = E = r E ∝r
r2 r2 R R
 Electric field due to uniformly charged nonconducting
sphere

(c) EF at the point inside the surface of the sphere Gaussian

surface
kq +
E = r E ∝r + +
R + +
+ +
In term of charge density + +
+
+ + +
kq +
E = r +
R + + +
+
+ + R
+
1 4 r +
E = ρ πR +
4πϵ 3 R +
+
ρr
E =
3ϵ
 Electric field due to uniformly charged nonconducting
sphere

For nonconducting sphere

2 2 3

Variation of E with r

max

out 2
Q. Electric field at a distance 30cm from a uniformly charged
sphere of radius 15cm is E. Find electric field at a distance
(a) r = 5cm (b) r = 15cm from center

Sol.
kQ kQ kQ
E = = = E given or 2 = 4E
30 4R R
r = 30cm
(a) 5cm (r < R),

kQ kQ r 5 + + +
E = 3 (r) = 2 = 4E = 4/3E + +
R R R 15
+ + + + +
+ + ++
+ +
+ + +
+ + R +
+ +
+ +
Sol.

(b) r = 15cm from center r = 15 cm = R

(b) 15 cm (r = R), + + +
2 + +
+ + + + +
+ + ++
+ +
+ + +
+ +R +
+ +
+ +
Electric field intensity due to uniformly charged
infinite sheet
Electric field intensity due to uniformly charged
infinite sheet
Large sheet of charge density

Q
+ ds E

+ (2) E
+ A
ds
(3) + (1)
E r P
ds + E
+

Gaussian +
cylinder σ
Electric field intensity due to uniformly charged
infinite sheet
According to Gauss’s theorem
qenclosed
E. ds =
∈0
qenclosed
Eds cos0° + Eds cos90° + Eds cos0° =
∈0
qenclosed
2E ds =
∈0
σA
2EA =
∈0
σ
E= n
2∈0

∵ E is independent of distance r, thus E is uniform for all nearly

points
Electric field intensity due to uniformly charged
infinite sheet

Variation of E with r
Electric field intensity due to large (infinite)
conducting sheet

Large conducting sheet of charge density 𝛔

σ
+ + ds
+
+ + + (2)
+ E=0 E
A ds
+ + (1)
+ (3) E
+ Gaussian
+ + cylinder
Electric field intensity due to large (infinite)
conducting sheet

Large conducting sheet of charge density 𝝈

Electric field intensity due to large (infinite)
conducting sheet
According to Gauss’s theorem
qenclosed
E. ds =
∈0
qenclosed
Eds cos0° + Eds cos90° + 0ds cos𝜃 =
∈0
qenclosed
E ds =
∈0

σA
EA =
∈0
σ
E= n
∈0
∵ E is independent of distance r, thus E is uniform for all nearly
points