Comprehension

A parallel plate capacitor of capacitance C has a dielectric slab between its plates. It is charged to a potential difference V by connecting it across a battery. The battery is then disconnected. If the dielectric slab is now withdrawn from the capacitor, how will the following be affected ?Justify your answer in each case.

Question: 1

Capacitance of the capacitor

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Removing the dielectric from a capacitor decreases its capacitance by the dielectric constant \(K\). If \[ C=KC_0, \] then after removing the dielectric, \[ \boxed{ C'=\frac{C}{K} } \] Remember that capacitance depends only on the geometry and the dielectric medium and is independent of charge and potential.

Updated On: Jul 2, 2026

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Solution and Explanation

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Question: 2

Energy stored in the capacitor

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If the battery is disconnected, then \[ \boxed{Q=\text{constant}} \] and therefore \[ \boxed{ U=\frac{Q^2}{2C} } \] must be used. Hence, when the dielectric is removed, \[ C \downarrow \quad\Rightarrow\quad U \uparrow \] and \[ \boxed{ U_f=K\,U_i } \] where \(K\) is the dielectric constant of the slab.

Updated On: Jul 2, 2026

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Question: 3

The potential difference between the plates of the capacitor.

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When the battery is disconnected: \[ \boxed{Q=\text{constant}} \] and \[ V=\frac{Q}{C}. \] Therefore, if the dielectric is removed, \[ C \downarrow \quad\Rightarrow\quad V \uparrow \] Specifically, \[ \boxed{ V'=KV } \] where \(K\) is the dielectric constant of the slab.

Updated On: Jul 2, 2026

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