JEE Main2018PhysicsElectrostaticsActual
Two identical conducting spheres A and B carry an equal charges. They are separated by a distance much larger than their diameters, and the force between them is F . A third identical conducting sphere, C , is uncharged. Sphere C is first touched to A , then to B , and then removed. As a result, the force between A and B would be equal to:
Options
- A3 F 4
- BF 2
- C3 F 8
- DF
Correct answer
C. 3 F 8
Step-by-step solution
F = k q 2 r 2 when A and C are touched charge on both will be q 2 . Then when B and C are touched, q B =   q 2 + q 2 = 3 q 4 F ′ = k q A q B r 2 =   k   × q 2   × 3 q 4 r 2 = 3 8 k q 2 r 2 = 3 8 F .