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## GR0877 Problem 67

PROBLEM STATEMENT: This problem is still being typed.

SOLUTION: (D) At any instant of time an electric field inside the parallel-plate capacitor is $\displaystyle E = \frac{\sigma}{\epsilon_0} = \frac{Q}{\epsilon_0 S}$, where $\sigma$ is a surface density of a charge $Q$ on a positive plate. Differentiating this with respect to $t$ gives $\displaystyle \frac{dE}{dt} = \frac{dQ/dt}{\epsilon_0 S} = \frac{I}{\epsilon_0 S} = \frac{9}{8.85 \cdot 10^{-12} \cdot (0.5)^2} \approx$ $\displaystyle \approx \frac{10^{12}}{0.25} = 4 \cdot 10^{12} \left(\frac{\text{V}}{\text{m} \cdot \text{s}}\right)$.

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