If $\epsilon_0$ denotes the permittivity of free space and $\Phi_{\mathrm{E}}$ is the flux of the electric field through the area bounded by the closed surface, then dimension of $\left(\epsilon_0 \frac{\mathrm{~d} \phi_{\mathrm{E}}}{\mathrm{dt}}\right)$ are that of :

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Aicharya · Accepted Answer

We know that formula for displacement current is given by $\mathrm{id}=\varepsilon_0 \frac{\mathrm{~d} \phi_{\varepsilon}}{\mathrm{dt}}$

If $\epsilon_0$ denotes the permittivity of free space and $\Phi_{\mathrm{E}}$ is the flux of the electric…

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