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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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).
