Question

Find the electric field inside and outside a hollow plastic ball of radius $\mathrm{R}$ that has charge$\mathrm{Q}$ uniformly distributed on its outer surface.

Short answer

Electric field inside a hollow plastic ball is $0\mathrm{nC}$.

Electric field outside a hollow plastic ball is$\frac{1}{4{\mathrm{\pi ϵ}}_{\mathrm{o}}}\frac{\mathrm{Q}}{{\mathrm{r}}^2}$.

Step-by-step solution

Figure for Electric field inside a hollow plastic ball  

Figure is,

Calculation for electric field inside a hollow plastic ball  

Electric flux

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

localid="1648763653778" $\mathrm{E}=\frac{{\mathrm{Q}}_{\text{in}}}{{\mathrm{ϵ}}_{\mathrm{o}}\mathrm{A}}...1$

The charge is spread on the outside surface of a plastic sphere, which is not conductive.

As a result, the sphere's net charge is zero.

The quantity of charge in the bulk determines the electric field.

As a result of the net charge being zero.

Inside the sphere, the electric field is zero.

${\mathrm{E}}_{\text{in}}=0\mathrm{N}/\mathrm{C}$

Figure for Electric field outside a hollow plastic ball  

Figure is,

Calculation for electric field outside a hollow plastic ball  

The Gaussian surface enclosed net charge for a point outside the spher$+\mathrm{Q}$.

The enclosed surface's area is

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

This charge is enclosedlocalid="1648918575953" ${\mathrm{Q}}_{\text{in}}$in the gaussian surface is,

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

Substitute all values in equation $1$,

We get,

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

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

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