A parallel plate capacitor consisting of two circular plates of radius 10 cm is being charged by a constant current of 0.15 A. If the rate of change of potential difference between the plates is 7× 10⁸ V/s then the integer value of the distance between the parallel plates is ( Take, ε₀=9× 10⁻¹²F/m, π=22/7) ____________. Numerical answer type. Work it out, then open the answer.

Answer: 1320. I = CdV/dt = (ε₀A)/(d)dV/dt ⇒ d = (ε₀A)/(I)dV/dt A = π r² = 22/7×(0.1)² = 0.22/7 m² d = 9×10⁻¹²×(0.22/7)×7×10⁸0.15 = 1.98×10⁻⁴0.15 = 1.32×10⁻³ m d = 1320

A parallel plate capacitor consisting of two circular plates of… | JEE Main 2025 (29 Jan, Shift 2)

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Capacitors in DC Circuits & RC

A parallel plate capacitor consisting of two circular plates of radius 10 cm is being charged by a constant current of 0.15 A. If the rate of change of potential difference between the plates is $7\times 10^{8}\text{ V/s}$ then the integer value of the distance between the parallel plates is $($ Take, $\epsilon_0=9\times 10^{-12}\frac{\text{F}}{\text{m}},\ \pi=\frac{22}{7})$ ____________ $\mu\text{m}$ .

Numerical answer type. Work it out, then open the answer.

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Answer: 1320

$I = C\dfrac{dV}{dt} = \dfrac{\varepsilon_0A}{d}\dfrac{dV}{dt} \Rightarrow d = \dfrac{\varepsilon_0A}{I}\dfrac{dV}{dt}$
$A = \pi r^2 = \dfrac{22}{7}\times(0.1)^2 = \dfrac{0.22}{7}\text{ m}^2$
$d = \dfrac{9\times10^{-12}\times(0.22/7)\times7\times10^{8}}{0.15} = \dfrac{1.98\times10^{-4}}{0.15} = 1.32\times10^{-3}\text{ m}$
$d = 1320\ \mu\text{m}$

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