Two circular metal plates each of radius ' r ' are kept parallel to each other distance ' d ' apart. The capacitance of the capacitor formed is ' C₁ '. If the radius of each of the plates is increased to √(2) times the earlier radius and their distance of separation decreased to half the initial value, the capacitance now becomes ' C₂ '. The ratio of the capacitances C₁: C₂ is (a) 1: 1 (b) 1: 2 (c) 1: 4 (d) 4: 1

Answer: (c). C = (ε₀ π r²)/(d) ∝ r²/d (C₂)/(C₁) = ((√(2))²)/(1/2) = 4 So the answer is (c) C₁: C₂ = 1: 4.

Two circular metal plates each of radius ' r ' are kept parallel to… | MHT CET 2024 (3 May, Shift 1)

Q6MHT CET 2024 · 3 MAY · SHIFT 1Easy

Parallel Plate Capacitor

Two circular metal plates each of radius ' $r$ ' are kept parallel to each other distance ' $d$ ' apart. The capacitance of the capacitor formed is ' $C_1$ '. If the radius of each of the plates is increased to $\sqrt{2}$ times the earlier radius and their distance of separation decreased to half the initial value, the capacitance now becomes ' $C_2$ '. The ratio of the capacitances $C_1 : C_2$ is

Show answer and solution

Answer: (c)

$C = \frac{\varepsilon_0 \pi r^2}{d} \propto \frac{r^2}{d}$
$\frac{C_2}{C_1} = \frac{(\sqrt{2})^2}{1/2} = 4$
So the answer is (c) $C_1 : C_2 = 1 : 4$ .

Report a mistake

Parallel Plate Capacitor question from MHT CET 2024 (3 May, Shift 1)

More questions on Parallel Plate Capacitor