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

Edited By admin | Updated on Sep 25, 2023 25:23 PM | #NEET

Quick Facts

  • Electric potential due to continuous charge distribution(II) is considered one the most difficult concept.

  • 30 Questions around this concept.

Solve by difficulty

A solid conducting sphere having a charge Q is surrounded by an uncharged concentric conducting hollow spherical shell. Let the potential difference between the surface of the solid sphere and that of the outer surface of the hollow shell be V.  If the shell is now given a charge of –3Q, the new potential difference between the same two surfaces is

A thin spherical shell of radius R has charge Q spread uniformly over its surface. Which of the following graphs most closely represents the electric field$E(r)$ produced by the shell in the range  $0 \leq r<\infty$   where  $r$   is the distance from the center of the shell?

A conducting sphere of radius R is given a charge Q. Consider three points B at the surface, A at the center, and C at a distance R/2 from the center. The electric potential at these points is such that.

A point charge q is placed at a distance of r from the center of an uncharged conducting sphere of radius R (<r). The potential at any point on the sphere is

A non-conducting solid sphere of radius  R is uniformly charged. The magnitude of the electric field due to the sphere at a distance r from its centrer

A conducting sphere of radius R is given a charge Q. The electric potential and the electric field at the centre of the sphere respectively are:

Concepts Covered - 1

Electric potential due to continuous charge distribution(II)

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

 

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Electric potential due to continuous charge distribution(II)

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