In a cathode-ray tube (CRT), an electron travels in a vacuum and enters a region between two "deflection" plates which have equal and opposite charges. The dimensions of each plate are L = 13 cm by d = 3 cm, and the gap between them is h = 2.5 mm (not to scale in the diagram).


During a 0.001 s interval while it is between the plates, the change of the momentum of the electron Δ
is < 0, 6.40e-17, 0 > kg m/s.

What is the electric field between the plates?
Hint: remember the Momentum Principle (the relationship between Impulse and change in momentum.)



What is the charge (both magnitude and sign) of the upper plate?

My attempt:

ΔP = FΔT

F = ΔP/T = <0,6.4E-17,0> / <0.001> = <0,6.4E-14,o>

F = EQ

E = [(Q/A)(S/R)]/2ε

F = [(Q/A)(S/R)]/2ε * Q

<0,6.4E-14,0> = [Q^2 / (.13*.3)](0.0025m/R)] / 2ε

I don't know what "R" is supposed to be. I know that:
Q = charge
A = Area
S = Distance of the 2 plates from each other.
ε = 8.85E-12 F/M

But I don't know what "R" is. Once I find "Q", I should be able to find E simply by using F=QE. I don't think I'm even using the right equation for this.

Can someone show me how to do this one, too? Thanks.

Never mind. I got it (or more or less someone else showed me).

F = EQ

E = F/Q

F = deltaP / t

E = (1/q)(deltaP/t) = (1/- 1.6x10^-19)(6.4x10^-17)/.001 = <0, -4x10^5, 0> N/C

E = D/e(o) = Q/Ae(o) = Q/Lde(o)

Q = ELde(o) = (4x10^5)(.13)(.03)(8.85x10^-12) = 1.38 x 10^-8 C