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  • Solve it, A solid conducting sphere having a charge Q is surrounded by an uncharged concentric conducting hollow spherical shell. Let the potential differencebetween the surface of the solid sphere and that of the outer surface of the hollow shell be

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 two surfaces is

  • Option 1)

    V

  • Option 2)

    2V

  • Option 3)

    4V

  • Option 4)

    -2V

 

Answers (3)

best_answer

As we learned 

 

Potential Due to 3 Concentric Spheres -

The figure shows three conducting concentric shell of radii a, b and c (a < b < c) having charges Qa, Qb and Qc respectively

- wherein

 

Potential at A;

V_{A}= \frac{1}{4\pi \varepsilon _{0}}\left [ \frac{Q_{a}}{a}+\frac{Q_{b}}{b} +\frac{Q_{c}}{c}\right ]

Potential at B;

V_{B}= \frac{1}{4\pi \varepsilon _{0}}\left [ \frac{Q_{a}}{b}+\frac{Q_{b}}{b} +\frac{Q_{c}}{c}\right ]

Potential at C;

V_{C}= \frac{1}{4\pi \varepsilon _{0}}\left [ \frac{Q_{a}}{c}+\frac{Q_{b}}{c} +\frac{Q_{c}}{c}\right ]

 

 

If a and b are radii of spheres and spherical shell respectively, potential at their surfaces 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}

and so according to the given problem.

V=V_{Sphere}- V_{Shell}=\frac{Q}{4\pi \epsilon _{0}}\left [ \frac{1}{a}-\frac{1}{b} \right ]\; \; \; \; \; \cdots (i)

Now when the shell is given a charge –3Q the potential at its surface and also inside will change by 
V_{0}=\frac{1}{4\pi \varepsilon _{0}}\left [ -\frac{3Q}{b} \right ]

So that nowV_{Sphere}=\frac{1}{4\pi \varepsilon _{0}}\left [ \frac{Q}{a}-\frac{3Q}{b} \right ]  and   V_{Shell}=\frac{1}{4\pi \varepsilon _{0}}\left [ \frac{Q}{b}-\frac{3Q}{b} \right ] hence 

V_{Sphere}-V_{Shell}=\frac{Q}{4\pi \varepsilon _{0}}\left [ \frac{1}{a}-\frac{1}{b} \right ]=V
 

 

 


Option 1)

V

Option 2)

2V

Option 3)

4V

Option 4)

-2V

Posted by

SudhirSol

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2V

Posted by

Tanya Arora

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