Question

A spherical metal shell $A$ of radius $R_A$ and a solid metal sphere $B$ of radius $R_8(<R_A)$ are kept far apart and each is given charge $+Q$. Now they are connected by a thin metal wire. Then

• A. $E_A^{inside}=0$
• B. $Q_A>Q_B$
• C. $\frac{{\sigma }_A}{{\sigma }_B}=\frac{R_B}{R_A}$
• D. $E_A^{onsurface}<E_B^{onsurface}$

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A $E_A^{inside}=0$
B $\frac{{\sigma }_A}{{\sigma }_B}=\frac{R_B}{R_A}$
C $E_A^{onsurface}<E_B^{onsurface}$
D $Q_A>Q_B$

Electric field inside a spherical metallic shell with charge on surface $=0$

$\therefore \left(a\right)$ is correct

On connecting Both with wise

Electric potential will be equal say $V$

$\therefore \frac{1}{4\pi Co}\frac{Q_A}{R_A}=\frac{1}{4\pi Co}\frac{Q_B}{R_B}=V$

as $R_A>R_B\therefore Q_A>Q_B$

Hence $\left(b\right)$ is correct

as $\frac{{\sigma }_A}{{\sigma }_B}=\frac{Q_B}{4\pi R_B^2}=\frac{R^{2}B}{R^{2}A}\times \frac{4\pi Co{R}_{A}V}{4\pi Co{R}_{A}R}$

$\frac{\sigma A}{\sigma B}=\frac{R_B}{R_A}$

Hence $\left(C\right)$ is correct

Also $E_A=\frac{{\sigma }_A}{{\sigma }_B}=\frac{R_B}{R_A}<1\therefore E_A<E_B$

Hence $\left(d\right)$ is correct