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Gauss Law and Applications

1. Statement: The total normal electric flux φe over a closed surface is 1ε times
0

the total charge Q enclosed within the surface.

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⎛1⎞
φe = ⎜ ⎟Q
⎝ ε0 ⎠

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2. Gauss Law is applicable for any distribution of charges and any type of closed

n.
surface, but it is easy to solve the problem of high symmetry.

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3. At any point over the spherical Gaussian surface, net electric field is the vector

at
sum of electric fields due to + q1 , − q1 and q 2 .
uc
ed
hi

4. Applications of Gauss Theorem

a) Electric field at a point due to a line charge

A thin straight wire over which ‘q’ amount of charge be uniformly distributed. l
.s

be the linear charge density i.e, charge present per unit length of the wire.
w

q
w

E=
2π ∈0 rl
w

λ
E=
2π ∈0 r

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m
This implies electric field at a point due to a line charge is inversely proportional
to the distance of the point from the line charge.

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b) Electric field intensity at a point due to a thin infinite charged sheet

n.
‘q’ amount of charge be uniformly distributed over the sheet. Charge present per

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unit surface area of the sheet is σ .

at
uc
ed
hi

q
E=
2 A ∈0
ks

q q
E= where σ =
2 ∈0
a

A
.s

E is independent of the distance of the point from the charged sheet.

c) Electric field intensity at a point due to a thick infinite charged sheet

‘q’ amount of charge be uniformly distributed over the sheet. Charge present per
w

unit surface area of the sheet be σ .

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q σ
E= =
A ∈0 ∈0

Electric field at a point due to a thick charged sheet is twice that produced by the
thin charged sheet of same charge density.

d) Electric intensity due to two thin parallel charged sheets

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Two charged sheets A and B having uniform charge densities σ A and σ B
respectively.

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In region I

n.
1
E= (σ A + σ B )

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2 ∈0

at
In region II uc
ed
hi
ks

1
EII = (σ A − σ B )
2 ∈0
a
.s

In region III
w

1
EIII = (σ A + σ B )
2 ∈0
w

e) Electric field due to two oppositely charged parallel thin sheets

1
EI = − [σ + ( −σ )] = 0
2 ∈0

1 σ
EII = [σ − ( −σ )] =
2 ∈0 ∈0

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1
EIII = (σ − σ ) = 0
2 ∈0

f) Electric field due to a charged Spherical shell

‘q’ amount of charge be uniformly distributed over a spherical shell of radius ‘R’

q
σ = Surface charge density, σ =
4π R 2

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i) When point ‘P’, lies outside the shell

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1 q
E= × 2

n.
4π ∈0 r

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This is the same expression as obtained for electric field at a point due to a point

at
charge. Hence a charged spherical shell behaves as a point charge concentrated
at the centre of it.
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1 σ .4π R 2 q σ .R 2
E= ∵σ = E=
ed

4π ∈0 r 2
4π r 2 ∈0 r 2

σ
ii) When point ‘P’, lies on the shell: E =
hi

∈0
ks

iii) When Point ‘P’ lies inside the shell

E=0
.s
w
w
w

The electric intensity at any point due to a charged conducting solid sphere is
same as that of a charged conducting spherical shell.

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g) Electric Potential (V) due to a spherical charged conducting shell (Hollow
sphere)

q
R r

m
r

co
n.
1
i) When point ( P3 ) lies outside the sphere ( r > R ) , the electric potential, V =
q
4πε 0 r

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1
ii) When point ( P2 ) lies on the surface ( r = R ) , V =
q

at
uc 4πε 0 R

1
iii) When point ( P1 ) lies inside the surface ( r < R ) , V =
q
4πε 0 R
ed

Note: The electric potential at any point inside the sphere is same and is equal to
that on the surface.
hi
a ks
.s
w

Note: The electric potential at any point due to a charged conducting sphere is
w

same as that of a charged conducting spherical shell.

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