Two small equally charged identical conducting balls are suspended from long threads from the same point. The charges and masses of the balls are such that they are in equilibrium. The distance between them is a = 
cm (the length of the threads L >> a) .One of the ball is discharged. After sometime both balls comes to rest in equilibrium. What will be the distance b (in cm) between the balls when equilibrium is restored?
Text Solution
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(3)
Sol. As the charge on one of the balls is removed so electrostatic force between the balls is zero.
The balls will first go down and due to contact with each other, charge on one ball is equally distributed on both balls and then the balls get separated due to electrostatic repulsion.
At equilibrium:
Here F = Force (electrostatic) between two balls = 
By force balance, T sin θ = F and T cos θ = mg
⇒ mg tan θ = F ........ [for small angle tan θ ≈ sin θ = 
]
⇒ mg tan θ = 
⇒ or kq 2 = 
Now the balls are discharged and charge on each ball = 
. & the distance between two ball = b.
By equation , mg tan θ = F 1 = 
& mg 
= 
By putting value of kq 2 , 
= 
⇒ b = 
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