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

$\frac{F}{4}$

$\frac{3 F}{4}$

$\frac{F}{8}$

$\frac{3 F}{8}$

Solution

$\frac{3 F}{8}$

Let the spherical conductors Band C have same charge as q. The electric force between them is

$F = \frac{1}{4 {πε}_{o}} \frac{q^{2}}{r^{2}}$

r - being the distance between them.

When third uncharged conductor A is brought in contact with B, then charge on each conductor.

$q_{A} = q_{B} = \frac{q_{A} + q_{B}}{2} = \frac{0 + q}{2} q_{A} = q_{B} = \frac{q}{2}$

When this conductor A is now brought in contact with C, then charge on each conductor

$q_{A} = q_{C} = \frac{q_{A} + q_{B}}{2} = \frac{(\frac{q}{2}) + q}{2} = \frac{3 q}{4}$

Hence, electric force acting between B and C is

$F' = \frac{1}{4 {πε}_{o}} \frac{q_{B} q_{C}}{r^{2}} = \frac{1}{4 {πε}_{o}} \frac{(\frac{q}{2}) (\frac{3 q}{4})}{r^{2}} F' = \frac{3}{8} [\frac{1}{4 {πε}_{o}} \frac{q^{2}}{r^{2}}] F' = \frac{3 F}{8}$

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Electric Charges And Fields

Some More Questions From Electric Charges and Fields Chapter

a) An electrostatic field line is a continuous curve. That is, a field line cannot have sudden breaks. Why not?
b) Explain why two field lines never cross each other at any point?

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