CBSE 2026 · Region 1 · Set 1 · Q23 · 3 marks
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 If the dielectric slab is now withdrawn from the capacitor, how will the following be affected ?(a)Capacitance of the capacitor,(b)Energy stored in the capacitor, and(c)The potential difference between the plates of the capacitor. Justify your answer in each case.
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 If the dielectric slab is now withdrawn from the capacitor, how will the following be affected ?
(a)
Capacitance of the capacitor,
(b)
Energy stored in the capacitor, and
(c)
The potential difference between the plates of the capacitor. Justify your answer in each case.
Marking-scheme solution
(a)
$\displaystyle \mathrm{C}=\mathrm{K} \dfrac{\varepsilon_{0} \mathrm{A}}{\mathrm{d}}$ Capacitance of capacitor with dielectric slab
$\displaystyle \mathrm{C}^{\prime}=\dfrac{\varepsilon_{0} \mathrm{A}}{\mathrm{d}}$ Capacitance of a capacitor without dielectric slab
$\displaystyle \mathrm{C}^{\prime}=\dfrac{\mathrm{C}}{\mathrm{K}}$ (Capacitance decreases)
(b)
As charge Q is constant
$\displaystyle \mathrm{U}=\dfrac{\mathrm{Q}^{2}}{2 \mathrm{C}}$
$\displaystyle \mathrm{U}^{\prime}=\dfrac{\mathrm{Q}^{2}}{2 \mathrm{C}^{\prime}}=\mathrm{KU}$ (Energy stored in the capacitor increases)
(c)
$\displaystyle \mathrm{V}=\dfrac{\mathrm{Q}}{\mathrm{C}}$
$\displaystyle \mathrm{V}^{\prime}=\dfrac{\mathrm{Q}}{\mathrm{C}^{\prime}}=\mathrm{KV}$ (Potential difference between the plates increases)
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CBSE Class 12 Physics past-paper question from the 2026board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.