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 of the either ball from vertical when each thread makes a small angle '$\theta$' with the vertical?

• A. x = $\left(\frac{q^2l^2}{2\pi\epsilon_0m^2g}\right)^{1/3}$
• B. x = $\frac{1}{2}\left(\frac{q^2l}{2\pi\epsilon_0mg}\right)^{1/2}$
• C. x = $\frac{1}{2}\left(\frac{q^2l}{2\pi\epsilon_0mg}\right)^{1/3}$
• D. x = $\left(\frac{q^2l^2}{2\pi\epsilon_0m^2g^2}\right)^{1/3}$

Answer 1

Answer: (C)

x = $\frac{1}{2}\left(\frac{q^2l}{2\pi\epsilon_0mg}\right)^{1/3}

Answer 2

Let the mass of each ball be $m$ and the charge be $q$. The length of the thread is $l$. Let the equilibrium separation between the two balls be $x$. When the thread makes a small angle $\theta$ with the vertical, the horizontal distance of each ball from the vertical line passing through the suspension point is $l \sin\theta$. Thus, the separation between the two balls is $x = 2l \sin\theta$.

Each ball is in equilibrium under the action of three forces:

1. Tension $T$ in the thread.
2. Gravitational force $mg$ acting vertically downwards.
3. Electrostatic force $F_e$ due to the other ball, acting horizontally away from the other ball.

Resolving the tension into vertical and horizontal components, we have:

Vertical equilibrium: $T \cos\theta = mg$

Horizontal equilibrium: $T \sin\theta = F_e$

Dividing the second equation by the first, we get:

$\tan\theta = \frac{F_e}{mg}$

The electrostatic force between the two charges is given by Coulomb's law:

$F_e = \frac{1}{4\pi\epsilon_0} \frac{q \cdot q}{x^2} = \frac{q^2}{4\pi\epsilon_0 x^2}$

Substituting this into the equilibrium equation:

$\tan\theta = \frac{q^2}{4\pi\epsilon_0 x^2 mg}$

We are given that the angle $\theta$ is small. For small angles, $\tan\theta \approx \sin\theta$.

Also, from the geometry, the horizontal distance of each ball from the vertical is $x/2$. So, $\sin\theta = \frac{x/2}{l} = \frac{x}{2l}$.

Using the small angle approximation $\tan\theta \approx \sin\theta$:

$\frac{x}{2l} \approx \frac{q^2}{4\pi\epsilon_0 x^2 mg}$

Now, we solve for $x$:

$x \cdot x^2 = \frac{q^2}{4\pi\epsilon_0 mg} \cdot 2l$

$x^3 = \frac{2q^2 l}{4\pi\epsilon_0 mg}$

$x^3 = \frac{q^2 l}{2\pi\epsilon_0 mg}$

Taking the cube root of both sides:

$x = \left(\frac{q^2 l}{2\pi\epsilon_0 mg}\right)^{1/3}$

The question asks for the equilibrium separation of either ball from vertical, which is $x/2$.

Therefore, $x/2 = \frac{1}{2}\left(\frac{q^2l}{2\pi\epsilon_0mg}\right)^{1/3}$