Electrostatics 3 Question 17
17. A spherical metal shell $A$ of radius $R_{A}$ and a solid metal sphere $B$ of radius $R_{B}\left(<R_{A}\right)$ are kept far apart and each is given charge $+Q$. Now they are connected by a thin metal wire. Then
(a) $E_{A}^{\text {inside }}=0$
(b) $Q_{A}>Q_{B}$
(c) $\frac{\sigma_{A}}{\sigma_{B}}=\frac{R_{B}}{R_{A}}$
(d) $E_{A}^{\text {on surface }}<E_{B}^{\text {on surface }}$
(2011)
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Answer:
Correct Answer: 17. (a, b, c, d)
Solution:
- Inside a conducting shell electric field is always zero. Therefore, option (a) is correct. When the two are connected, their potentials become the same.
$ \therefore \quad V_{A}=V_{B} \quad \text { or } \quad \frac{Q_{A}}{R_{A}}=\frac{Q_{B}}{R_{B}} \quad V=\frac{1}{4 \pi \varepsilon_{0}} \frac{Q}{R} $
Since, $\quad R_{A}>R_{B} \quad \therefore Q_{A}>Q_{B}$
$\therefore$ Option (b) is correct.
Potential is also equal to, $V=\frac{\sigma R}{\varepsilon_{0}}, V_{A}=V_{B}$
$\therefore \quad \sigma_{A} R_{A}=\sigma_{B} R_{B} \quad$ or $\quad \frac{\sigma_{A}}{\sigma_{B}}=\frac{R_{B}}{R_{A}} \quad$ or $\quad \sigma_{A}<\sigma_{B}$
$\therefore$ Option (c) is correct.
Electric field on surface, $E=\frac{\sigma}{\varepsilon_{0}}$ or $E \propto \sigma$
Since, $\sigma_{A}<\sigma_{B} \quad \therefore \quad E_{A}<E_{B}$
$\therefore$ Option (d) is also correct
$\therefore$ Correct options are (a), (b), (c) and (d).
