Subjective Type

A particle having a charge of $$2.0\times 10^{-4}$$C is placed directly below and at a separation of $$10$$cm from the bob of a simple pendulum at rest. The mass of the bob is $$100$$g. What charge should the bob be given so that the string becomes loose?

Solution

Mass of the bob$$=100$$g $$=0.1$$kg
So Tension in the string $$=0.1\times 9.8=0.98$$N.
For the Tension to be $$0$$, the charge below should repet the first bob.
$$\Rightarrow F=\dfrac{kq_1q_2}{r^2}$$

At equilibrium, the forces acting on the particle is:
$$T-mg+F=0$$

When spring becomes loose, the tension will be zero.
$$\Rightarrow 0=mg-F$$
$$F=mg$$

$$\Rightarrow 0.98=\dfrac{9\times 10^9\times 2\times 10^{-4}\times q_2}{(0.01)^2}$$

$$\Rightarrow q_2=\dfrac{0.98\times 1\times 10^{-4}}{9\times 2\times 10^5}=0.054\times 10^{-9}N$$.

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