© All Rights reserved @ LearnWithDash
Step-by-Step Solution
Step 1: Identify the forces
Each tennis ball of mass $m$ carries charge $q$ and is suspended by a thread of length $l$. In equilibrium, each ball is at rest. The forces acting on each ball are:
Weight ($mg$): Acting vertically downward.
Coulomb’s repulsive force ($F_e$): Acting horizontally between the two charged balls, given by $F_e = \dfrac{k q^2}{x^2}$, where $k = \dfrac{1}{4 \pi \varepsilon_0}$ and $x$ is the separation between them.
Tension ($T$): Acting along the thread.
Step 2: Set up force balance equations
Resolve the tension $T$ into vertical and horizontal components:
Vertical component: $T \cos \theta = mg$
Horizontal component: $T \sin \theta = \dfrac{k q^2}{x^2}$
Hence, we get:
$T = \dfrac{mg}{\cos \theta}$
and
$\dfrac{mg}{\cos \theta} \sin \theta = \dfrac{k q^2}{x^2}$
Step 3: Relate angle $\theta$ to the separation $x$
For small angles $\theta$, we can use the approximation $\sin \theta \approx \tan \theta \approx \theta$ and the geometry: the horizontal displacement is roughly $\dfrac{x}{2}$ (each ball is separated by $x$, so from the centerline to one ball is $x/2$). Thus:
$\sin \theta \approx \dfrac{x}{2l}.$
From the force balance in the horizontal direction:
$\tan \theta = \dfrac{\text{horizontal force}}{\text{vertical force}}
= \dfrac{k q^2 / x^2}{mg} \approx \dfrac{k q^2}{mg \, x^2}.
$
Since $\tan \theta \approx \sin \theta \approx \dfrac{x}{2l}$, we get:
$\dfrac{x}{2l} = \dfrac{k q^2}{mg \, x^2}.
$
Step 4: Solve for $x$
Rearrange the above equation:
$x^3 = 2l \, \dfrac{k q^2}{mg} = 2l \, \dfrac{1}{4 \pi \varepsilon_0} \, \dfrac{q^2}{mg}.
$
Hence,
$x = \left( 2l \, \dfrac{1}{4 \pi \varepsilon_0} \, \dfrac{q^2}{mg} \right)^{1/3}.
$
Noting that $k = \dfrac{1}{4 \pi \varepsilon_0}$, we can rewrite the separation $x$ more neatly as:
$x = \left(\dfrac{q^2 \, l}{2 \pi \varepsilon_0 \, m g}\right)^{1/3}.
Therefore, the equilibrium separation between the two charged tennis balls is
$x = \left(\dfrac{q^2 \, l}{2 \pi \varepsilon_0 \, m g}\right)^{1/3}.
Final Answer: $x = \left(\dfrac{q^2 l}{2\pi \varepsilon_0 mg}\right)^{\tfrac{1}{3}}$
