Two identical pith balls are charged by rubbing one against the other. They are suspended from a horizontal rod through two strings of length 20 cm each, the separation between the suspension points being 5 cm. In equilibrium, the separation between the balls is 3 cm. Find the mass of each ball and the tension in the strings. The charge on each ball has a magnitude 2.0 × 10^{−8} C.

#### Solution

Let the tension in the string be T and the force of attraction between the two balls be F.

From the free-body diagram of the balls, we get

Tcosθ = mg ...(1)

Tsinθ = F ...(2)

From ∆ABC,

\[\sin\theta = \frac{1}{20}\]

\[\cos\theta = \sqrt{1 - \left( \frac{1}{20} \right)^2}\]

From equation (1) and (2),

\[\tan\theta = \frac{F}{\text{mg}}\]

\[ \therefore \text{m} = \frac{F}{g\tan\theta}\]

\[ \Rightarrow m = \left( \frac{1}{4\pi \epsilon_0}\frac{q^2}{r^2} \right) \times \left( \frac{1}{g\tan\theta} \right)\]

\[ \Rightarrow m = \frac{9 \times {10}^9 \times \left( 2 \times {10}^{- 8} \right)^2}{9 \times {10}^{- 4} \times 0 . 98 \times \tan\theta}\]

\[\Rightarrow m = 0 . 0082 \text{ kg = 8 . 2 g}\]

\[T = \frac{F}{\sin\theta} = \frac{9 \times {10}^9 \times \left( 2 \times {10}^{- 8} \right)^2}{9 \times {10}^{- 4} \times \frac{1}{20}}\]

\[ \Rightarrow T = 8 . 2 \times {10}^2 \]N