Q.

A particle A having a charge q is fixed on a vertical (insulated) wall. A second particle B of mass m, charge Q is suspended by a silk thread of length l from a point P on the wall that is at a distance l above the A. The angle in degree made by the thread with the vertical, when B stays in equilibrium ( Qq=4πε0mgl2) is 200×n. Find n.

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answer is 3.

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Detailed Solution

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Let the angle made by the string be θ,

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The distance between the charges is given by x=2lsinθ2

Since the body is in equilibrium,

Fsin180-θ=mgsin90+θ2

Qq4πε0×14l2sin2θ2×1sinθ=mgcosθ2

θ=2sin1Qq32πεomgl213
⇒  θ=2sin112  θ=2π6
⇒  θ=60

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A particle A having a charge q is fixed on a vertical (insulated) wall. A second particle B of mass m, charge Q is suspended by a silk thread of length l from a point P on the wall that is at a distance l above the A. The angle in degree made by the thread with the vertical, when B stays in equilibrium ( Qq=4πε0mgl2) is 200×n. Find n.