Guassian infinite sheet and charge with kinetic energy

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Homework Help Overview

The problem involves two infinite sheets of charge, one positive and one negative, separated by a distance of 2 meters. A charge of 3 microcoulombs is released from the positive plate and reaches the negative plate with a kinetic energy of 0.6 joules. The objective is to determine the electric field between the plates.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the assumption of constant charge density and electric field between the sheets. Questions arise regarding the relationship between energy, charge, and electric field, particularly how to connect kinetic energy and electric potential.

Discussion Status

Some participants have provided guidance on focusing on relevant equations and questioning the appropriateness of certain formulas in this context. There is an ongoing exploration of how voltage relates to the electric field, with participants expressing varying levels of understanding.

Contextual Notes

There is mention of confusion regarding the application of specific equations and the physical meaning of the variables involved. Participants are encouraged to clarify their assumptions and the context of the equations they are using.

scholio
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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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