Question
In the figure below, two tiny conducting balls of identical massm and identical charge q hang from nonconductingthreads of length L. If one of them is discharged (losesits charge q to, say, the ground) find(a) the new equilibrium separation x,using the given values of L and m and(b) the computed value of |q|. Assumethat is so small that tan can bereplaced by its approximate equal, sin . Usepermittivity constant 0 and g(free-fall acceleration).
Hint:If one loses its charge, isthere then a force keeping them apart? If they touch, how muchcharge does each get? Can you then recalculate the separation usingthis new charge?
Explanation / Answer
(a)the gravitational force acting between the two tinyconducting balls when they are in equilibrium is Fg = m * g * sin g = 9.8 m/s2 the electric force acting between the two balls is Fe = k * (q * q/x12) k = (1/4o) = 9 * 109Nm2/C2 the gravitaional and electric forces balance eachothertherefore we get Fg = Fe or m * g * sin = k * (q * q/x12) =k * (q2/x12) or x1 = (k * q2/m * g *sin)1/2 = q * (k/m * g *sin)1/2 (b)the garvitational force between the two tiny balls is Fg = G * (m * m/x2) ---------(1) the electric force between the two tiny balls is Fe = k * (q *q/x2) ---------(2) from equations (1) and (2) we get G * (m * m/x2) = k * (q *q/x2) or q = (G/k)1/2 * m G = 6.67 * 10-11Nm2/kg2 or m * g * sin = k * (q * q/x12) =k * (q2/x12) or x1 = (k * q2/m * g *sin)1/2 = q * (k/m * g *sin)1/2 (b)the garvitational force between the two tiny balls is Fg = G * (m * m/x2) ---------(1) the electric force between the two tiny balls is Fe = k * (q *q/x2) ---------(2) from equations (1) and (2) we get G * (m * m/x2) = k * (q *q/x2) or q = (G/k)1/2 * m G = 6.67 * 10-11Nm2/kg2