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Physics · 2024 · 3 marks
CBSE 2024 · Region 4 · Set 3 · Q23
Two conducting spherical shells A and B of radii R and 2 R are kept far apart and charged to the same charge density σ. They are connected by a wire.…
Two conducting spherical shells A and B of radii R and $\displaystyle 2$ R are kept far apart and charged to the same charge density $\displaystyle \sigma$. They are connected by a wire. Obtain an expression for final potential of shell A.
Marking-scheme solution
Charge on spherical shell A:
$\displaystyle q_{A}=4 \pi R^{2} \sigma$
Charge on spherical shell B:
$\displaystyle q_{B}=4 \pi(2 R)^{2} \sigma=16 \pi R^{2} \sigma$
After connecting by a wire, their potentials will become equal after sharing of charge:
$\displaystyle \frac{1}{4 \pi \varepsilon_{0}} \frac{q_{A}^{\prime}}{R}=\frac{1}{4 \pi \varepsilon_{0}} \frac{q_{B}^{\prime}}{2 R}$
$\displaystyle q_{B}^{\prime}=2 q_{A}^{\prime}$
From conservation of charge:
$\displaystyle q_{A}+q_{B}=q_{A}^{\prime}+q_{B}^{\prime}$
$\displaystyle 4 \pi R^{2} \sigma+16 \pi R^{2} \sigma=3 q_{A}^{\prime}$
$\displaystyle q_{A}^{\prime}=\frac{20 \pi R^{2} \sigma}{3}$
Final potential of shell A:
$\displaystyle V_{A}=\frac{1}{4 \pi \varepsilon_{0}} \frac{q_{A}^{\prime}}{R}$
$\displaystyle V_{A}=\frac{1}{4 \pi \varepsilon_{0}} \frac{20 \pi R^{2} \sigma}{3 R}$
$\displaystyle V_{A}=\frac{5 \sigma R}{3 \varepsilon_{0}}$
OR
Therefore the potential of shell A:
$\displaystyle V_{A}=V_{\text {common }}=\frac{q_{A}+q_{B}}{C_{A}+C_{B}}$
$\displaystyle =\frac{4 \pi R^{2} \sigma+16 \pi R^{2} \sigma}{4 \pi \varepsilon_{0} R+4 \pi \varepsilon_{0}(2 R)}$
$\displaystyle =\frac{5 \sigma R}{3 \varepsilon_{0}}$
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