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Physics · 2025 · 5 marks
CBSE 2025 · Region 7 · Set 1 · Q31
(a) (i) A parallel plate capacitor with plate area A and plate separation d has a capacitance C_0. A slab of dielectric constant K having area A and…
(i)
A parallel plate capacitor with plate area A and plate separation $\displaystyle d$ has a capacitance $\displaystyle C_{0}$. A slab of dielectric constant $\displaystyle K$ having area $\displaystyle A$ and thickness $\displaystyle \left(\frac{d}{4}\right)$ is inserted in the capacitor, parallel to the plates. Find the new value of its capacitance.
(ii)
You are provided with a large number of $\displaystyle 1 \mu \mathrm{~F}$ identical capacitors and a power supply of $\displaystyle 1200$ V . The dielectric medium used in each capacitor can withstand up to $\displaystyle 200$ V only. Find the minimum number of capacitors and their arrangement, required to build a capacitor system of equivalent capacitance of $\displaystyle 2 \mu \mathrm{~F}$ for use with this supply.
(i)
An electric dipole of dipole moment p consists of point charges q and -q , separated by $\displaystyle 2$ a . Derive an expression for electric potential in terms of its dipole moment at a point at a distance x (>> a) from its centre and lying (I) along its axis, and (II) along its bisector line.
(ii)
An electric dipole of dipole moment $\displaystyle \overrightarrow{\mathrm{p}}=(0.8 \hat{\mathrm{i}}+0.6 \hat{\mathrm{j}}) 10^{-29} \mathrm{Cm}$ is placed in an electric field $\displaystyle \overrightarrow{\mathrm{E}}=1.0 \times 10^{7} \hat{\mathrm{k}} \frac{\mathrm{V}}{\mathrm{m}}$. Calculate the magnitude of the torque acting on it and the angle it makes with the x-axis, at this instant.
Marking-scheme solution
(a)(i)
\[C_0 = \frac{\varepsilon_0 A}{d}\]
\[C = \frac{\varepsilon_0 A}{(d-t) + \dfrac{t}{K}}\]
$\displaystyle t = d/4$
\[C = \frac{\varepsilon_0 A}{\left(d - \dfrac{d}{4}\right) + \dfrac{d}{4K}} \;=\; \frac{\varepsilon_0 A}{d\left(\dfrac{3}{4} + \dfrac{1}{4K}\right)}\]
\[= C_0 \frac{4K}{(3K+1)}\]
Alternatively: When dielectric is inserted, the electric field between the plates is $\displaystyle E = E_0/K$
The potential difference will be
\[V = E_0\left(\frac{3d}{4}\right) + E\left(\frac{d}{4}\right)\]
\[= E_0\left(\frac{3d}{4}\right) + \frac{E_0}{K}\left(\frac{d}{4}\right)\]
\[= V_0\left(\frac{3}{4} + \frac{1}{4K}\right)\]
\[V = V_0\left(\frac{3K+1}{4K}\right)\]
\[C = \frac{Q_0}{V} = \left(\frac{4K}{3K+1}\right)\frac{Q_0}{V_0}\]
\[C = C_0\left(\frac{4K}{3K+1}\right)\]
(ii)
Each capacitance can withstand 200V
No. of capacitors in each row $\displaystyle = \dfrac{1200}{200} = 6$
Net capacitance of each row $\displaystyle = 1/6\ \mu F$
Number of rows $\displaystyle =\ n$
\[C_{eq} = C_1 + C_2 + \_\_\_\_\_\_\_ + C_n\]
\[C_{eq} = \frac{1}{6} + \frac{1}{6} + - - - - - n\]
\[2 = \frac{n}{6}\]
\[\therefore n = 12\]
Total no. of capacitors in the arrangement $\displaystyle = 6 \times 12$
$\displaystyle = 72$
OR(b) I. Along its axis
\[V_- = \frac{-kq}{x+a}\]
\[V_+ = \frac{kq}{x-a}\]
\[V = V_- + V_+\]
\[= kq\left(\frac{-1}{x+a} + \frac{1}{x-a}\right)\]
\[= kq\frac{2a}{\left(x^2 - a^2\right)} = \frac{kp}{x^2 - a^2}\]
$\displaystyle x \gg a \quad \therefore V = \dfrac{kp}{x^2}$
II. Along the bisector line
\[V_- = \frac{kq}{\sqrt{x^2 + a^2}}\]
\[V_+ = \frac{-kq}{\sqrt{x^2 + a^2}}\]
\[V = V_- + V_+\]
\[= 0\](ii)\[\vec{\tau} = \vec{p} \times \vec{E}\]
\[= (0.8\hat{i} + 0.6\hat{j}) \times 10^{-29} \times (1 \times 10^{7})\hat{k}\]
\[= \left[0.8\left(-\hat{j}\right) + 0.6\hat{i}\right] \times 10^{-22}\]
\[\tau = \left[\sqrt{(0.8)^2 + (0.6)^2}\right] \times 10^{-22}\]
\[= 10^{-22}\ Nm\]
\[\tan\alpha = \frac{|0.8|}{0.6}\]
\[\alpha = \tan^{-1}\left(\frac{4}{3}\right)\]
\[\alpha = 53^{0}\]
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