Solution to 2008 Problem 67

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.

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