Question

Prove that the electric potential at the centre of a uniformly charged non-conducting sphere is 1.5 times more than that on the surface.

Answer 1

Electric Potential due to Charged Non-conducting Sphere Consider a non-conducting sphere of radius R be charged by a charge q. The electric field intensity at the points outside the sphere, on the surface and inside the sphere is as follows:
Outside, $$ \quad \vec{E}=\dfrac{1}{4 \pi \varepsilon_{0}} \dfrac{q}{r^{2}} \hat{\mathrm{r}}(r>R) $$
On the surface, $$ \vec{E}=\dfrac{1}{4 \pi \varepsilon_{0}} \dfrac{q}{R^{2}} \hat{r}(r=R) $$
Inside, $$ \quad \vec{E}=\dfrac{1}{4 \pi \varepsilon_{0}} \dfrac{q}{R^{3}} \hat{r}(r<R) $$
The potential at a point in electric field intensity $$ \vec{E} $$ is calculated by the help of following relation
$$ -\int_{\infty}^{r} \vec{E} \cdot \overrightarrow{d r} $$
The electric potential is calculated at the observation point in different conditions:
(a) At a point outside the non-conducting sphere
By the definition of electric potential
$$V=-\int_{\infty}^{r} \vec{E} \cdot \vec{d} r$$
The electric field at point $$ P $$.
But,$$\vec{E}=\dfrac{1}{4 \pi \varepsilon_{0}} \dfrac{q}{r^{2}} \hat{r}$$
Thus,$$V=-\int_{\infty}^{r} \dfrac{1}{4 \pi \varepsilon_{0}} \dfrac{q}{r^{2}} \hat{r} \hat{d} r$$
or $$V=-\dfrac{q}{4 \pi \varepsilon_{0}} \int_{\infty}^{r} \dfrac{1}{r^{2}} d r(\vec{r} \vec{d} r=d r)$$
or $$ \quad V=\dfrac{-q}{4 \pi \varepsilon_{0}}\left[\dfrac{-1}{r}\right]_{\infty}^{r} $$
or $$V=\dfrac{q}{4 \pi \varepsilon_{0}}\left[\dfrac{1}{r}-\dfrac{1}{\infty}\right] \quad\left(\because \dfrac{1}{\infty}=0\right)$$
or $$V=\dfrac{q}{4 \pi \varepsilon_{0} r} \ldots (1)$$
Thus, for external points on non-conducting sphere,
$$V \propto \dfrac{1}{r}$$
(b) At a point on the surface of non-conducting spherePutting $$ r=R $$ in equation (1)
$$ V_{S}=\dfrac{1}{4 \pi \varepsilon_{0}} \dfrac{q}{R} \ldots$$(2)
(c) At a point inside the non-conducting sphere $$ (r<R) $$
For a point inside the non-conducting sphere, the electric field intensity E does not depend uniformly from infinity to r distance, i.e., from infinity to surface, it is inversely proportional to the square of the distance rand it is directly proportional to the distance from the surface to the centre r. Thus, integral is divided into two parts:
Thus, $$ \quad V=\left[-\int_{\infty}^{R} \vec{E} \cdot \vec{d} r\right]+\left[-\int_{R}^{r} \vec{E} \cdot \vec{d} r\right] $$
or, $$ \quad V=\dfrac{1}{4 \pi \varepsilon_{0}} \dfrac{q}{R}-\int_{R}^{r} \dfrac{1}{4 \pi \varepsilon_{0}} \dfrac{q}{r^{3}} r \hat{r} \vec{d} r $$
$$ \left[\because \vec{E}=\dfrac{1}{4 \pi \varepsilon_{0}} \dfrac{q}{R^{3}} r \hat{r}\right] $$
or, $$ \quad V=\dfrac{1}{4 \pi \varepsilon_{0}} \dfrac{q}{R}-\dfrac{1}{4 \pi \varepsilon_{0}} \dfrac{q}{R^{3}}\left(\dfrac{r^{2}}{2}\right)_{R}^{r} $$
or, $$ \quad V=\dfrac{1}{4 \pi \varepsilon_{0}} \dfrac{q}{R}-\dfrac{1}{4 \pi \varepsilon_{0}} \dfrac{q}{R^{3}}\left[\dfrac{r^{2}}{2}-\dfrac{R^{2}}{2}\right] $$
or, $$ V=\dfrac{1}{4 \pi \varepsilon_{0}} q\left[\dfrac{1}{R}-\dfrac{r^{2}}{2 R^{3}}+\dfrac{1}{2 R}\right] $$
or, $$ V=\dfrac{1}{4 \pi \varepsilon_{0}} q\left[\dfrac{3 R^{2}-r^{2}}{2 R^{3}}\right] $$
or, $$ V=\dfrac{q}{4 \pi \varepsilon_{0}}\left[\dfrac{3 R^{2}-r^{2}}{2 R^{3}}\right] \ldots$$(3)
For a point at centre, putting $$ r=0 $$ in eqn. (3)
$$\begin{array}{l}V_{\text {cente }}=\dfrac{3}{2}\left(\dfrac{q}{4 \pi \varepsilon_{0} R}\right)\ldots(4) \\\mathrm{or \quad}V_{\text {centre }}=\dfrac{3}{2} V_{s}=1.5 \mathrm{V}_{s}\end{array}$$
Thus, the electric potential at centre of a charged non-conducting sphere is 1.5 times that on its surface. It is clear that the electric potential decreases with $$ r^{2} $$ from centre to surface in a charged non-conducting sphere. After that, it decreases as per the law of $$ r^{-1} $$ and becomes zero at infinity. It is shown in a graph infigure (3.16)