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two spherical conductors B &C having equal radii and carrying equal charges with them repel eachother with a force F when kept at a small distance apart . a third spherical conductor A having same charge as that of B but uncharged is brought in contact with B and then with C and finally removed away from both . what is the new force of repulsion between B & C ?

  1. two spherical conductors B &C having equal radii and carrying equal charges with them repel eachother with a force F when kept at a small distance apart . a third spherical conductor A having same charge as that of B but uncharged is brought in contact with B and then with C and finally removed away from both . what is the new force of repulsion between B & C ?

Grade:12th pass

1 Answers

Arun
25750 Points
3 years ago
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

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