Guassian infinite sheet and charge with kinetic energy

AI Thread Summary
Two infinite sheets of charge, one positive and one negative, are separated by 2 meters, with a 3 microcoulomb charge released from the positive plate, gaining 0.6 joules of kinetic energy upon reaching the negative plate. The electric field between the plates can be determined using the relationship between electric potential (V) and electric field (E), where V = E * d. The user initially struggles with the correct application of equations related to energy, charge, and capacitance, leading to confusion about how they interrelate. It is emphasized that only one relevant equation should be used to simplify the problem, focusing on the physical context rather than memorizing formulas. Understanding the specific situation is crucial for correctly applying the equations to find the electric field.
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Homework Statement



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

Homework 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


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
 
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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.
 
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?
 
scholio said:
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?

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.
 
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