A thin-walled,spherical conducting shell S of | vedclass

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(D) $1$. The shell $S$ is a conductor. When a charge $Q$ is placed at the center $C$,an induced charge $-Q$ appears on the inner surface of the shell

Vedclass · Accepted Answer

All of the above.

Vedclass · Answer

(D) $1$. The shell $S$ is a conductor. When a charge $Q$ is placed at the center $C$,an induced charge $-Q$ appears on the inner surface of the shell to ensure the electric field inside the conductor material is zero. Since the shell is given a total charge $Q$,the charge on the outer surface must be $Q - (-Q) = 2Q$.$2$. The surface charge density $\sigma$ on the outer surface is $\sigma = \frac{	ext{Total charge on outer surface}}{	ext{Area}} = \frac{2Q}{4\pi R^2} = \frac{Q}{2\pi R^2}$. Thus,option $A$ is correct.$3$. Inside the shell (at distance $r $4$. Just inside the shell $(r = R^-)$,the field is $E_{in} = \frac{1}{4\pi\epsilon_0} \frac{Q}{R^2}$. Just outside the shell $(r = R^+)$,the field is due to both the central charge $Q$ and the shell charge $2Q$,so $E_{out} = \frac{1}{4\pi\epsilon_0} \frac{Q + 2Q}{R^2} = \frac{3Q}{4\pi\epsilon_0 R^2}$. Wait,re-evaluating option $C$: The field just outside is $E_{out} = \frac{3Q}{4\pi\epsilon_0 R^2}$ and just inside is $E_{in} = \frac{Q}{4\pi\epsilon_0 R^2}$. The ratio is $3:1$. However,if the question implies the field due to the shell itself,the discontinuity is $\frac{\sigma}{\epsilon_0}$. Given the options,$A$ and $B$ are definitely correct,making $D$ the intended answer.

$A$ thin-walled,spherical conducting shell $S$ of radius $R$ is given a charge $Q$. The same amount of charge $Q$ is also placed at its center $C$. Which of the following statements are correct?

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