JEE Main · 2004 · 24 Apr 2004 · Offline · Electrostatics

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Two spherical conductor 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 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

(A) $F /4$
(B) $3 F /4$
(C) $F /8$
(D) $3 F /8$ .✓ Correct

Answer: (D)

Solution

To begin with, B and C each carry charge

q

and repel with

F = \frac{1}{4 π ε _{0}} \frac{q ^{2}}{d ^{2}}

. Touching an identical uncharged conductor to B shares that charge equally and leaves B with

\frac{q}{2}

. Bringing the same conductor, now carrying

\frac{q}{2}

, into contact with C makes the total

q + \frac{q}{2} = \frac{3 q}{2}

split equally, so C is left with

\frac{3 q}{4}

. The new force is

\frac{1}{4 π ε _{0}} \frac{( q /2 ) ( 3 q /4 )}{d ^{2}} = \frac{3}{8} F

JEE Main · 2004 · 24 Apr 2004 · Offline · Single correct option

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JEE Main · 2004 · 24 Apr 2004 · Offline · Electrostatics

Solution

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