Question

A hollow metal sphere has inner radius$\mathrm{a}$and outer radius . The hollow sphere has charge$+2\mathrm{Q}$. A point charge$+\mathrm{Q}$sits at the center of the hollow sphere.

a. Determine the electric fields in the three regions $\mathrm{r}\le \mathrm{a}$,$\mathrm{a}<\mathrm{r}<\mathrm{b}$, and $\mathrm{r}\ge \mathrm{b}$.

b. How much charge is on the inside surface of the hollow sphere$?$On the exterior surface$?$

Short answer

a.Three regions have electric fields${\mathrm{E}}_{\mathrm{r}\le \mathrm{a}}=\frac{1}{4{\mathrm{\pi ϵ}}_{\mathrm{o}}}\frac{\mathrm{Q}}{{\mathrm{r}}^2}$,${\mathrm{E}}_{\mathrm{a}<\mathrm{r}<\mathrm{b}}=0$and ${\mathrm{E}}_{\mathrm{r}\ge \mathrm{b}}=\frac{1}{4{\mathrm{\pi ϵ}}_{\mathrm{o}}}\frac{3\mathrm{Q}}{{\mathrm{r}}^2}$.

b.There is electricity in the interior cavity is$-\mathrm{Q}$and the external surface is$+3\mathrm{Q}$.

Step-by-step solution

Calculation for electric field in Er≤a(part a)

(a).

The electric flux is the quantity of electricity that flows through a closed surface.

Gauss's law states that the electric field passing through a surface is proportional to the charge within it.

The electric flux ${\mathrm{\Phi }}_{\mathrm{e}}$is,

${\mathrm{\Phi }}_{\mathrm{e}}=\mathrm{EA}=\frac{{\mathrm{Q}}_{\text{in}}}{{\mathrm{ϵ}}_{\mathrm{o}}}$

$\mathrm{E}=\frac{{\mathrm{Q}}_{\text{in}}}{{\mathrm{ϵ}}_{\mathrm{o}}\mathrm{A}}$

For distancelocalid="1648728179590" $\mathrm{r}\le \mathrm{a}$,the charge enclosed islocalid="1648728194274" $+\mathrm{Q}$.

Charge${\mathrm{Q}}_{\text{in}}$in the sphere is,

${\mathrm{Q}}_{\text{in}}=+\mathrm{Q}$

The area is,

$\mathrm{A}=4{\mathrm{\pi r}}^2$.

So,

${\mathrm{E}}_{\mathrm{r}\le \mathrm{a}}=\frac{{\mathrm{Q}}_{\text{in}}}{{\mathrm{ϵ}}_{\mathrm{o}}\mathrm{A}}$

$=\frac{\mathrm{Q}}{{\mathrm{ϵ}}_{\mathrm{o}}\left(4{\mathrm{\pi r}}^2\right)}$

$=\frac{1}{4{\mathrm{\pi ϵ}}_{\mathrm{o}}}\frac{\mathrm{Q}}{{\mathrm{r}}^2}$

Calculation of electric field inEa<r<bandEr≥bpart(a) solution

(a).

For distance $\mathrm{a}<\mathrm{r}<\mathrm{b}$is inside the conductor.

Inside the conductor, there is no electric field.

${\mathrm{E}}_{\mathrm{a}<\mathrm{r}<\mathrm{b}}=0$

For distance localid="1648915677849" $\mathrm{r}\ge \mathrm{b}$,

The charge enclosed is localid="1648915682412" $+\mathrm{Q}$andlocalid="1648915687074" $+2\mathrm{Q}$.

So,

${\mathrm{Q}}_{\text{in}}=2\mathrm{Q}+\mathrm{Q}=3\mathrm{Q}$

The area is,

$\mathrm{A}=4{\mathrm{\pi r}}^2$.

$=\frac{3\mathrm{Q}}{{\mathrm{ϵ}}_{\mathrm{o}}\left(4{\mathrm{\pi r}}^2\right)}$

$=\frac{1}{4{\mathrm{\pi ϵ}}_{\mathrm{o}}}\frac{3\mathrm{Q}}{{\mathrm{r}}^2}$

Calculation for charge on the inner and outer surface of sphere (part b)

(b).

The cavity's positive charge $+\mathrm{Q}$causes a negative charge of the same magnitude on the cavity's inner surface.

As a result, the charge on the cavity's inner surface is the same magnitude as the charge within, but in the opposite direction.

Inner surface charge is,

$-\mathrm{Q}$

The sphere has chargelocalid="1648731642219" $+2\mathrm{Q}$.

And the inner surface of the sphere has chargelocalid="1648731651493" $-\mathrm{Q}$.

On the outside surface,

The total charge becomes,

$2\mathrm{Q}-(-\mathrm{Q})=+3\mathrm{Q}$