Question

Consider a uniformly charged sphere of radius R and total cAlC charge Q. The electric field $E_{\text{out}}$ outside the sphere$(r\ge R)$ is simply that of a point charge Q. In Chapter 24, we used Gauss's law to find that the electric field $E_{\text{in}}$ inside the sphere$(r\le R)$ is radially outward with field strength

$E_{\text{in}}=\frac{1}{4\pi {ϵ}_0}\frac{Q}{R^3}r$

a. The electric potential $V_{\text{out}}$outside the sphere is that of a point charge Q. Find an expression for the electric potential$V_{\text{in}}$at position r inside the sphere. As a reference, let $V_{\text{in}}=V_{\text{out}}$at the surface of the sphere.

b. What is the ratio $V_{\text{center}}/V_{\text{surface}}?$

c. Graph V versus r for 0 $\le$r $\le$3 R.

Short answer

The electric potential at a position inside the sphere,

$V_{in}=\frac{kQ}{2R}\left(3-\frac{r^2}{R^2}\right)$

The ratio

$V_{\text{cemes}}/V_{\text{surface}}=\frac{3}{2}$

Graph of

$V\text{versus}r\text{for}0\le r\le 3R\text{is drawn}$

Step-by-step solution

part (a) step 1 : given information

The electric field, $E_{\text{out}}$and electric potential, $V_{\text{out}}$outside the sphere $(r\ge R)$is that of a point charge Q. The electric field inside the sphere is given by,

$E_{in}=\frac{1}{4\pi {\in }_0}\frac{Q}{R^3}r$

$V_{\text{in}}=V_{\text{out}}$at the surface of the sphere

Formula Used:

If $\overrightarrow{E}$ represents electric field and V represents potential associated with it then they are related by $V=-\int Edz$

We have to find the potential at a point inside the sphere.

$E_{\text{out}}(r\ge R)=\frac{1}{4\pi {ϵ}_0}\frac{Q}{r^2}=\frac{kQ}{r^2}$

$E_{\text{in}}(r\le R)=\frac{1}{4\pi {ϵ}_0}\frac{Q}{R^3}r=\frac{kQ}{R^3}r$

Also,

$V_{\text{out}}(r\ge R)=\frac{1}{4\pi {ϵ}_0}\frac{Q}{r}=\frac{kQ}{r}$

- Since $V=-\int Edz$the potential inside the sphere is given by,

$$\begin{gathered} V_{\text{in}}=-{\int }_{=}^R{E}_{ow}dr-{\int }_{R^{\text{'}}}^r{E}_{in}dr=-{\int }_{=}^R\frac{kQ}{r^2}dr-{\int }_R^r\frac{kQ}{R^3}rdr \\ {\left.V_{\text{in}}=kQ\mid \frac{1}{r}\right]}_{-}^R-\frac{kQ}{R^3}{\left[\frac{r^2}{2}\right]}_R^r=\frac{kQ}{R}-\frac{kQ}{2R^3}\left(r^2-R^2\right) \end{gathered}$$

Therefore, $V_{in}=\frac{kQ}{2R}\left(3-\frac{r^2}{R^2}\right)$

part (b) step 1 : given information

From the expression for potential inside the sphere,

$V_{\text{center}}(r=0)=\frac{3kQ}{2R}$

$V_{\text{surface}}(r=R)=\frac{kQ}{R}$

Therefore, $\frac{V_{\text{center}}}{V_{\text{suffure}}}=\frac{3}{2}$

part (c) step 1 : given information

Graph of $V\text{versus}r\text{for}0\le r\le 3RV\left(r\right)$