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(A) For an isolated charged parallel plate capacitor, the charge $q$ on the plates remains constant.The electric field $E$ between the plates is given

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(A) For an isolated charged parallel plate capacitor, the charge $q$ on the plates remains constant.The electric field $E$ between the plates is given by $E = \frac{\sigma}{\varepsilon_{0}} = \frac{q}{A \varepsilon_{0}}$, which is constant since $q$, $A$, and $\varepsilon_{0}$ are constant.The electrostatic force of attraction between the plates is $F = qE = \frac{q^2}{2A \varepsilon_{0}}$.Since the force is constant, the acceleration $a$ of the plates is constant $(a = F/m)$.The separation $d$ between the plates at time $t$ is given by $d(t) = d_{0} - \frac{1}{2} a t^2$, where $d_{0}$ is the initial separation.The potential difference $V$ is given by $V = E \cdot d(t) = E (d_{0} - \frac{1}{2} a t^2)$.This equation $V(t) = E d_{0} - (\frac{E a}{2}) t^2$ represents a downward-opening parabola, which corresponds to Graph $A$.

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