Question

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 and C 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

• A. $\frac{F}{4}$
  
• B. $\frac{3F}{4}$
  
• C. $\frac{F}{8}$
  
• D. $\frac{3F}{8}$

Answer 1

The correct option is D $\frac{3F}{8}$

Initially $F=k.\frac{Q^2}{r^2}$ (fig. A). Finally when a third spherical conductor comes in contact alternately with B and C then removed, so charges on B and C are Q / 2 and 3Q / 4 respectively (fig. B)

after conduction charge on B:

$Q^{\prime }=\frac{Q}{2}$

after conduction charge on C:

$Q^{\prime }=\frac{3Q}{2}\times \frac{1}{2}=\frac{3Q}{4}$

Now force $F^{\prime }=k.\frac{\left(\frac{Q}{2}\right)\left(\frac{3Q}{4}\right)}{r^2}=\frac{3}{8}F$