(a) Using Gauss law, derive expression for electric field due to a spherical shell of uniform charge distribution $\mathrm{\sigma }$ and radius R at a point lying at a distance x from the centre of shell, such that (i) 0 < x < R, and (ii) x > R.
(b) An electric field is uniform and acts along + x direction in the region of positive x. It is also uniform with the same magnitude but acts in – x direction in the region of negative x. The value of the field is E = 200 N/C for x > 0 and E =–200 N/C for x < 0. A right circular cylinder of length 20 cm and radius 5 cm has its center at the origin and its axis along the x-axis so that one flat face is at x = + 10 cm and the other is at x = –10 cm.

OR
Find:
(i) The net outward flux through the cylinder.
(ii) The net charge present inside the cylinder.

(a) Find the expression for the potential energy of a system of two point charges q1 and q2 located at $\to$ 1 r and $\to$ 2 r , respectively in an external electric field $\to$ E .
(b) Draw equipotential surfaces due to an isolated point charge (– q) and depict the electric field lines.
(c) Three point charges + 1 $\mathrm{\mu }$C, – 1 $\mathrm{\mu }$C and + 2 C are initially infinite distance apart. Calculate the work done in assembling these charges at the vertices of an equilateral triangle of side 10 cm.

(a) Derive an Expression for the Electric Field Due to a Spherical Shell

Using Gauss's Law:

Gauss's law states: ∮E·dA = qin/ε0

Case (i): 0 < x < R (Inside the Shell)
For a spherical shell of uniform charge distribution, any charge resides on the surface. Inside the shell, the enclosed charge qin is zero. Hence: ∮E·dA = 0 
Therefore: E = 0

Case (ii): x > R (Outside the Shell)
The total charge enclosed by a Gaussian surface at a distance x is: q = σ · 4πR²
Substituting into Gauss's law: ∮E·dA = q/ε0
Since dA = 4πx², we have: E · 4πx² = σ · 4πR² / ε0
Simplifying: E = σR² / (ε0x²)

(b) Electric Field Flux and Charge Inside a Cylinder

Given Data: Electric field magnitude: E = 200 N/C, cylinder length: L = 20 cm, radius: r = 5 cm

(i) Derive an Expression for the Net Outward Flux:
The net outward flux through the cylinder is given by: Φ = E · A + E · A = 2E · A
Where A = πr² is the area of one flat face of the cylinder.
Substituting values: Φ = 2 · 200 · π · (0.05)²
Φ = 3.14 N·m²/C

(ii) Derive an Expression for the Net Charge:
From Gauss's law: q = ε0 · Φ
Substituting ε0 = 8.854 × 10⁻¹² C²/N·m² and Φ = 3.14: q = 8.854 × 10⁻¹² × 3.14
q ≈ 2.78 × 10⁻¹¹ C

(c) Work Done in Assembling Charges

To assemble three charges at the vertices of an equilateral triangle of side r = 0.1 m, the work done is given by:

W = k[q₁q₂/r₁₂ + q₁q₃/r₁₃ + q₂q₃/r₂₃]
Substituting k = 9 × 10⁹ N·m²/C², q₁ = 1 μC = 10⁻⁶ C, q₂ = -1 μC = -10⁻⁶ C, and q₃ = 2 μC = 2 × 10⁻⁶ C:

W = (9 × 10⁹) · [(10⁻⁶ × -10⁻⁶ / 0.1) + (10⁻⁶ × 2 × 10⁻⁶ / 0.1) + (-10⁻⁶ × 2 × 10⁻⁶ / 0.1)]
W = 9 × 10⁻² [(-1) + 2 - 2]
W = -9 × 10⁻² J

(d) Equipotential Surfaces and Electric Field Lines

Equipotential Surfaces for an Isolated Point Charge (-q):

• The surfaces are concentric spheres centered around the charge.
• Electric field lines are radial and point inward (towards the charge).