Two pitch balls each weighing \( 19.6\times10^{-4}\,\mathrm{kg} \) are suspended from the same point 'O' by two silk threads each 1 m long. Balls hold identical charge and repel each other making angle \( 90^\circ \) at point 'O' between threads. The magnitude of charge is :
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( 2.06\times10^{-6}\,\mathrm{C} \)
Each ball is in equilibrium under three forces: its weight \( mg \) downwards, the tension \( T \) along the thread and the Coulomb repulsion \( F \) along the line joining the balls (horizontal).
The angle between the threads is \( 90^\circ \), so each thread makes \( 45^\circ \) with the vertical.
\( T\cos45^\circ = mg \)
\( T\sin45^\circ = F \)
Dividing the two equations:
\( F = mg\tan45^\circ = mg \)
\( \displaystyle mg = 19.6\times10^{-4} \times 9.8 = 1.92\times10^{-2}\,\mathrm{N} \)
The separation of the balls (from the figure) is
\( \displaystyle r = 2 \times 1 \times \sin45^\circ = \sqrt{2}\,\mathrm{m} \)
Using Coulomb's law with \( 1/4\pi\varepsilon_0 = 9\times10^{9}\,\mathrm{N\,m^2\,C^{-2}} \):
\( \displaystyle F = \frac{1}{4\pi\varepsilon_0}\,\frac{q^{2}}{r^{2}} \;\Rightarrow\; q^{2} = \frac{F\,r^{2}}{9\times10^{9}} \)
\( \displaystyle q^{2} = \frac{1.92\times10^{-2} \times 2}{9\times10^{9}} = 4.27\times10^{-12}\,\mathrm{C^2} \)
\( \displaystyle q = \sqrt{4.27\times10^{-12}} \approx 2.06\times10^{-6}\,\mathrm{C} \)