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  • Solve! Two spherical conductors B and C having equal radii and carrying equal charges in them repel each other with a force F when kept apart at some distance .A third spherical conductor having same radius as that of B but uncharged is brought in

Two spherical conductors B  and C   having equal radii and carrying equal charges in them repel each other with a force  F   when kept apart at some distance .A  third spherical conductor having same radius as that of  B   but uncharged is brought in contact with  B   then brought in contact with   C  and  finally removed away from both , The new force of repulsion between  B  and  C   is  :

  • Option 1)

    F/4

  • Option 2)

    3F/4

  • Option 3)

    F/8

  • Option 4)

    3F/8

 

Answers (2)

best_answer

As we learnt in

Magnitude of the Resultant force -

F_{net}=\sqrt{F_{1}^{2}+F_{2}^{2}+2F_{1}F_{2}\cos \Theta }

- wherein

 

 

 

  Initially,\: \: F= \frac{1}{4\pi \varepsilon _{0}}\frac{q^{2}}{d^{2}}\cdots \cdots \cdots (i)

when the third equal conductor touches B,the charge of  B is shared equally between them

\therefore \: \: \: charge \: \: on \: \: B= \frac{q}{2}= charge on third conductor Now this third conductor with charge \left ( \frac{q}{2} \right ) touches  C  their total charge \left ( q+\frac{q}{2} \right ) is equally shared between them

\therefore \: \: \: \: charge\: on\:

C= \frac{3q}{4}= Charge of third conductor

\therefore            New force between  B and C 

= \frac{1}{4\pi \varepsilon _{0}d^{2}}\left ( \frac{q}{2} \times \frac{3q}{4}\right )= \frac{3}{8}F

 


Option 1)

F/4

Incorrect

Option 2)

3F/4

Incorrect

Option 3)

F/8

Incorrect

Option 4)

3F/8

Correct

Posted by

Aadil

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