Derive expression for speed

Question

Derive a symbolic expression for the speed v at which a particle of charge -Q and mass m must move to hold itself in a circular orbit of radius r centered on a massless particle of charge 4Q that cannot move. What happens to this equation if both particles have the same algebraic sign?

Answer 1

Since electrostatic attraction is the only force that can keep one charge in orbit around the other in this problem, set the coulomb force between the charges equal to the centripetal force, and solve for v.

 

k(-Q)(4Q) / r² = -mv² / r  [the minus sign on the right makes the force attractive toward the point where r = 0]

 

4kQ² / r = mv²

 

v = √(4kQ² / mr) = 2Q√(k / mr)  [Use the absolute value of Q, since we just want a magnitude for v.]

 

If both particles have the same algebraic sign, the electrostatic force would be repulsive, and hence could not act as a centripetal force to keep one particle in orbit around the other.