Q21P - Introduction to Electrodynamics

Question

Find the potential inside and outside a uniformly charged solid sphere whose radius is  and whose total charge is . Use infinity as your reference point. Compute the gradient of  in each region, and check that it yields the correct field. Sketch $\mathit{V}\mathbf{\left(}\mathit{r}\mathbf{\right)}$ .

Step-by-Step Solution

Answer

The potential inside and outside the sphere are $\frac{q}{4\pi {\epsilon }_0}\left(\frac{1}{r}\right)\text{ and }\frac{q}{4\pi {\epsilon }_0}\frac{1}{2R}\left(3-\frac{r^2}{R^2}\right)$ respectively.

The gradient of in each region are $-\frac{q}{4\pi {\epsilon }_0}\frac{1}{r^2}\hat{r}\text{ and }-\frac{q}{4\pi {\epsilon }_0}\frac{r}{R^3}\hat{r}$respectively.

Identification of the given data

The given data can be listed below as:

• The radius of the uniformly charged solid sphere is $R$.
• The charge of the solid sphere is $q$ .

Significance of the potential

The potential is described as the work done to move a charge from one point to another point. The electrostatic forces are applied in the potential of a sphere.

Determination of the potential inside an outside of a uniformly charged solid sphere

The equation of the electric field outside the sphere is expressed as:

  $E_1=\frac{1}{4\pi {\epsilon }_0}\frac{q}{r^2}\hat{r}$                                                                                                              …(i)

 

Here, $E_2$ is the electric field outside the sphere, $\frac{1}{4\pi {\epsilon }_0}$ is the electric field constant, $q$ is the charge of the field, $R$ is the radius of the electric field and $\widehat{r }$ is the position vector.

 

The equation of the electric field inside the sphere is expressed as:

 

   $E_2=\frac{1}{4\pi {\epsilon }_0}\frac{q}{r^2}\stackrel{ ^}{rr}$                                                                                                         …(ii)

 

Here, $E_2$ is the electric field outside the sphere, $\frac{1}{4\pi {\epsilon }_0}$ is the electric field constant, $q$ is the charge of the field, $R$ is the radius of the sphere and $\hat{r}$ is the position vector.

 $V_1\left(r\right)=-{\int }_{\infty }^r{E}_1·dl$

The equation of the potential inside the sphere is expressed as:

 

$V_1\left(r\right)=-{\int }_{\infty }^r{E}_1·dl$ 

 

Here, $V_1\left(r\right)$ is the potential inside the sphere and $dl$ is the change in the length.

Substitute the value of the equation (i) in the above equation.

 

$$\begin{gathered} V_1\left(r\right)=-{\int }_{\infty }^r\left(\frac{1}{4\pi {\epsilon }_0}\frac{q}{r^2}\hat{r}\right)dr \\ =\frac{q}{4\pi {\epsilon }_0}{\left(\frac{1}{r}\right)}_{\infty }^r \\ =\frac{q}{4\pi {\epsilon }_0}{\left(\frac{1}{r}\right)}_{\infty }^{} \end{gathered}$$

 

The equation of the potential outside the sphere is expressed as:

 

 $V_2\left(r\right)=-{\int }_{\infty }^r{E}_2·dl$

 

Here, $V_2\left(r\right)$ is the potential outside the sphere and $dl$ is the change in the length.

 

Substitute the value of the equation (ii) in the above equation.

 $$\begin{gathered} V_2\left(r\right)=-{\int }_{\infty }^R\left(\frac{1}{4\pi {\epsilon }_0}\frac{q}{{\overline{r}}^2}\right)dr-{\int }_R^r\left(\frac{1}{4\pi {\epsilon }_0}\frac{q}{R^2}\overline{r}\right)dr \\ =\frac{q}{4\pi {\epsilon }_0}\left(\frac{1}{R}-\frac{1}{R^3}\left(\frac{r^2-R^2}{2}\right)\right) \\ =\frac{q}{4\pi {\epsilon }_0}\frac{1}{2R}\left(3-\frac{r^2}{R^2}\right) \end{gathered}$$

 

Thus, the potential inside and outside the sphere are $\frac{q}{4\pi {\epsilon }_0}\left(\frac{1}{r}\right)\text{ and }\frac{q}{4\pi {\epsilon }_0}\frac{1}{2R}\left(3-\frac{r^2}{R^2}\right)$ respectively.

Determination of the gradient of and sketching the function V ( r )

The equation of the potential gradient inside the sphere is expressed as:

 

 $\nabla V_1=\frac{\partial }{\partial r}\left(V_1\left(r\right)\right)$

 

Here, $\nabla V_1$ is the potential gradient inside the sphere and $\frac{\partial }{\partial r}$ is the derivative with respect to the radius $r$ .

 

Substitute the values in the above equation.

 $$\begin{gathered} \nabla V_1=\frac{q}{4\pi {\epsilon }_0}\frac{\partial }{\partial r}\left(\frac{1}{r}\right)\hat{r} \\ =-\frac{q}{4\pi {\epsilon }_0}\frac{\partial }{\partial r}\frac{1}{r^2}\hat{r} \end{gathered}$$

 

The equation of the potential gradient outside the sphere is expressed as:

 

 $\nabla V_2=\frac{\partial }{\partial r}\left(V_2\left(r\right)\right)$

 

Here, $\nabla V_2$ is the potential gradient outside the sphere and $\frac{\partial }{\partial r}$ is the derivative with respect to the radius $r$ .

 

Substitute the values in the above equation.

$$\begin{gathered} \nabla V_2=\frac{q}{4\pi {\epsilon }_0}\frac{1}{2R}\frac{\partial }{\partial r}\left(3-\frac{r^2}{R^2}\right)\hat{r} \\ =-\frac{q}{4\pi {\epsilon }_0}\frac{1}{2R}\left(-\frac{2r}{R^2}\right)\hat{r} \\ =-\frac{q}{4\pi {\epsilon }_0}\frac{r}{R^3}\hat{r} \end{gathered}$$

The function $V\left(r\right)$ has been sketched below:

 

 

In the above diagram, it has been observed that with the increase in the distance $r$, the function$V\left(r\right)$significantly decreases.

 

Thus, the gradient of $V$ in each region are $-\frac{q}{4\pi {\epsilon }_0}\frac{1}{r^2}\hat{r}\text{ and }-\frac{q}{4\pi {\epsilon }_0}\frac{r}{R^3}\hat{r}$ respectively.

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