Two small spheres each having the charge +Q a | vedclass

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(A) In the absence of gravity,the only force acting on the spheres is the electrostatic repulsive force between them. Since both spheres have the same

Vedclass · Accepted Answer

$180^\circ, \frac{1}{4\pi \varepsilon_0} \frac{Q^2}{(2L)^2}$

Vedclass · Answer

(A) In the absence of gravity,the only force acting on the spheres is the electrostatic repulsive force between them. Since both spheres have the same charge $+Q$,they repel each other. To maximize the distance between them and reach an equilibrium position,the threads will stretch out in opposite directions,forming a straight line. Thus,the angle between the two threads is $180^\circ$. The distance between the two charges is $r = L + L = 2L$. According to Coulomb's law,the electrostatic force $F$ between them is $F = \frac{1}{4\pi \varepsilon_0} \frac{Q \cdot Q}{(2L)^2} = \frac{1}{4\pi \varepsilon_0} \frac{Q^2}{(2L)^2}$. This force is equal to the tension $T$ in each thread.

Two small spheres each having the charge $+Q$ are suspended by insulating threads of length $L$ from a hook. This arrangement is taken into space where there is no gravitational effect. Then the angle between the two suspensions and the tension in each will be:

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