Electrostatics Question 216
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 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 [AIEEE 2004]
Options:
A) $ F/4 $
B) $ 3F/4 $
C) $ F/8 $
D) $ 3F/8 $
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Answer:
Correct Answer: D
Solution:
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) Now force $ F’=k.\frac{( \frac{Q}{2} )( \frac{3Q}{4} )}{r^{2}}=\frac{3}{8}F $
BETA
