CBSE 2026 · Region 1 · Set 3 · Q26 · 3 marks
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.
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 2026board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.