Question

$\mathrm{FIGURE}\mathrm{P}24.38$shows a solid metal sphere at the center of a hollow metal sphere. What is the total charge on (a) the exterior of the inner sphere, (b) the inside surface of the hollow sphere, and (c) the exterior surface of the hollow sphere?

Short answer

a.Charge on the exterior of the inner sphere is$-10.7\mathrm{nC}$.

b.Charge on the inside surface of the hollow sphere is role="math" localid="1648744482331" $+10.7\mathrm{nC}$.

c.Charge on the exterior surface of the hollow sphere is$48.2\mathrm{nC}$.

Step-by-step solution

Calculation for charge on the outside of the inner sphere (part a)

(a).

The amount of electrical field that travels through a closed surface is stated because the electric flux.

The electric field through a surface is expounded to the charge inside the surface, in line with Gauss's law.

Electric flux is,

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

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

localid="1648745447364" ${\mathrm{Q}}_{\text{in}}=4{\mathrm{\pi ϵ}}_{\mathrm{o}}{\mathrm{Er}}^2.....1$

The gaussian surface's radius is localid="1648747531003" $8\mathrm{cm}$.

Because the electrical field is contained within the hollow sphere, it's negative.

$\mathrm{E}=-15000\mathrm{N}/\mathrm{C}$.

Substitute values in equation localid="1648747540022" $1$,

We get,

${\mathrm{Q}}_{\text{in}}=4{\mathrm{\pi ϵ}}_{\mathrm{o}}{\mathrm{Er}}^2$

$=4\mathrm{\pi }\left(8.85\times {10}^{-12}{\mathrm{C}}^2/\mathrm{N}·{\mathrm{m}}^2\right)(-15000\mathrm{N}/\mathrm{C})(0.08\mathrm{m}{)}^2$

$=-10.7\times {10}^{-9}\mathrm{C}=-10.7\mathrm{nC}$

Calculation for charge on the within surface of the hollow sphere (part b)

(b).

The electric charge on the inner sphere's external surface generates a electric charge of the identical magnitude on the hollow sphere's interior surface.

The charge on the hollow sphere's inner surface is the image of the charge inside the small sphere, except it points within the other direction.

So,

${\mathrm{Q}}_{\text{in}}=-(-10.7\mathrm{nC})=+10.7\mathrm{nC}$

Calculation for Charge on the outside surface of the hollow sphere (part c)

(c).

The radius of the gaussian surface is $\mathrm{r}=17\mathrm{cm}$

The electric field emanating from the hollow spherical is positive.

$\mathrm{E}=15000\mathrm{N}/\mathrm{C}$

Substitute values in equation localid="1648747517181" $1$,

We get,

${\mathrm{Q}}_{\mathrm{in}}=4{\mathrm{\pi ϵ}}_{\mathrm{o}}{\mathrm{Er}}^2$

$=4\mathrm{\pi }\left(8.85\times {10}^{-12}{\mathrm{C}}^2/\mathrm{N}·{\mathrm{m}}^2\right)(15000\mathrm{N}/\mathrm{C})(0.17\mathrm{m}{)}^2$

$=48.2\times {10}^{-9}\mathrm{C}=48.2\mathrm{nC}$