Potential Due To Hollow Conductiong, Solid Conducting, Hollow Non Conducting MCQ - Practice Questions with Answers

Electric Potential due to Hollow conducting, Hollow non-conducting, Solid conducting Sphere-

In the case of Hollow conducting, Hollow non-conducting, Solid conducting Spheres charges always resides on the surface of the sphere.

If the charge on a conducting sphere of radius R is Q. And we want to find V at point P at distance r from the center of the sphere.

• Outside the sphere (P lies outside the sphere. I.e r>R)

$
\begin{aligned}
& E_{\text {out }}=\frac{1}{4 \pi \epsilon_0} \frac{Q}{r^2}=\frac{\sigma R^2}{\epsilon_0 r^2} \\
& V(r)=-\int_{r=\infty}^{r=r} \vec{E} \cdot d \vec{r}=\frac{1}{4 \pi \varepsilon_0} \frac{\vec{Q}}{r}
\end{aligned}
$

- Inside the sphere ( P lies inside the sphere. I.e $\mathrm{r}<\mathrm{R}$ )

$
E_{i n}=0
$

$V_{\text {in }}=$ constant and it is given as

$
\begin{aligned}
\boldsymbol{V}(\boldsymbol{r}) & =-\int_{r=\infty}^{r=r} \overrightarrow{\boldsymbol{E}} \cdot d \vec{r}=-\int_{\infty}^R \boldsymbol{E}_r(d \boldsymbol{r})-\int_R^r \boldsymbol{E}_r(\boldsymbol{d} \boldsymbol{r})=\frac{1}{4 \pi \boldsymbol{\varepsilon}_0} * \frac{\boldsymbol{q}}{\boldsymbol{R}}+0 \\
\Rightarrow V(r) & =\frac{1}{4 \pi \varepsilon_0} * \frac{\boldsymbol{q}}{\boldsymbol{R}}
\end{aligned}
$

- At the surface of Sphere (I.e at $r=R$ )

$
\begin{aligned}
& E_s=\frac{1}{4 \pi \epsilon_0} \frac{Q}{R^2}=\frac{\sigma}{\epsilon_0} \\
& V_s=\frac{1}{4 \pi \epsilon_0} \frac{Q}{R}=\frac{\sigma R}{\epsilon_0}
\end{aligned}
$

• The graph between (E vs r)  and (V vs r) is given below

Electric Potential due to Uniformly charged Non conducting sphere-

Suppose charge Q is uniformly distributed in the volume of a non-conducting sphere of Radius R.

And we want to find V at point P at distance r from the center of the sphere.

• Outside the sphere (P lies outside the sphere. I.e r>R)

$
\begin{array}{ll}
E_{\text {out }}=\frac{1}{4 \pi \epsilon_0} \frac{Q}{r^2} & V_{\text {out }}=\frac{1}{4 \pi \epsilon_0} \frac{Q}{r} \\
E_{\text {out }}=\frac{\rho R^3}{3 \epsilon_0 r^2} & V_{\text {out }}=\frac{\rho R^3}{3 \epsilon_0 r}
\end{array}
$

- Inside the sphere ( P lies inside the sphere. I.e $\mathrm{r}<\mathrm{R}$ )

$
\begin{aligned}
& E_{\text {in }}=\frac{1}{4 \pi \epsilon_0} \frac{Q r}{R^3} \quad V_{\text {in }}=\frac{Q}{4 \pi \epsilon_0} * \frac{3 R^2-r^2}{2 R^3} \\
& E_{\text {in }}=\frac{\rho r}{3 \epsilon_0} \quad V_{\text {in }}=\frac{\rho\left(3 R^2-r^2\right)}{6 \epsilon_0}
\end{aligned}
$

- At the surface of Sphere (I.e at $r=R$ )

$
\begin{array}{ll}
E_s=\frac{1}{4 \pi \epsilon_0} \frac{Q}{R^2} & V_s=\frac{1}{4 \pi \epsilon_0} \frac{Q}{R} \\
E_s=\frac{\rho R}{3 \epsilon_0} & V_s=\frac{\rho R^2}{3 \epsilon_0}
\end{array}
$

Note - If P lies at the centre of the uniformly charged non-conducting sphere (I.e at $\mathrm{r}=0$ )

$
\begin{aligned}
& V_{\text {centre }}=\frac{3}{2} \times \frac{1}{4 \pi \epsilon_0} \frac{Q}{R}=\frac{3}{2} V_s \\
& \text { i.e } \quad V_c>V_s
\end{aligned}
$
 

• The graph between (E vs r)  and (V vs r) is given below