Question

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?

Short answer

1. The radial distance inside the ball where the magnitude of the electric field is 1/4 of the maximum electric field is$0.25R$.
2. The radial distance outside the ball where the magnitude of the electric field is 1/4 of the maximum electric field is $2.00R$.

Step-by-step solution

The given data

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

Understanding the concept of Gauss law

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)

a) Calculation of the radial distance inside the ball

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 $0.25R$.

b) Calculation of the radial distance outside the ball

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 $2.00R$.