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]

Hint:

In electrostatic equilibrium problems involving suspended charges, the condition $\tan\theta = F_e/mg$ is almost always the starting point. For small angles, the approximation $\tan\theta \approx \sin\theta$ is very useful and often intended for simpler calculations.

Answer 1

Correct Answer: 20

Let's analyze the forces acting on one sphere in equilibrium. The forces are: Tension (T), Weight (mg), and Electrostatic Force ($F_e$).
Let $\theta$ be the angle the thread makes with the vertical. The distance between the spheres is $d = 0.2$ m, and the length of the thread is $L=0.5$ m.
From the geometry, $\sin\theta = \frac{d/2}{L} = \frac{0.1}{0.5} = \frac{1}{5} = 0.2$.
Since $\theta$ is small, we can use the approximation $\tan\theta \approx \sin\theta = 0.2$.
In equilibrium, the horizontal components of forces balance: $T\sin\theta = F_e$.
The vertical components of forces balance: $T\cos\theta = mg$.
Dividing the two equations gives $\tan\theta = \frac{F_e}{mg}$.
The electrostatic force is $F_e = k \frac{q^2}{d^2}$, where $k = 9 \times 10^9$ N·m²/C².
The weight is $mg = (10 \times 10^{-6} \text{ kg}) \times (10 \text{ m/s}^2) = 10^{-4}$ N.
Using the small angle approximation: $F_e = mg \tan\theta \approx mg \sin\theta$.
$F_e \approx 10^{-4} \times 0.2 = 2 \times 10^{-5}$ N.
Now we can find the charge q:
$q^2 = \frac{F_e d^2}{k} = \frac{(2 \times 10^{-5}) \times (0.2)^2}{9 \times 10^9} = \frac{2 \times 10^{-5} \times 0.04}{9 \times 10^9} = \frac{8 \times 10^{-7}}{9 \times 10^9} = \frac{8}{9} \times 10^{-16}$.
$q = \sqrt{\frac{8}{9}} \times 10^{-8} = \frac{2\sqrt{2}}{3} \times 10^{-8}$ C.
$q \approx \frac{2 \times 1.414}{3} \times 10^{-8} \approx 0.9428 \times 10^{-8}$ C.
We are given that $q = \frac{a}{21} \times 10^{-8}$ C.
$\frac{a}{21} = 0.9428 \implies a = 21 \times 0.9428 \approx 19.8$.
Rounding off to the nearest integer, the value of a is 20.