A displacement current of 1.0 A is established in the space between the two parallel plates of 1 μF capacitor. The rate of change of voltage across the capacitor is:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( 10^{6}\,\mathrm{V\,s^{-1}} \)
The displacement current between the plates of a charging capacitor equals the rate at which the charge on the plates changes, \( I_D = dQ/dt \). With \( Q = CV \) and a fixed capacitance C:
\( \displaystyle I_D = C\,\frac{dV}{dt} \)
\( \displaystyle \frac{dV}{dt} = \frac{I_D}{C} = \frac{1.0\,\mathrm{A}}{1\times10^{-6}\,\mathrm{F}} = 10^{6}\,\mathrm{V\,s^{-1}} \)
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 – …
