Physics-Conservation of energy and charge

Question

A small ball has a mass of 5.0 x 10-8 kg and a charge of 0.0 x 10-5 C. It enters a box with an initial speed of 1.2 x 104 m/s. The point where the charge leaves the box is at a voltage 350 kV higher than the entry point. [Hint: Use energy conservation]

Answer 1

What you'd want to do is 1) calculate the kinetic energy at the entry point, 2) calculate the change in kinetic energy due to the voltage difference, 3) calculate the resulting kinetic energy at the exit point, and 4) calculate the resulting speed at the exit point.

 

First, calculate the kinetic energy of the ball when it enters the box:

 

K.E = 1/2 mv² = 1/2 (5.0 x 10^-8 kg) (1.2 x 10⁴ m/s)² = 0.5 * 5.0 x 10^-8 * 1.44 x 10⁸) J = 3.6 J

 

In order to determine the change in kinetic energy, you would multiply the charge on the ball times the voltage difference, with a positive charge being repulsed by the higher voltage at the exit point (lose kinetic energy) and a negative charge being attracted to the exit point (gain kinetic energy).

 

However, you say that the charge on the ball is 0.0 x 10^-5 C, which is 0!  That means that there is no force between the ball and the exit point, so the kinetic energy at the exit point will be the same as it is at the entry point.  That means that the speed will not change.

 

Are you sure you stated the charge correctly???