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  • Need explanation for: 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 ho

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

  • Option 1)

    V

  • Option 2)

    2V

  • Option 3)

    4V

  • Option 4)

    -2V

 

Answers (1)

best_answer

As we learned

 

Charged conducting sphere -

If charge on a conducting sphere of radius R is Q.

- wherein

 

 V_{inside}=V_{centre}=V_{surface}=\frac{1}{4\pi \varepsilon _{0}}\cdot \frac{q}{R},V_{outside} = \frac{1}{4\pi \varepsilon _{0}}\cdot \frac{q}{r}

If a and b are the radii of sphere and spherical shell respectively, then potential at their surface will be

V_{sphere}=\frac{1}{4\pi \varepsilon _{0}}\cdot \frac{Q}{a}  and  V_{shell}=\frac{1}{4\pi \varepsilon _{0}}\cdot \frac{Q}{b}

\therefore V=V_{sphere}-V_{shell}=\frac{1}{4\pi \varepsilon _{0}}\cdot \left [ \frac{Q}{a} -\frac{Q}{b}\right ]

Now when the shell is given charge (–3Q), then the potential will be

{V}'_{sphere}=\frac{1}{4\pi \varepsilon _{0}}\left [ \frac{Q}{b}+\frac{-3Q}{b} \right ]

\therefore {V}'_{sphere}- {V}'_{shell}=\frac{1}{4\pi \varepsilon _{0}}\left [ \frac{Q}{a}-\frac{Q}{b} \right ]=V

 


Option 1)

V

Option 2)

2V

Option 3)

4V

Option 4)

-2V

Posted by

Avinash

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