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

Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main

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Electric potential due to continuous charge distribution(II) is considered one the most difficult concept.
39 Questions around this concept.

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EASY

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 centre of the shell?

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A thin spherical conducting shell of radius R has a charge $q$ . Another charge $Q$ is placed at the centre of the shell. The electrostatic potential at a point P at a distance $R/2$ from the centre of the shell is:

$\frac{2Q}{4\pi \varepsilon _{0}R}$

$\frac{2Q}{4\pi \varepsilon _{0}R}-\frac{2q}{4\pi \varepsilon _{0}R}$

$\frac{2Q}{4\pi \varepsilon _{0}R}+\frac{q}{4\pi \varepsilon _{0}R}$

$\frac{\left ( q+Q \right )}{4\pi \varepsilon _{0}}\frac{2}{R}$

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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)

$\dpi{100} E_{out}=\frac{1}{4\pi \epsilon _{0}}\frac{Q}{r^{2}}=\frac{\sigma R^{2}}{\epsilon _{0}r^{2}}$

$V(r)=-\int_{r=\infty}^{r=r} \vec{E}.d \vec{r} =\frac{1}{4 \pi \varepsilon_{0}} \frac{\vec{Q}}{r}$

Inside the sphere (P lies inside the sphere. I.e r<R )

$E_{in}=0$

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

$\boldsymbol{V}(\boldsymbol{r})=-\int_{r=\infty}^{r=r} \overrightarrow{\boldsymbol{E}} . d \vec{r} =-\int_{\infty}^{R} \boldsymbol{E}_{r}(d \boldsymbol{r})-\int_{R}^{r} \boldsymbol{E}_{r}(\boldsymbol{d} \boldsymbol{r})=\frac{1}{4 \pi \boldsymbol{\varepsilon}_{0}} *\frac{\boldsymbol{q}}{ {\boldsymbol{R}}}+0 \\ \\ \Rightarrow V(r)=\frac{1}{4 \pi \boldsymbol{\varepsilon}_{0}} *\frac{\boldsymbol{q}}{ {\boldsymbol{R}}}$

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

$E_{s}=\frac{1}{4\pi \epsilon _{0}}\frac{Q}{R^{2}}=\frac{\sigma }{\epsilon _{0}}$

$V_{s}=\frac{1}{4\pi \epsilon _{0}}\frac{Q}{R}=\frac{\sigma R}{\epsilon _{0}}$

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)

$E_{out}=\frac{1}{4\pi \epsilon _{0}}\frac{Q}{r^{2}}$ $V_{out}=\frac{1}{4\pi \epsilon _{0}}\frac{Q}{r}$

$E_{out}=\frac{\rho R^{3}}{3 \epsilon _{0}r^{2}}$ $V_{out}=\frac{\rho R^{3}}{3 \epsilon _{0}r}$

Inside the sphere (P lies inside the sphere. I.e r<R )

$E_{in}=\frac{1}{4\pi \epsilon _{0}}\frac{Qr}{R^{3}}$ $\dpi{100} V_{in}=\frac{Q}{4\pi \epsilon _{0}}*\frac{3R^{2}-r^{2}}{2R^{3}}$

$\dpi{100} E_{in}=\frac{\rho r}{3 \epsilon _{0}}$ $V_{in}=\frac{\rho \left ( 3R^{2}-r^{2} \right )}{6 \epsilon _{0}}$

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

$E_{s}=\frac{1}{4\pi \epsilon _{0}}\frac{Q}{R^{2}}$ $V_{s}=\frac{1}{4\pi \epsilon _{0}}\frac{Q}{R}$

$\dpi{100} E_{s}=\frac{\rho R}{3 \epsilon _{0}}$ $V_{s}=\frac{\rho R^{2}}{3 \epsilon _{0}}$

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

$V_{centre}=\frac{3}{2}\times \frac{1}{4\pi \epsilon _{0}}\frac{Q}{R}=\frac{3}{2}V_{s}$

i.e $\dpi{100} V_{c}> V_{s}$

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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