A particle carrying a charge of 2.0 × 10^ -4 C is positioned directly below the resting bob of a simple pendulum, at a separation of 10 cm . If the mass of the bob is 100 g , determine the charge that must be given to

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Quantrex Academy · Accepted Answer

Correct answer: C. 5.4 × 10^ -9 C

For the string to become loose, the tension in the string must be zero. This happens when the upward electrostatic force balances the downward gravitational force.

F_{e} = mg

\dfrac{1}{4\pi\epsilon_{0}} \dfrac{q_{1} q_{2}}{r^2} = mg

Substituting the given values:

9 \times 10^9 \times \dfrac{2.0 \times 10^{-4} \times q_{2}}{(0.1)^2} = 0.1 \times 9.8

18 \times 10^7 \times q_{2} = 0.98

q_{2} = \dfrac{0.98}{18 \times 10^7} = 5.44 \times 10^{-9} \text{ C}

The required charge is approximately

5.4 \times 10^{-9} \text{ C}

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