A spherical ball of charged particles has a uniform charge density. In terms of the ball’s radius R, at what radial distances

(a) inside and

(b) outside the ball is the magnitude of the ball’s electric field equal to$\frac{\mathbf{1}}{\mathbf{4}}$of the maximum magnitude of that field?

The radius r of the ball is $R$, that has a uniform charge density.

Using the concept of the electric field, we can get the equation of maximum, internal, and external fields. Thus, using these values, we can get the radial distance r for the given cases.

Formula:

The electric field at a point due to charged particle, $E\left(r\right)=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{q}{r^2}$ . (i)

The field maximum occurs at the outer surface. Thus, the maximum field, for charge q is given using equation (i) as:

${\left(E_{max}\right)}_{r=R}=\frac{\left|q\right|}{4{\mathrm{\pi \epsilon }}_0{\mathrm{R}}^2}$

Now, the internal electric field is given using equation (i) as:

$$\begin{gathered} E_{internal}=\frac{1}{4}{E}_{max} \\ \frac{\left|q\right|r}{4\pi {\epsilon }_0{R}^3}=\frac{1}{4}\frac{\left|q\right|}{4\pi {\epsilon }_0{R}^2} \\ =\left(R/4\right) \\ =0.25R \end{gathered}$$

Hence, the value of the radial distance is $0.25R$.

For the radial distance outside the sphere, we take the maximum field and field equation of (i) as follows:

$$\begin{gathered} E_{external}=\frac{1}{4}{E}_{max} \\ \frac{\left|q\right|}{4\pi {\epsilon }_0{r}^2}=\frac{1}{4}\left(\frac{\left|q\right|}{4\pi {\epsilon }_0{R}^2}\right) \\ r=2.00R \end{gathered}$$

Hence, the value of the radial distance is $2.00R$.