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Physics · 2025 · 5 marks
CBSE 2025 · Region 1 · Set 2 · Q32
(a) (i) Two point charges 5 μ C and -1 μ C are placed at points ( -3 cm, 0,0) and ( 3 cm, 0,0 ) respectively. An external electric field E =A/(r^2) r…
(a)(i)
Two point charges $\displaystyle 5 \mu \mathrm{C}$ and $\displaystyle -1 \mu \mathrm{C}$ are placed at points ( -$\displaystyle 3$ cm, $\displaystyle 0,0)$ and ( $\displaystyle 3 \mathrm{~cm}, 0,0$ ) respectively. An external electric field $\displaystyle \overrightarrow{\mathrm{E}}=\frac{\mathrm{A}}{\mathrm{r}^{2}} \hat{\mathrm{r}}$ where $\displaystyle \mathrm{A}=3 \times 10^{5} \mathrm{Vm}$ is switched on in the region. Calculate the change in electrostatic energy of the system due to the electric field.
(ii)
A system of two conductors is placed in air and they have net charge of $\displaystyle +80 \mu \mathrm{C}$ and $\displaystyle -80 \mu \mathrm{C}$ which causes a potential difference of $\displaystyle 16$ V between them.
(1)
Find the capacitance of the system.
(2)
If the air between the capacitor is replaced by a dielectric medium of dielectric constant $\displaystyle 3$, what will be the potential difference between the two conductors?
(3)
If the charges on two conductors are changed to $\displaystyle +160 \mu \mathrm{C}$ and $\displaystyle -160 \mu \mathrm{C}$, will the capacitance of the system change? Give reason for your answer.
(b)(i)
Consider three metal spherical shells $\displaystyle \mathrm{A}, \mathrm{B}$ and C , each of radius R . Each shell is having a concentric metal ball of radius $\displaystyle \mathrm{R} / 10$. The spherical shells $\displaystyle \mathrm{A}, \mathrm{B}$ and C are given charges $\displaystyle +6 \mathrm{q},-4 \mathrm{q}$, and $\displaystyle 14 \mathrm{q}$ respectively. Their inner metal balls are also given charges $\displaystyle -2 \mathrm{q},+8 \mathrm{q}$ and -$\displaystyle 10$ q respectively. Compare the magnitude of the electric fields due to shells $\displaystyle \mathrm{A}, \mathrm{B}$ and C at a distance $\displaystyle 3$ R from their centres.
(ii)
A charge $\displaystyle -6 \mu \mathrm{C}$ is placed at the centre B of a semicircle of radius $\displaystyle 5$ cm , as shown in the figure. An equal and opposite charge is placed at point D at a distance of $\displaystyle 10$ cm from B . A charge $\displaystyle +5 \mu \mathrm{C}$ is moved from point ' C ' to point ' A ' along the circumference. Calculate the work done on the charge. 
Marking-scheme solution
(i)
$\displaystyle \vec{E}=\frac{3 \times 10^{5}}{r^{2}} \hat{r}$ (Given), $\displaystyle d V=-\vec{E} \cdot d \vec{r}$
$\displaystyle V=3 \times 10^{5} / r$
Electrostatic energy of the system in the absence of the field:
$\displaystyle U_{i}=\frac{K q_{1} q_{2}}{r_{12}}$
Electrostatic energy in the presence of the field:
$\displaystyle U_{f}=\frac{K q_{1} q_{2}}{r_{12}}+q_{1} V\left(\vec{r}_{1}\right)+q_{2} V\left(\vec{r}_{2}\right)$
$\displaystyle \Delta U=U_{f}-U_{i}=q_{1} V\left(\vec{r}_{1}\right)+q_{2} V\left(\vec{r}_{2}\right)$
$\displaystyle \Delta U=\frac{5 \times 10^{-6} \times 3 \times 10^{5}}{3 \times 10^{-2}}-\frac{1 \times 10^{-6} \times 3 \times 10^{5}}{3 \times 10^{-2}}$
$\displaystyle =40 \mathrm{~J}$
(ii)(1)
$\displaystyle C=\frac{Q}{V}=\frac{80}{16}=5 \mu \mathrm{F}$
(2)
$\displaystyle C^{\prime}=K C$
$\displaystyle =3 \times 5 \mu \mathrm{F}=15 \mu \mathrm{F}$
$\displaystyle V^{\prime}=\frac{Q}{C^{\prime}}=\frac{80 \mu \mathrm{C}}{15 \mu \mathrm{F}}=5.33 \mathrm{~V}$
(3)
No.
The capacitance of the system depends on its geometry.
(i)
Total charge for A = Total charge for B = Total charge for C $\displaystyle =+4 q$
As $\displaystyle E=\frac{k Q}{r^{2}}$
Since $\displaystyle Q=4 q$ and $\displaystyle r=3 R$:
$\displaystyle E=\frac{k(4 q)}{9 R^{2}}=\frac{4 k q}{9 R^{2}}$
$\displaystyle \therefore E_{A}=E_{B}=E_{C}$
(ii)
$\displaystyle V_{C}=\left[\frac{k \times 6 \times 10^{-6}}{5 \times 10^{-2}}-\frac{k \times 6 \times 10^{-6}}{5 \times 10^{-2}}\right]=0$
$\displaystyle V_{A}=\left[\frac{k \times 6 \times 10^{-6}}{15 \times 10^{-2}}-\frac{k \times 6 \times 10^{-6}}{5 \times 10^{-2}}\right]$
$\displaystyle =\frac{k \times 6 \times 10^{-6}}{10^{-2}}\left[\frac{1-3}{15}\right]$
$\displaystyle =-\frac{9 \times 10^{9} \times 6 \times 10^{-6} \times 2}{15 \times 10^{-2}}$
$\displaystyle =-7.2 \times 10^{5} \mathrm{~V}$
$\displaystyle W=q\left[V_{A}-V_{C}\right]$
$\displaystyle =5 \times 10^{-6}\left[-7.2 \times 10^{5}-0\right]$
$\displaystyle W=-3.6 \mathrm{~J}$
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