A spherical conductor of diameter 3 mm is kept in a uniform electric field of intensity 4 × 10⁷ N/C. The maximum charge on the conductor is ((1)/(4πε₀) = 9 × 10⁹ SI units) (a) 2 × 10⁻⁸ C (b) 10⁻⁸ C (c) 4 × 10⁻⁸ C (d) 3 × 10⁻⁸ C

Answer: (b). The given field is the largest field the surrounding air can bear, so the surface field must not exceed it: (Q)/(4πε₀ r²) ≤ E. r = 1.5 mm = 1.5 × 10⁻³ m. Qmax = (E r²)/(9 × 10⁹) = 4 × 10⁷ × 2.25 × 10⁻⁶9 × 10⁹ = 10⁻⁸ C So the answer is (b) 10⁻⁸ C.

A spherical conductor of diameter 3 mm is kept in a uniform electric… | MHT CET 2026 (12 May, Shift 2)

Q13MHT CET 2026 · 12 MAY · SHIFT 2Medium

Electric Flux & Gauss's Law

A spherical conductor of diameter $3$ mm is kept in a uniform electric field of intensity $4 \times 10^{7}$ N/C. The maximum charge on the conductor is $\left(\frac{1}{4\pi\epsilon_0} = 9 \times 10^{9}\text{ SI units}\right)$

Show answer and solution

Answer: (b)

The given field is the largest field the surrounding air can bear, so the surface field must not exceed it: $\frac{Q}{4\pi\varepsilon_0 r^2} \le E$ .
$r = 1.5$ mm $= 1.5 \times 10^{-3}$ m.
$Q_{max} = \frac{E r^2}{9 \times 10^9} = \frac{4 \times 10^7 \times 2.25 \times 10^{-6}}{9 \times 10^9} = 10^{-8}$ C
So the answer is (b) $10^{-8}$ C.

Report a mistake

Electric Flux & Gauss's Law question from MHT CET 2026 (12 May, Shift 2)

More questions on Electric Flux & Gauss's Law