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Physics · 2026 · 3 marks
Electrostatic Potential and CapacitanceEffect of Dielectric on Capacitance3 marksUnderstandshort answer
CBSE 2026 · Region 1 · Set 1 · Q23
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…
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 If the dielectric slab is now withdrawn from the capacitor, how will the following be affected ?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 2026 board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.