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Physics · 2026 · 3 marks
CBSE 2026 · Region 1 · Set 3 · Q26
A parallel plate capacitor of capacitance C is charged to V volt by a battery. After sometime the battery is disconnected and the distance between…
A parallel plate capacitor of capacitance C is charged to V volt by a battery. After sometime the battery is disconnected and the distance between the plates is doubled. A slab of dielectric constant $\displaystyle \mathrm{k}=1.8$ is then introduced to completely fill the space between the plates. How will the following be affected ?
(a)
The capacitance of the capacitor.
(b)
The electric field between the plates of the capacitor.
(c)
The energy stored in the capacitor. Justify your answer in each case.
Marking-scheme solution
(a)
For parallel plate capacitor
$\displaystyle \mathrm{C}_{0}=\dfrac{\varepsilon_{0} \mathrm{~A}}{\mathrm{~d}}$
If the distance between the plate is doubled and dielectric slab of dielectric constant $\displaystyle \mathrm{K}=1.8$ is introduced.
The new capacitance $\displaystyle \mathrm{C}=\dfrac{\mathrm{K} \varepsilon_{0} \mathrm{~A}}{2 \mathrm{~d}}$
$\displaystyle \mathrm{C}=\dfrac{\mathrm{K}}{2} \mathrm{C}_{\mathrm{o}}$
$\displaystyle \mathrm{C}=\dfrac{1.8}{2} \mathrm{C}_{\mathrm{o}}$
$\displaystyle \mathrm{C}=0.9\ \mathrm{C}_{\mathrm{o}}$
New capacitance decreases
(b)
$\displaystyle \because$ Battery is disconnected, charge on capacitor remains same.
New potential on capacitor is
$\displaystyle \mathrm{V}^{\prime}=\dfrac{\mathrm{Q}}{\mathrm{C}^{\prime}}$
$\displaystyle \mathrm{V}^{\prime}=\dfrac{\mathrm{Q}}{0.9 \mathrm{C}_{\mathrm{o}}}$
Electric field between the plates of capacitor $\displaystyle \mathrm{E}^{\prime}=\dfrac{\mathrm{V}^{\prime}}{\mathrm{d}^{\prime}}$
$\displaystyle \mathrm{E}^{\prime}=\dfrac{\mathrm{Q}}{0.9 \mathrm{C} \times 2 \mathrm{~d}}$
$\displaystyle \mathrm{E}^{\prime}=\dfrac{1}{1.8} \dfrac{\mathrm{Q}}{\mathrm{Cd}}$
$\displaystyle \mathrm{E}^{\prime}=\dfrac{\mathrm{E}}{1.8}$
Electric field decreases
(c)
$\displaystyle \because$ Energy stored in the capacitor $\displaystyle \mathrm{U}=\dfrac{1}{2} \dfrac{\mathrm{Q}^{2}}{\mathrm{C}}$
$\displaystyle \mathrm{U}^{\prime}=\dfrac{1}{2} \dfrac{\mathrm{Q}^{2}}{\mathrm{C}^{\prime}}$
$\displaystyle \mathrm{U}^{\prime}=\dfrac{1}{2} \dfrac{\mathrm{Q}^{2}}{\left(0.9 \mathrm{C}_{\mathrm{o}}\right)}$
$\displaystyle \mathrm{U}^{\prime}=\dfrac{\mathrm{U}}{0.9}$
Energy stored 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.