Two identical metallic spheres A and B carry charges $+ Q$ and $- 2 Q$ respectively. The force between them is 'F' newton, when they are separated by a distance 'd' in air. The spheres are allowed to touch each other and are moved back to their initial position. The force between them now is:

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Toppr Admin · Accepted Answer

Intial force Fi-k-Q-2Q-d2Fi-2k-xA0-Q2d2Now when spheres are touched- charge distribution changes to make potential equal on both spheres-x2234-Qf-Q2 -as both the spheres are identical-

Two identical metallic spheres A and B carry charges +Q and −2Q respectively. The force between them is 'F' newton, when they are separated by a distance 'd' in air. The spheres are allowed to touch each other and are moved back to their initial position. The force between them now is:2FF8F48F

intial force Fi=k(Q)((2Q)d2Fi=2k Q2d2Now when spheres are touched, charge distribution changes to make potential equal on both spheres∴Qf=Q2 (as both the spheres are identical)

intial force $F_{i} = \frac{k (Q) ((2 Q)}{d^{2}}$
$F_{i} = \frac{2 k Q^{2}}{d^{2}}$
Now when spheres are touched, charge distribution changes to make potential equal on both spheres
$∴ Q_{f} = \frac{Q}{2}$ (as both the spheres are identical)