Question

Help Please!!!

Answer 1

I need help solving the attached question.

Answer 2

Can anyone help please?

Answer 3

The electric field inside a capacitor is uniform, and is zero outside. Assuming that the dimensions of those plates are such that we can treat them as ideal capacitors, you can find the change in kinetic energy by multiplying the charge of the particle by the electric potential through which it passes. For example, once it passes all the way through the first capacitor, it will be moving with a velocity equal to:
\[ \frac{1}{2} mv^2 = qV \rightarrow v = \sqrt{\frac{2qV}{m}}\]
where m is the mass of the test particle, V is the potential difference of the first plate (V = 300 V) and q is the charge of the particle.

Then it will coast at the same speed (because there's no electric field) until it enters the other capacitor. It will then slow down until it passes through an electric potential of 300 V (which, since the field is uniform, will be 3/5 of the way through), stop, and go back the other way.

Answer 4

Does that mean that on the return trip the particle will stop at "x"?

Answer 5

Yes indeed it does.

Answer 6

@Jemurray3 Brilliant answer !!

Answer 7

Thank you very much for your help @Jemurray3!!!:D

Answer 8

Thank you, I'm glad I could help!