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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.

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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 =================================== 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.