A capacitor is being charged. The electric flux through the gap between its plates changes at the rate of \(1.13\times10^{11}\ \mathrm{V\,m\,s^{-1}}\). The displacement current in the gap is (Take \(\varepsilon_0=8.85\times10^{-12}\ \mathrm{C^{2}\,N^{-1}\,m^{-2}}\))
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(1.0\ \mathrm{A}\)
There is no conduction current in the gap. The changing electric flux there gives a displacement current, \( i_d=\varepsilon_0\,d\Phi_E/dt \).
\( \frac{d\Phi_E}{dt}=1.13\times10^{11}\ \mathrm{V\,m\,s^{-1}} \)
\( \varepsilon_0=8.85\times10^{-12}\ \mathrm{C^{2}\,N^{-1}\,m^{-2}} \)
\( \begin{aligned} i_d &= \varepsilon_0\frac{d\Phi_E}{dt} \\ &= 8.85\times10^{-12}\times1.13\times10^{11} \\ &= 1.0\ \mathrm{A} \end{aligned} \)
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)). So we have”
\( \displaystyle i=i_c+i_d=i_c+\varepsilon_0\frac{d\Phi_E}{dt} \qquad (8.5) \)