Two small balls,each having mass m and charge | vedclass

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Question

(A) At equilibrium,the forces acting on each ball are tension $T$,electrostatic force $F_e$,and weight $mg$.Resolving forces: $F_e = T \sin 	heta$ $(

Vedclass · Accepted Answer

${\left( {\frac{{{q^2}L}}{{2\pi {\varepsilon _0}mg}}} \right)^{\frac{1}{3}}}$

Vedclass · Answer

(A) At equilibrium,the forces acting on each ball are tension $T$,electrostatic force $F_e$,and weight $mg$.Resolving forces: $F_e = T \sin 	heta$ $(i)$ and $mg = T \cos 	heta$ (ii).Dividing $(i)$ by (ii),we get $	an 	heta = \frac{F_e}{mg} = \frac{q^2}{4\pi \varepsilon_0 x^2 mg}$.Given $	an 	heta \approx \sin 	heta = \frac{x/2}{L} = \frac{x}{2L}$.Equating the two expressions for $	an 	heta$: $\frac{x}{2L} = \frac{q^2}{4\pi \varepsilon_0 x^2 mg}$.Rearranging for $x$: $x^3 = \frac{2q^2 L}{4\pi \varepsilon_0 mg} = \frac{q^2 L}{2\pi \varepsilon_0 mg}$.Therefore,$x = {\left( {\frac{{{q^2}L}}{{2\pi {\varepsilon _0}mg}}} ight)^{1/3}}$.

Two small balls,each having mass $m$ and charge $q$,are suspended from a common point by insulating strings of length $L$ as shown in the figure. Assuming $\theta$ is very small such that $\tan \theta \approx \sin \theta$,find the value of $x$ at equilibrium.

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