Question

Two identical tennis balls each having mass 'm' and charge 'q' are suspended from a fixed point by threads of length 'l'. What is the equilibrium separation when each thread makes a small angle '$\theta$' with the vertical ?

• A. $x = \left(\frac{q^2 l^2}{2\pi\epsilon_0 m^2 g}\right)^{1/3}$
• B. $x = \left(\frac{q^2 l}{2\pi\epsilon_0 mg}\right)^{1/3}$
• C. $x = \left(\frac{q^2 l}{2\pi\epsilon_0 mg}\right)^{1/2}$
• D. $x = \left(\frac{q^2 l^2}{2\pi\epsilon_0 m^2 g^2}\right)^{1/3}$

Hint:

For equilibrium problems involving suspended charges, the condition $\tan\theta = F_e / mg$ is almost always the starting point. Remember to use the small angle approximation ($\tan\theta \approx \sin\theta$) when specified.

Answer 1

The Correct Option is B

Consider one of the tennis balls in equilibrium. The forces acting on it are:
1. Tension (T) along the thread.
2. Gravitational force (mg) acting vertically downwards.
3. Electrostatic repulsive force ($F_e$) acting horizontally.
For equilibrium, the net force is zero. Resolving the tension T into components:
$T \cos\theta = mg$ (vertical equilibrium)
$T \sin\theta = F_e$ (horizontal equilibrium)
Dividing the second equation by the first:
$\frac{T \sin\theta}{T \cos\theta} = \frac{F_e}{mg} \implies \tan\theta = \frac{F_e}{mg}$
The electrostatic force between the two charges separated by distance x is:
$F_e = \frac{1}{4\pi\epsilon_0} \frac{q^2}{x^2}$
From the geometry of the setup, if x is the separation, the distance from the center to one ball is $x/2$.
$\sin\theta = \frac{x/2}{l} = \frac{x}{2l}$
For a small angle $\theta$, we can use the approximation $\tan\theta \approx \sin\theta$.
Therefore, $\tan\theta \approx \frac{x}{2l}$.
Now, equate the two expressions for $\tan\theta$:
$\frac{x}{2l} = \frac{F_e}{mg} = \frac{q^2}{4\pi\epsilon_0 x^2 mg}$
Rearrange the equation to solve for x:
$x^3 = \frac{2l \cdot q^2}{4\pi\epsilon_0 mg} = \frac{q^2 l}{2\pi\epsilon_0 mg}$
Taking the cube root of both sides:
$x = \left(\frac{q^2 l}{2\pi\epsilon_0 mg}\right)^{1/3}$