guassian infinite sheet and charge with kinetic energy

scholio

1. The problem statement, all variables and given/known data

two infinite sheets of charge (one positive, one negative) are arranged parallel to each other and separated by a distance of 2m. a charge of 3 microcoulombs is realeased at the positive plate. when it reaches the negative plate it has a kinetic energy of 0.6 joules. what is the electric field between the plates

2. Relevant equations

electric field of charge sheet E = sigma/2(epsilon_o) where sigma = charge density, epsilon_o = constant 8.85*10^-12

energy U = 1/2 CV^2 where C is capacitance, v is electric potential

C = Q/V where Q is charge, V is electric potential


3. The attempt at a solution

do i assume sigma as constant?
do i assume both sheets have the same electric field, so electric field between the sheets is the sum of the two?

how does the energy play a role in determining the electric field, the equation for electric field is already specified, and can be solved in terms of constants, assuming sigma is constant too.

if i use the energy and capacitance equation i get:
U = 1/2[(Q/V)(V^2)]
0.6 = QV/2

i am not sure how Q, V relate to electric field, i'm lost actually...

help appreciated

Dick

Yes, yes. Constant sigma. Constant E field. Now start throwing out equations you don't need. Don't stop too soon. You only need one.

scholio

still somewhat lost, i did find in actuality that using :

U = 1/2[(Q/V)(V^2)]
0.6 = QV/2

V = 400000 volts when Q = +3*10^-6 coulombs

but how does V relate to electric field E?

Dick

You are CONSTANTLY using WRONG equations. Take U = 1/2[(Q/V)(V^2)]. What kind of a situation does that apply to? What's U, is it kinetic energy? What's Q? What sort of charge is it? Can it be the 3*10^(-6) coulombs? If you have a formula involving U,Q and V, that doesn't mean you can use it in EVERY problem that has the letters U,Q and V in it. Don't only remember equations, remember what sort of physical setup they apply to.