CBSE 2025 · Region 4 · Set 1 · Q29 · 4 marks
A parallel plate capacitor consists of two conducting plates kept generally parallel to each other at a distance. When the capacitor is charged, the charge resides on the inner surfaces of the plates and an electric field is set up between them. Thus, electrostatic energy is stored in the capacitor. The figure shows three large square metallic plates, each of side ' $\displaystyle \mathrm{L}$ ' held parallel and equidistant from each other. The space between $\displaystyle \mathrm{P}_{1}$ and $\displaystyle \mathrm{P}_{2}$ and $\displaystyle \mathrm{P}_{2}$ and $\displaystyle \mathrm{P}_{3}$ is completely filled with mica sheets of dielectric constant ' K '. The plate $\displaystyle \mathrm{P}_{2}$ is connected to point A and other plates $\displaystyle \mathrm{P}_{1}$ and $\displaystyle \mathrm{P}_{3}$ are connected to point B . Point A is maintained at a positive potential with respect to point B and the potential difference between A and B is V .
(i)The capacitance of the system between A and B will be :(A)$\displaystyle \frac{\varepsilon_{0} \mathrm{KL}^{2}}{\mathrm{~d}}$(B)$\displaystyle \frac{\varepsilon_{0} \mathrm{KL}^{2}}{2 \mathrm{~d}}$(C)$\displaystyle \frac{2 \varepsilon_{0} \mathrm{KL}^{2}}{\mathrm{~d}}$(D)$\displaystyle \frac{2 \varepsilon_{0} \mathrm{Kd}}{\mathrm{L}^{2}}$(ii)The charge on plate $\displaystyle \mathrm{P}_{1}$ is :(A)$\displaystyle \frac{\varepsilon_{0} \mathrm{VKL}^{2}}{2 \mathrm{~d}}$(B)$\displaystyle \frac{\varepsilon_{0} \mathrm{VKL}^{2}}{\mathrm{~d}}$(C)$\displaystyle \frac{2 \varepsilon_{0} \mathrm{VKL}^{2}}{\mathrm{~d}}$(D)$\displaystyle \frac{\varepsilon_{0} \mathrm{VKL}^{2}}{4 \mathrm{~d}}$(iii)The electric field in the region between $\displaystyle \mathrm{P}_{1}$ and $\displaystyle \mathrm{P}_{2}$ is :(A)$\displaystyle \frac{\mathrm{V}}{\mathrm{d}}$(B)$\displaystyle \frac{2 \mathrm{~V}}{\mathrm{~d}}$(C)$\displaystyle \frac{\mathrm{V}}{2 \mathrm{~d}}$(D)$\displaystyle \frac{\mathrm{d}}{\mathrm{V}}$(iv)The separation between the plates of same area ( $\displaystyle \mathrm{L}^{2}$ ) of a parallel plate air capacitor having capacitance equal to that of this system, will be :(A)$\displaystyle \frac{\mathrm{d}}{\mathrm{K}}$(B)$\displaystyle \frac{2 \mathrm{~d}}{\mathrm{~K}}$(C)$\displaystyle \frac{\mathrm{d}}{2 \mathrm{~K}}$(D)$\displaystyle \frac{\mathrm{d}}{4 \mathrm{~K}}$If the source of potential difference applied between A and B is removed, and then A and B are connected by a conducting wire, the net charge on the system will be :(A)$\displaystyle \frac{\varepsilon_{0} \mathrm{VKL}^{2}}{4 \mathrm{~d}}$(B)$\displaystyle \frac{\varepsilon_{0} \mathrm{VKL}^{2}}{2 \mathrm{~d}}$(C)$\displaystyle \frac{\varepsilon_{0} \mathrm{VKL}^{2}}{\mathrm{~d}}$(D)Zero
A parallel plate capacitor consists of two conducting plates kept generally parallel to each other at a distance. When the capacitor is charged, the charge resides on the inner surfaces of the plates and an electric field is set up between them. Thus, electrostatic energy is stored in the capacitor. The figure shows three large square metallic plates, each of side ' $\displaystyle \mathrm{L}$ ' held parallel and equidistant from each other. The space between $\displaystyle \mathrm{P}_{1}$ and $\displaystyle \mathrm{P}_{2}$ and $\displaystyle \mathrm{P}_{2}$ and $\displaystyle \mathrm{P}_{3}$ is completely filled with mica sheets of dielectric constant ' K '. The plate $\displaystyle \mathrm{P}_{2}$ is connected to point A and other plates $\displaystyle \mathrm{P}_{1}$ and $\displaystyle \mathrm{P}_{3}$ are connected to point B . Point A is maintained at a positive potential with respect to point B and the potential difference between A and B is V .
