If $\sigma$ is the surface charge density on the positive plate, then the electric field between the plates will be
$\begin{align*}E =\frac{\sigma}{\epsilon_0}\end{align*}$
$\sigma = Q/A$ where $Q$ is the charge on the positive plate and $A$ is the area of either plate. $A = \left(0.5 \;\mathrm{m} \right)^2 = 0.25 \;\mathrm{m}^2$ and $Q = \left(9 \mbox{ A}\right) t + Q_0$ . So,
$\begin{align*}\frac{d E}{dt} &= \frac{d}{dt}\left(\frac{\sigma}{\epsilon_0} \right) \\&= \frac{d}{dt}\left(\frac{Q}{A...$
Therefore, answer (D) is correct.

Solution to 2008 Problem 67