Chargeis uniformly distributed in a sphere of radius R.

(a) What fraction of the charge is contained within the radius is r = R/2.00?

(b) What is the ratio of the electric field magnitude at $\mathbf{r}\mathbf{=}\mathbf{R}\mathbf{/}\mathbf{2}\mathbf{.}\mathbf{00}$to that on the surface of the sphere?

• The radius of the sphere R.
• Charge on the sphere Q.

To explain the electrostatic force between the two charges, we assume that the charges create an electric field around them. The magnitude of electric field E generated by the electric charge q at a distance r is given as,

$\mathrm{E}=\frac{\mathrm{q}}{4{\mathrm{\pi \epsilon }}_0{\mathrm{r}}^2}$

Using the concept of the electric field to determine the magnitude of the electric field.

The given situation is explained in the below diagram:

Since the volume contained within a radius of$R/2$ is one-eighth the volume contained within a radius of R, the charge at is$0<r<R/2isQ/8$ . Therefore, the fraction is,

$\frac{1}{8}=0.125$

Therefore, the fraction of the charge contained within the radius$r=R/2.00$ is 0.125.

The electric field produced at the surface by the charged sphere of radius R and charge Q is,

$E=\frac{Q}{4{\mathrm{\pi \epsilon }}_0{\mathrm{R}}^2}$

At $r=R/2.00$, the magnitude of the field is: $E_{R/2}$

role="math" localid="1657346499822" $$\begin{gathered} E_{R/2}=\frac{Q/8}{4{\mathrm{\pi \epsilon }}_0{\left(R/2\right)}^2} \\ =\frac{1}{2}\frac{Q}{4{\mathrm{\pi \epsilon }}_0{\mathrm{R}}^2} \end{gathered}$$

$$\begin{gathered} \frac{E_{R/2}}{E}=\frac{1}{2}\frac{Q}{4{\mathrm{\pi \epsilon }}_0{\mathrm{R}}^2} \\ \frac{Q}{4{\mathrm{\pi \epsilon }}_0{\mathrm{R}}^2} \\ =\frac{1}{2} \end{gathered}$$

Thus, The ratio of the electric field at R/2 and at the surface is$\frac{1}{2}$ .