Two identical small spheres, each of mass \( 0.1 \) g carrying equal and similar charges, are suspended by two non-conducting threads of equal length. At equilibrium position, the charges are \( 10 \) cm apart as shown in figure. The charge on either ball is approximately
(Take: \( g = 10\,\mathrm{m\,s^{-2}} \))
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( \pm33 \) nC
Each sphere is in equilibrium under three forces: its weight \( mg \) downwards, the Coulomb repulsion \( F \) along the line joining the spheres (horizontal), and the tension \( T \) along the thread.
From the figure each thread makes \( 45^\circ \) with the horizontal, so it also makes \( \theta = 45^\circ \) with the vertical. Resolving the tension:
\( T\cos\theta = mg \)
\( T\sin\theta = F \)
\( \tan\theta = F/mg \)
\( m = 0.1\,\mathrm{g} = 1\times10^{-4}\,\mathrm{kg} \)
\( r = 10\,\mathrm{cm} = 0.1\,\mathrm{m} \)
\( F = mg\tan45^\circ = (1\times10^{-4})(10)(1) = 1\times10^{-3}\,\mathrm{N} \)
\( \displaystyle F = \frac{kq^{2}}{r^{2}} \Rightarrow q = r\sqrt{\frac{F}{k}} \)
\( \displaystyle q = 0.1\sqrt{\frac{1\times10^{-3}}{9\times10^{9}}}\,\mathrm{C} = 0.1\times3.33\times10^{-7}\,\mathrm{C} \)
\( q = 3.33\times10^{-8}\,\mathrm{C} \approx 33\,\mathrm{nC} \)
The charges are only known to be alike, so each is \( +33\,\mathrm{nC} \) or \( -33\,\mathrm{nC} \); hence \( \pm33\,\mathrm{nC} \).