Between the plates of a parallel plate capaci | vedclass

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(B) When the capacitor is disconnected from the battery, the charge $Q$ on the plates remains constant.The capacitance $C$ of a parallel plate capacit

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(B) When the capacitor is disconnected from the battery, the charge $Q$ on the plates remains constant.The capacitance $C$ of a parallel plate capacitor with a dielectric is given by $C = \frac{K \epsilon_0 A}{d}$.As the dielectric plate of length $L$ is drawn out by a distance $x$, the capacitor can be considered as two capacitors in parallel: one with dielectric (length $L-x$) and one with air (length $x$).The equivalent capacitance $C(x)$ decreases as $x$ increases.Since $V = \frac{Q}{C}$, and $Q$ is constant, the potential difference $V$ increases as $C$ decreases.Specifically, $C(x) = \frac{\epsilon_0 w}{d} (L-x)K + \frac{\epsilon_0 w}{d} x = \frac{\epsilon_0 w}{d} [K(L-x) + x]$.As $x$ increases from $0$ to $L$, $C(x)$ decreases linearly, which means $V(x) = \frac{Q}{C(x)}$ will increase non-linearly (specifically, it follows a hyperbolic curve $V \propto \frac{1}{a-bx}$).Once the dielectric is completely removed $(x=L)$, the capacitance becomes constant $(C_{air} = \frac{\epsilon_0 A}{d})$, and thus the potential difference $V$ becomes constant.Graph $B$ correctly shows $V$ increasing non-linearly and then becoming constant.

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