1. By Squadron Leader Zahid Mir CS&IT Department , Superior University
PHY-AP Applications of Gauss’ Law By
Squadron Leader Zahid Mir
CS&IT Department , Superior University

2. Infinite Line of Charge
A long thin wire of negligible thickness and carrying charge is called infinite line of charge.

3. Infinite Line of Charge
Consider a infinite line charge of uniform linear charge density λ
λ = dq/ds
We choose a cylinder Gaussian surface ( circular cylinder ) of radius ‘r’ and length ‘h’, closed at each end by plane caps normal to axis.
The electric field is radial i.e along the radius of the cylinder.

4. Infinite Line of Charge
The total flux diverging out through the imaginary cylinder is given by;
ϕE = ϕE1 + ϕE2
ϕE =ʃ E . dA + ʃ E . dA +ʃ E . dA
ʃ E . dA = ʃ E . dA = since θ1 = θ3 =900
ϕE =ʃ E . dA = ʃ EdACosθ but θ2 = 00 S1 S2 S3 S1 S3 S2 S2

5. Infinite Line of Charge
ϕE =E ʃdA
But ʃdA = (2πr)h =Area of surface S2
ϕE =E (2πr)h
Now the amount of charge enclosed
by the imaginary cylinder = λ h
So according to Gauss’ Law ϕE = (charge enclosed) / εo
E (2πr)h = λ h / εo
E = λ / 2πrεo S2 S2

6. Infinite Sheet of Charge
A long thin sheet of negligible thickness and carrying charge is called infinite sheet of charge.

7. Infinite Sheet of Charge
Consider a infinite sheet of charge having uniform surface charge density Ϭ
Ϭ = dq/da
We choose an imaginary cylinder at right angle to the sheet of charge.
From symmetry, we conclude that E points at right angles to the end caps and away from the plane.
Since E does not pierce the cylindrical surface, there is no contribution to the flux from the curved wall of the cylinder.
We also assume that the end caps are equidistant from the sheet.

8. Infinite Sheet of Charge
Flux through each cap is;
ϕE =ʃ E . dA = ʃ EdACosθ here θ = 00
ϕE = E A
So total flux through two caps is
ϕE = 2E A
Now the amount of charge enclosed
by the imaginary cylinder = Ϭ A
Now according to Gauss’ Law; ϕE = (charge enclosed) / εo
2EA = Ϭ A / εo
E = Ϭ /2εo S S