Question

Two metal spheres of identical mass m = 4.40 g are suspended by light strings 0.500 m in length. The left-hand sphere carries a charge of 0.745 µC, and the right-hand sphere carries a charge of 1.47 µC. What is the equilibrium separation between the centers of the two spheres?

Answer 1

let 2*d is the equilibrium sepration.

let T is the tension in the strings.

let theta is the angle made by the strings with vertical axis.

Fnety = 0

T*cos(theta) - m*g = 0

T*cos(theta) = m*g ----(1)

Fnetx = 0

T*sin(theta) - Fe = 0

T*sin(theta) = Fe ---(2)

from equations 1 and 2

tan(theta) = Fe/(m*g)

m*g*tan(theta) = Fe

d/sqrt(d^2 + L^2) = k*q1*q2/(2*d)^2

4.4*10^-3*9.8*d/sqrt(0.5^2 - d^2) = 9*10^9*1.47*10^-6*0.745*10^-6/(2*d)^2

==> d = 0.286 m

so, equilibrium sepration = 2*d

= 2*0.286

= 0.572 m