Electromagnetism

Question

Two metal plates are squares of 12 cm by 12 cm, separated by 1.0 mm of air. Wires connected to each plate are also connected to a 9.0 V battery. After a long time, the battery is disconnected and the two plates are arranged to be perpendicular instead of parallel and almost touching each other along one edge (an “L” shape). A proton (charge +1.6 x 10-19 C, mass 1.67 x 10-27 kg) is released from rest from the middle of the positive plate. When and where does the proton hit the negative plate?

Answer 1

I get where the problem is coming from: each plate will put forward a constant electric field, which will accelerate the plate differently in each dimension: horizontally it will accelerate to the right, and vertically it will accelerate downward. (Ignore gravitational forces.)

First, find the charge on each plate using Q=CV and C=(epsilon-0)A/d (I'd treat the dielectric constant of air as 1).

Second, calculate the surface charge density of the plate by dividing the charge by the area in m^2.

Third, find the electric field of each plate using E=(surface charge density)/(2 x epsilon-0)

Fourth, calculate the acceleration for each plate using E=F/q and a=F/m.

Fifth, use kinematics to find the time that the proton strikes the horizontal plate (s=0.5at^2 should do the trick), keeping in mind that the proton is 6 cm above the horizontal plate.

Sixth, use kinematics again to find the horizontal location where the proton strikes the plate (the same equation should work).

When I did it, I got an unsatisfactory answer of 6 cm, which made total sense when I thought about it: the fields are identical, so the proton could only get 6 cm forward if it only took the time to accelerate down at the exact same rate. It actually made me feel good to do the work. I highly recommend it.