A system of two charged particles A and B having charges +q and -q and mass m and 2m respectively, are connected with an ideal massless inextensible string which lies along the axis of a uniform charged fixed non-conducting ring as shown. If gravity is neglected and ignore the force between A and B, find the tension in the string at this instant.

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k q Q 22 R 2

k q Q 32 R 2

k q Q 62 R 2

answer is D.

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Detailed Solution

As they are connected by a string, both the particles move with same acceleration.

So, $\frac{T + q E}{2 m} = \frac{q E - T}{m} \Rightarrow T + q E = 2 q E - 2 T \Rightarrow T = \frac{q E}{3}$
$T = \frac{qkQR}{3 {(R^{2} + R^{2})}^{\frac{3}{2}}} = \frac{kQqR}{3 {(2 R^{2})}^{\frac{3}{2}}}$

$\Rightarrow T = \frac{kQqR}{3 [2 \sqrt{2}] R^{3}} = \frac{kQq}{6 \sqrt{2} R^{2}}$

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