Question

Two spheres are located 0.60 m from one another. Sphere A with a charge of + 2 x 10^–4 C, is fixed in position, but sphere B with a charge of +4 x 10^–5 C is free to move. Spheres A and B each have a mass of 10 g. How far is sphere B (assume vi=0) when it reaches a speed of 80 m/s from sphere A? Please show and explain as much work as possible. Thank you!

Answer 1

Using energy conservation:

KEi + EPEi = KEf + EPEf

KEi = Initial kinetic energy of both spheres = 0 J, since intially both particle were at rest

EPEi = initial electric potential energy of system = k*q1*q2/ri

EPEf = final electric potential energy of system = k*q1*q2/rf

m1 = mass of sphere A, m2 = mass of sphere B

KEf = final KE of system = (1/2)*m2*V^2 (Since sphere A is in fixed position)

m2 = m1 = 10 gm = 0.01 kg

q1 = charge on sphere A = 2*10^-4 C

q2 = charge on sphere B = 4*10^-5 C

ri = Initial distance between both spheres = 0.60 m

rf = final distance between both spheres = ?

V = final speed of sphere B = 80 m/s

So Using given values:

0 + k*q1*q2/ri = (1/2)*m2*V^2 + k*q1*q2/rf

1/rf = 1/ri - m2*V^2/(2*k*q1*q2)

1/rf = 1/0.60 - 0.01*80^2/(2*9*10^9*2*10^-4*4*10^-5)

1/rf = 1.222

rf = 1/1.222 = 0.8183 m

In two significant figures:

So final distance between both spheres = 0.82 m

Let me know if you've any query.