For a parallel plate capacitor of capacitance 2 μF to establish an instantaneous displacement current of 2.0 mA in the space between its plates, the rate of change of voltage required is:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( 10^{3}\,\mathrm{V\,s^{-1}} \)
The displacement current between the plates of a capacitor is \( I_d = \varepsilon_0\,\frac{d\Phi_E}{dt} \). For a parallel plate capacitor \( \Phi_E = EA = \frac{V}{d}A \), so with \( C = \varepsilon_0A/d \):
\( \displaystyle I_d = \varepsilon_0\frac{A}{d}\frac{dV}{dt} = C\,\frac{dV}{dt} \)
Putting \( I_d = 2.0\,\mathrm{mA} = 2.0\times10^{-3}\,\mathrm{A} \) and \( C = 2\,\mu\mathrm{F} = 2\times10^{-6}\,\mathrm{F} \):
\( \displaystyle \frac{dV}{dt} = \frac{I_d}{C} = \frac{2.0\times10^{-3}}{2\times10^{-6}}\,\mathrm{V\,s^{-1}} = 10^{3}\,\mathrm{V\,s^{-1}} \)
Note: NTA picked this question from an NCERT Exemplar problem and changed the figures. Do not ignore the Exemplar while practising numericals.
NCERT Exemplar:
Electromagnetic Waves – …
