A parallel plate capacitor has square plates of side 20 cm. While the capacitor is being charged, the electric field between its plates increases at the rate of \( 5\times10^{12}\,\mathrm{V\,m^{-1}\,s^{-1}} \). The displacement current between the plates is (Take \( \varepsilon_0=8.85\times10^{-12}\,\mathrm{C^{2}\,N^{-1}\,m^{-2}} \))
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 1.77 A
The displacement current is \( \varepsilon_0 \) times the rate of change of the electric flux through the gap. The field between the plates is uniform over the plate area \( A \), so \( \Phi_E=EA \) and \( i_d=\varepsilon_0A\,dE/dt \).
The plate area is:
\( \begin{aligned} A &= (0.20\,\mathrm{m})^{2} \\ &= 0.04\,\mathrm{m^{2}} \end{aligned} \)
The displacement current is:
\( \begin{aligned} i_d &= \varepsilon_0A\,\frac{dE}{dt} \\ &= 8.85\times10^{-12}\,\mathrm{C^{2}\,N^{-1}\,m^{-2}}\times0.04\,\mathrm{m^{2}}\times5\times10^{12}\,\mathrm{V\,m^{-1}\,s^{-1}} \\ &= 1.77\,\mathrm{A} \end{aligned} \)
NCERT (2026-27):
Displacement Current – “… More precisely, the total current i is the sum of the conduction current denoted by ic, and the displacement current denoted by id (= ε0 (dΦE/dt)).”