Question

Two small spheres each of mass 10 mg are suspended from a point by threads 0.5 m long. They are equally charged and repel each other to a distance of 0.20 m. The charge on each of the sphere is \(\frac{a}{21} \times 10^{-8} C\). 
The value of 'a' will be ______. [Given g = 10 ms–2]

Answer 1

Given : \(m = 10\,mg = 10^{-5}\,kg,\; l = 0.5\,m,\; r = 0.20\,m,\; g = 10\,m/s^2\)

Let angle made by each thread with vertical be \(\theta\)

\(\begin{array}{l} 2l\sin\theta = 0.20 \Rightarrow \sin\theta = \frac{0.20}{1} = 0.2 \end{array}\)

\(\begin{array}{l} \cos\theta = \sqrt{1-0.2^2} \approx 0.98,\quad \tan\theta \approx \frac{0.2}{0.98} \approx 0.2 \end{array}\)

At equilibrium :

\(\begin{array}{l} T\cos\theta = mg,\quad T\sin\theta = F_e \end{array}\)

\(\begin{array}{l} F_e = mg\tan\theta \end{array}\)

\(\begin{array}{l} F_e = 10^{-5} \times 10 \times 0.2 = 2 \times 10^{-5}\,N \end{array}\)

Using Coulomb’s law :

\(\begin{array}{l} F_e = \frac{kq^2}{r^2} \end{array}\)

\(\begin{array}{l} 2 \times 10^{-5} = \frac{9\times10^9 \cdot q^2}{(0.20)^2} \end{array}\)

\(\begin{array}{l} q^2 = \frac{2\times10^{-5}\times0.04}{9\times10^9} = \frac{8\times10^{-7}}{9\times10^9} = \frac{8}{9}\times10^{-16} \end{array}\)

\(\begin{array}{l} q = \sqrt{\frac{8}{9}}\times10^{-8} \approx 0.94\times10^{-8}\,C \end{array}\)

Given \(q = \frac{a}{21}\times10^{-8}\)

\(\begin{array}{l} \frac{a}{21} = 0.94 \Rightarrow a \approx 20 \end{array}\)

Hence, \(a = \mathbf{20}\).