Q172 / 200 · MHT CET - Medical 2026 (25 April Shift 2) · Physics
A parallel plate capacitor with capacitance $1\ \mu\text{F}$ is being charged at a rate such that the potential difference across it changes at $10^6\ \text{V/s}$. The instantaneous displacement current between the plates is
- A$0\ \text{A}$
- B$0.1\ \text{A}$
- Correct answerC$1\ \text{A}$
- D$10^{-2}\ \text{A}$
Solution
The displacement current $I_d$ between the plates of a capacitor is given by
$I_d = \varepsilon_0 \dfrac{d\Phi_E}{dt}$
Since the electric flux $\Phi_E = EA = \dfrac{V}{d}A$, we have
$I_d = \varepsilon_0 \dfrac{A}{d} \dfrac{dV}{dt} = C \dfrac{dV}{dt}$
Given $C = 1\ \mu\text{F} = 10^{-6}\ \text{F}$ and $\dfrac{dV}{dt} = 10^6\ \text{V/s}$
$I_d = 10^{-6} \times 10^6 = 1\ \text{A}$
Answer: $1\ \text{A}$
$I_d = \varepsilon_0 \dfrac{d\Phi_E}{dt}$
Since the electric flux $\Phi_E = EA = \dfrac{V}{d}A$, we have
$I_d = \varepsilon_0 \dfrac{A}{d} \dfrac{dV}{dt} = C \dfrac{dV}{dt}$
Given $C = 1\ \mu\text{F} = 10^{-6}\ \text{F}$ and $\dfrac{dV}{dt} = 10^6\ \text{V/s}$
$I_d = 10^{-6} \times 10^6 = 1\ \text{A}$
Answer: $1\ \text{A}$
Hence, the correct answer is option: 'C'.