JEE Main · 2024 · 06 Apr 2024 · Shift 2 · Electrostatics

Two identical conducting spheres

P

and

S

with charge

Q

on each, repel each other with a force 16 N . A third identical uncharged conducting sphere R is successively brought in contact with the two spheres. The new force of repulsion between P and S is :

(A)1 N
(B)12 N
(C)4 N
(D)6 N✓ Correct

Answer: (D)

Solution

Two identical spheres

P, S

each carrying

Q

repel with

F = k \frac{Q ^{2}}{r ^{2}} = 16

N. Touching neutral sphere

R

P

leaves each with

Q /2

. Touching

R

(now

Q /2

) to

S

(still

Q

) leaves each with

\frac{Q /2 + Q}{2} = \frac{3 Q}{4}

. With

P

now at

Q /2

and

S

3 Q /4

F^{'} = k \frac{( Q /2 ) ( 3 Q /4 )}{r ^{2}} = \frac{3}{8} \cdot k \frac{Q ^{2}}{r ^{2}} = \frac{3}{8} \times 16 = 6 N

JEE Main · 2024 · 06 Apr 2024 · Shift 2 · Single correct option

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