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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 ( $Qq = 4 {πε}_{0} {mgl}^{2}$ ) is $20^{0} \times n$ . Find $n$ .

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answer is 3.

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Detailed Solution

Let the angle made by the string be $θ$ ,

The distance between the charges is given by $x = 2 l \sin (\frac{θ}{2})$

Since the body is in equilibrium,

$\frac{F}{\sin (180 - θ)} = \frac{mg}{\sin (90 + \frac{θ}{2})}$

$\Rightarrow \frac{Qq}{4 {πε}_{0}} \times \frac{1}{4 l^{2} \sin^{2} (\frac{θ}{2})} \times \frac{1}{sinθ} = \frac{mg}{\cos (\frac{θ}{2})}$

$θ = 2 \sin^{- 1} [{(\frac{Qq}{32 {πε}_{o} mg l^{2}})}^{\frac{1}{3}}]$
$\Rightarrow θ = 2 \sin^{- 1} (\frac{1}{2}) \Rightarrow θ = 2 (\frac{π}{6})$
$\Rightarrow θ = 60^{\circ}$

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