Chapter 2 Electrostatics: Problem 2.38

Problem 2.38: A metal sphere of radius $R$, carrying charge $q$, is surrounded by a
thick concentric metal shell (inner radius $a$, outer radius $b$, as in Fig. 2.48). The
shell carries no net charge.
(a) Find the surface charge density $\sigma$ at $R$, at $a$, and at $b$.
(b) Find the potential at the center, using infinity as the reference point.
(c) Now the outer surface is touched to a grounding wire, which drains off charge
and lowers its potential to zero (same as at infinity). How do your answers to
(a) and (b) change?

 

 

 

 

 

 

 

 

 

 

 

Solution:

Create a spherical guassian surface with radius $r$ such that $a\le r\le b$.

 

 

 

 

 

 

 

 

 

 

 

Since electric field inside a conductor is zero. 

Hence the inner part of the shell with inner radius a will get induced with a charge $-q$.

Now a make the shell electrically neutral, positive charge $+q$ will get induced on the outer surface of shell with outer radius $b$. 

 

$a)$ 

$${\sigma }_b=\frac{+q}{4\pi b^2}$$ 

$${\sigma }_a=\frac{-q}{4\pi a^2}$$ 

$${\sigma }_R=\frac{+q}{4\pi R^2}$$

$b)$ Let's denote the center of sphere with C so $$V_{(C)}-V_{\mathrm{\infty }}=-{\int }_{\mathrm{\infty }}^0\overrightarrow{E}\cdot \overrightarrow{dr}$$ 

$$V_{(C)}-0=-[{\int }_{\mathrm{\infty }}^b\frac{+q}{4\pi {ϵ}_0{r}^2}dr+{\int }_b^{a}0dr+{\int }_a^R\frac{+q}{4\pi {ϵ}_0{r}^2}dr+{\int }_R^{0}0dr]$$ 

$$V_{(C)}=-[\frac{+q}{4\pi {ϵ}_0}(\frac{-1}{r}{)}_{\mathrm{\infty }}^b+\frac{+q}{4\pi {ϵ}_0}(\frac{-1}{r}{)}_a^R]$$

$$V_{(C)}=\frac{+q}{4\pi {ϵ}_0}[\frac{1}{b}+\frac{1}{R}-\frac{1}{a}]$$ 

$c)$ If the outer surface of the shell is connected to ground, So the potential of the outer sphere will turn to 0.

Since charges in the sphere + charges on inner side of shell = $+q+(-q)=0$

Hence the ground will supply extra $-q$ to the outer side of shell to make its charge 0.

Hence the final charge distribution would be $${\sigma }_b=0$$ $${\sigma }_a=\frac{-q}{4\pi a^2}$$ $${\sigma }_R=\frac{+q}{4\pi R^2}$$

 Regarding Potential $$V_{(C)}-V_{\mathrm{\infty }}=-{\int }_{\mathrm{\infty }}^0\overrightarrow{E}\cdot \overrightarrow{dr}$$ $$V_{(C)}-0=-[{\int }_{\mathrm{\infty }}^{b}0dr+{\int }_b^{a}0dr+{\int }_a^R\frac{+q}{4\pi {ϵ}_0{r}^2}dr+{\int }_R^{0}0dr]$$ 

$$V_{(C)}=-[\frac{+q}{4\pi {ϵ}_0}(\frac{-1}{r}{)}_a^R]$$

$$V_{(C)}=\frac{+q}{4\pi {ϵ}_0}[\frac{1}{R}-\frac{1}{a}]$$

 

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