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$\mathit{V}\mathbf{\left(}\mathit{r}\mathbf{\right)}$.

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$q$ .

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.

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$\hat{r}$ is the position vector.

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

$E_2=\frac{1}{4\pi {\epsilon }_0}\frac{q}{r^2}\widehat{rr}$ …(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:

role="math" localid="1657772808865" $V_1\left(r\right)=-{\int }_{\infty }^r{E}_1·dl$

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.

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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