(i)
The capacitance of the system between A and B will be :
(A)
$\displaystyle \frac{\varepsilon_{0} \mathrm{KL}^{2}}{\mathrm{~d}}$
(B)
$\displaystyle \frac{\varepsilon_{0} \mathrm{KL}^{2}}{2 \mathrm{~d}}$
(C)
$\displaystyle \frac{2 \varepsilon_{0} \mathrm{KL}^{2}}{\mathrm{~d}}$
(D)
$\displaystyle \frac{2 \varepsilon_{0} \mathrm{Kd}}{\mathrm{L}^{2}}$
(ii)
The charge on plate $\displaystyle \mathrm{P}_{1}$ is :
(A)
$\displaystyle \frac{\varepsilon_{0} \mathrm{VKL}^{2}}{2 \mathrm{~d}}$
(B)
$\displaystyle \frac{\varepsilon_{0} \mathrm{VKL}^{2}}{\mathrm{~d}}$
(C)
$\displaystyle \frac{2 \varepsilon_{0} \mathrm{VKL}^{2}}{\mathrm{~d}}$
(D)
$\displaystyle \frac{\varepsilon_{0} \mathrm{VKL}^{2}}{4 \mathrm{~d}}$
(iii)
The electric field in the region between $\displaystyle \mathrm{P}_{1}$ and $\displaystyle \mathrm{P}_{2}$ is :
(A)
$\displaystyle \frac{\mathrm{V}}{\mathrm{d}}$
(B)
$\displaystyle \frac{2 \mathrm{~V}}{\mathrm{~d}}$
(C)
$\displaystyle \frac{\mathrm{V}}{2 \mathrm{~d}}$
(D)
$\displaystyle \frac{\mathrm{d}}{\mathrm{V}}$
(iv)
The separation between the plates of same area ( $\displaystyle \mathrm{L}^{2}$ ) of a parallel plate air capacitor having capacitance equal to that of this system, will be :
(A)
$\displaystyle \frac{\mathrm{d}}{\mathrm{K}}$
(B)
$\displaystyle \frac{2 \mathrm{~d}}{\mathrm{~K}}$
(C)
$\displaystyle \frac{\mathrm{d}}{2 \mathrm{~K}}$
(D)
$\displaystyle \frac{\mathrm{d}}{4 \mathrm{~K}}$
If the source of potential difference applied between A and B is removed, and then A and B are connected by a conducting wire, the net charge on the system will be :
(A)
$\displaystyle \frac{\varepsilon_{0} \mathrm{VKL}^{2}}{4 \mathrm{~d}}$
(B)
$\displaystyle \frac{\varepsilon_{0} \mathrm{VKL}^{2}}{2 \mathrm{~d}}$
(C)
$\displaystyle \frac{\varepsilon_{0} \mathrm{VKL}^{2}}{\mathrm{~d}}$
(D)
Zero
Marking-scheme solution
(C)
$\displaystyle \frac{2 \varepsilon_{0} K L^{2}}{d}$
(B)
$\displaystyle \frac{\varepsilon_{0} V K L^{2}}{d}$
(A)
$\displaystyle \frac{V}{d}$
(a)
(C) $\displaystyle \frac{d}{2 K}$
(D)
Zero
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