Question

A particle with a charge of 2.0 × 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?

Answer 1

Given:
Mass of the bob, m = 100 g = 0.1 kg
So, tension in the string, T = mg
⇒ T = 0.1 × 9.8 = 0.98 N
For the tension to be zero, the repelling force (F_e) on the bob = T
Magnitude of the charge placed below the bob, q = 2.0 × 10⁻⁴ C
Separation between the charges, r = 0.1 m
When the electrostatic force between the bob and the particle is balanced by the tension in the string then the string will become loose.
Let the required charge on the bob be q' .
$\Rightarrow F_e=\frac{1}{4\mathrm{\pi }{\epsilon }_0}\frac{qq\text{'}}{r^2}=T$
$$\begin{gathered} \Rightarrow \frac{9\times {10}^9\times q\text{'}\times 2\times {10}^{-4}}{{10}^{-2}}=0.98 \\ \Rightarrow q\text{'}=5.4\times {10}^{-9}\mathrm{C} \end{gathered}$$