Q7 · Electrostatics

Problem

Thin uniformly charged spherical shell — choose correct E(r) and V(r) graphs.

Key insight: Inside a conducting shell, E = 0 (electrostatic shielding). V is constant but non-zero inside.

Setup

flowchart TB
  subgraph inside["r < R"]
    Ei["E = 0"]
    Vi["V = kQ/R (constant)"]
  end
  subgraph outside["r > R"]
    Eo["E ∝ 1/r²"]
    Vo["V ∝ 1/r"]
  end

1. Inside (r < R)

E = 0, V = constant = kQ/R.

$$E = 0,\; V = \text{const}$$

2. Outside (r > R)

Same as point charge.

$$E \propto 1/r^2,\; V \propto 1/r$$

3. Graph

Option (D) shows this behavior.

Deep dive

Gauss’s law on a sphere inside the shell encloses zero charge → E = 0. Potential inside equals surface potential (no field means no change). Outside: E = kQ/r², V = kQ/r.

Common pitfalls

• Thinking V = 0 inside
• E discontinuous at surface (it’s continuous for V, not E)

Hints

• E = 0 inside shell; V constant inside; E ∝ 1/r² outside.

Answer

(D)