Two identical tennis balls each having mass m | vedclass

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(C) Let $T$ be the tension in the thread and $x$ be the separation between the balls.At equilibrium,the forces acting on each ball are: tension $T$,we

Vedclass · Accepted Answer

$x=\left(\frac{q^{2} l}{2 \pi \varepsilon_{0} mg}\right)^{1 / 3}$

Vedclass · Answer

(C) Let $T$ be the tension in the thread and $x$ be the separation between the balls.At equilibrium,the forces acting on each ball are: tension $T$,weight $mg$,and electrostatic force $F_e = \frac{1}{4 \pi \varepsilon_0} \frac{q^2}{x^2}$.Resolving the forces,we have:$T \cos 	heta = mg$ (vertical component)$T \sin 	heta = \frac{1}{4 \pi \varepsilon_0} \frac{q^2}{x^2}$ (horizontal component)Dividing the two equations: $	an 	heta = \frac{q^2}{4 \pi \varepsilon_0 x^2 mg}$.For small angles,$	an 	heta \approx \sin 	heta = \frac{x/2}{l} = \frac{x}{2l}$.Substituting this into the equation: $\frac{x}{2l} = \frac{q^2}{4 \pi \varepsilon_0 x^2 mg}$.Rearranging for $x^3$: $x^3 = \frac{q^2 l}{2 \pi \varepsilon_0 mg}$.Thus,$x = \left(\frac{q^2 l}{2 \pi \varepsilon_0 mg}ight)^{1/3}$.

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?

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