Physics - EM Waves Question with Solution | TestHub

$$J_d={ϵ}_0\frac{\partial E}{\partial t}$$

For a sinusoidal electric field $E=E_m\sin (\omega t)$, the time derivative is $\frac{\partial E}{\partial t}=E_m\omega \cos (\omega t)$.

So, the amplitude of the displacement current density is $J_{dm}={ϵ}_0{E}_m\omega$.

The total current amplitude ($I_m$) in the capacitor is related to the displacement current density amplitude by the area $S$:

$$I_m=J_{dm}S$$

$$I_m=({ϵ}_0{E}_m\omega )S$$

Now, we can solve for the electric field strength amplitude $E_m$:

$$E_m=\frac{I_m}{{ϵ}_0\omega S}$$

Substitute the given values:

$$E_m=\frac{1.0\times 10^{-3}A}{(8.85\times 10^{-12}F/m)(1.6\times 10^7{s}^{-1})(10^{-2}{m}^2)}$$

$E_m\approx 7\times 10^{2}V/m$.

The final answer is $\boxed{7V/cm}$.