Choose the correct dimension of \( \varepsilon_{o}\,\frac{d\varphi_{E}}{dt} \) from the following:
In the given expression symbols have their usual meaning.
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( \left[M^0L^0T^0A\right] \)
The quantity \( \varepsilon_0\,\frac{d\phi_E}{dt} \) is the displacement current of Maxwell, so its dimensions are those of a current. The same result follows from the dimensions of the factors.
Permittivity, from \( F = \frac{q_1q_2}{4\pi\varepsilon_0r^{2}} \):
\( \displaystyle [\varepsilon_0] = \frac{[A\,T]^{2}}{[M\,L\,T^{-2}][L^{2}]} = [M^{-1}L^{-3}T^{4}A^{2}] \)
Electric flux, \( \phi_E = EA \), with \( [E] = \frac{[M\,L\,T^{-2}]}{[A\,T]} = [M\,L\,T^{-3}A^{-1}] \):
\( [\phi_E] = [M\,L\,T^{-3}A^{-1}][L^{2}] = [M\,L^{3}T^{-3}A^{-1}] \)
\( \displaystyle \left[\frac{d\phi_E}{dt}\right] = [M\,L^{3}T^{-4}A^{-1}] \)
\( \displaystyle \left[\varepsilon_0\frac{d\phi_E}{dt}\right] = [M^{-1}L^{-3}T^{4}A^{2}][M\,L^{3}T^{-4}A^{-1}] = [M^{0}L^{0}T^{0}A] \)
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)).”