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A charge Q is distributed over two concentric hollow spheres of radii r and \left ( R> r \right )  such that the surface densities are equal. The potential at the common centre is

  • Option 1)

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

     

     

     

     

  • Option 2)

    \frac{Q}{R+r}

  • Option 3)

    Zero

  • Option 4)

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

 

Answers (2)

best_answer

As we learned

Potential Due to 2 Concentric Spheres -

If two concentric conducting shells of radii r1 and r2(r2 > r1) carrying uniformly distributed charges Q1 and Q2 respectively

- wherein

Potential at the surface of inner shell is

V_{1}= \frac{1}{4\pi \varepsilon _{0}}\cdot \frac{Q_{1}}{r_{1}}+ \frac{1}{4\pi \varepsilon _{0}}\cdot \frac{Q_{2}}{r_{2}}

and potential at the surface of outer shell

V_{1}= \frac{1}{4\pi \varepsilon _{0}}\cdot \frac{Q_{1}}{r_{1}}+ \frac{1}{4\pi \varepsilon _{0}}\cdot \frac{Q_{2}}{r_{2}}

 

 

 

 

 

 

If q1 and q2 are the charges on spheres of radius r and R respectively, in accordance with conservation of charge

Q=q_{1}+q_{2}\cdots \cdots \cdots \cdots \cdots (i)

and according to the given problem \sigma _{1}=\sigma _{2}

i.e \frac{q_{1}}{4\pi r^{2}}=\frac{q_{2}}{4\pi R^{2}}\Rightarrow \frac{q_{1}}{q_{2}}=\frac{r^{2}}{R^{2}}\cdots \cdots \cdots \cdots \cdots \cdots (ii)                                                   

So equation (i) and (ii) gives  q_{1}=\frac{Qr^{2}}{\left ( R^{2}+r^{2} \right )}\: and\; q_{2}=\frac{QR^{2}}{\left ( R^{2}+r^{2} \right )}

Potential at common centre     
 V=\frac{1}{4\pi \varepsilon _{0}}\left [ \frac{q_{1}}{r}+\frac{q_{2}}{R} \right ]=\frac{1}{4\pi \varepsilon _{0}}\left [ \frac{Qr}{\left ( R^{2}+r^{2} \right )}+ \frac{Qr}{\left ( R^{2}+r^{2} \right )}\right ]=\frac{1}{4\pi \varepsilon _{0}}\cdot \frac{Q\left ( R+r \right )}{\left ( R^{2}+r^{2} \right )}

 


Option 1)

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

 

 

 

 

Option 2)

\frac{Q}{R+r}

Option 3)

Zero

Option 4)

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

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