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 \(\frac{a}{21}\times10^{-8}\) C. The value of 'a' will be ______.
[Given g = 10 ms^–2]

Answer 1

Given: 

mass, m₁ = 10 mg

mass, m₂ = 10 mg

distance, d = 0.20 m

The free body diagram is written as;

With the help of a free body diagram we have;

T cosθ = mg = 10 × 10⁻⁶ × 10 = 10⁻⁴    ......(1)

T sinθ = F_e 

Here, F_e = k\(\frac{q^2}{r^2}\)

Tsinθ =  k\(\frac{q^2}{d^2}\)

⇒ Tsinθ =  k\(\frac{q^2}{d^2}\) = F

⇒ Tsinθ =  k\(\frac{9 \times 10^9 \times q^2}{(0.2)^2}\) = F

⇒ Tsinθ = k\(\frac{9 \times 10^9 \times q^2}{(0.04)}\) = F   .....(2)

Now, by dividing the equation (1) by (2) we have;

\(\frac{T_{\sin \theta}}{T_{\cos \theta}} = \frac{0.1}{\sqrt{24}}\)

⇒ tanθ = \(\frac{0.1}{\sqrt{24}} = \frac F{mg}\)   .....(3)

Now, using equation (3) in equation (2) we have;

q = \(\frac{2\sqrt{10}}{3\sqrt{24}} \) × 10⁻⁸

⇒ q = 0.95 × 10⁻⁸

⇒ q = \(\frac{20}{21}\) × 10⁻⁸

Hence, a = 20.

Answer 2

Answer is 20