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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.

  • 26 Questions around this concept.

Solve by difficulty

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 fieldE (r) produced by the shell in the range  0\leq r< \infty   where  r   is the distance from the centre of the shell?

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)

         \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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