Assuming that the spheres have a negligible diameter, determine the spring constant of the springs. Given that

Question

Two spheres are mounted on identical horizontal springs and rest on a frictionless table, as in the drawing. When the spheres are uncharged, the spacing between them is 0.0500 m, and the springs are unstrained. When each sphere has a charge of +1.97 C, the spacing doubles. Assuming that the spheres have a negligible diameter, determine the spring constant of the springs.

Answer 1

♠ old spacing is x1=0.05m, new spacing is x2=2*x1 = 0.10m;

therefore depression of each spring is x3=(x2-x1)/2 = 0.075m;    

♣ electrostatic force on each spring is

F= q1*q2/(4*pi *ε0*x3^2);

♣ mechanical force on each spring is

F= k*x3; and F=F; therefore  

q1*q2/(4*pi *ε0*x3^2) =k*x3, hence

k= q1*q2/(4*pi *ε0*x3^3) =

= 1.97^2 / (4*3.1416 *8.854e-12 *0.075^3) =

= 82.7e12 N/m;