Question

two identical small metallic sphere each of mass m, each carry charge q are suspended from a common point by silk thread each of length l in equilibrium separation between the balls is x and x<<l. now each sphere starts licking its charge at the rate dq/dt such that the two balls approach towards each other the relative velocity of approach is given by v=a/√ x  , where a is constant.find the value of dq/dt.

Answer 1

Suppose at t=0, the spheres are in equilibrium and at rest. Forces are balanced along vertical and horizontal.

T cos α = mg            T sin α = q²/(4π∈ x²)

tan α = α = sin α =  x/2l  = q²/(4π∈ mg x²)            as x << l

q² = (2π∈ mg/l) x³

q = √(2π∈mg/l)  x^3/2

assume that this equation is valid, when the sphere are moving.  Differentiate

dq/dt = √(2π∈mg/l) * 3/2 * √x * dx/dt
         = - 3a √(π∈mg/2l)                        as dx/dt = v = -  a/√x
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If more accurate expression is to be derived :

You have to do by find the acceleration and equating net force = mass * acceleration.  Then the value of q and dq/dt are nearly same expressions
compared to the above.

let c = 3a √(π∈mg/2l)  and b = a²l/2g

q = c √[x³ - b]

dq/dt  =  - c √[x³/(x³-b)]           
                        if b ≈ 0, then both methods give same answer.