A 250-g metal ball with a charge of 0.50 mC has a kinetic energy of 6.0 * 10⁸ J.It is traveling directly at an infinite plane of charge with a charge density of 33 C/m².If it is currently 2.0 m away from the plane of charge,how close will it come to the plane before stopping?

Question

Quizplus · Accepted Answer

First, we can use conservation of energy to find the initial velocity of the ball: 
KE = 1/2mv^2 
6.0 x 10^18 J = 1/2 (0.250 kg) v^2 
v = 6.95 x 10^7 m/s

Next, we can use the equation for electric potential energy to find the closest approach distance: 
U = kqQ/d 
where U is the potential energy, k is Coulomb's constant, q and Q are the charges, and d is the distance between them. 
We want to find d when U = KE, so: 
KE = kqQ/d 
d = kqQ/KE = (9 x 10^9 N m^2/C^2) (0.50 x 10^-3 C) (33 C/m^2) / (6.0 x 10^18 J) 
d = 1.36 m

Therefore, the closest approach distance is 1.36 m, which is answer choice C.

A 250-G Metal Ball with a Charge of 0