Guassian infinite sheet and charge with kinetic energy

In summary, the problem involves two infinite sheets of charge separated by 2m, a charge of 3 microcoulombs released at the positive plate, and a kinetic energy of 0.6 joules at the negative plate. The electric field between the plates can be calculated using the equation E = sigma/2(epsilon_o) where sigma is the charge density and epsilon_o is the constant 8.85*10^-12. The energy and capacitance equations are not applicable in this problem. The correct equation to use is E = sigma/2(epsilon_o).
  • #1
scholio
160
0

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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  • #2
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.
 
  • #3
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?
 
  • #4
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.
 

1. What is a Gaussian infinite sheet?

A Gaussian infinite sheet is a hypothetical surface with an infinite extent in two dimensions and a constant charge density. This means that the charge is uniformly distributed across the surface and does not vary with position.

2. What is the charge on a Gaussian infinite sheet?

The charge on a Gaussian infinite sheet is represented by the symbol σ (sigma) and is measured in units of coulombs per square meter. It is equal to the product of the charge density (ρ) and the area of the sheet (A), so σ = ρA.

3. How does a Gaussian infinite sheet affect electric fields?

A Gaussian infinite sheet creates an electric field that is perpendicular to the surface and uniform in magnitude. The strength of the electric field is directly proportional to the charge density, so a higher charge density will result in a stronger electric field.

4. What is the kinetic energy associated with a charge on a Gaussian infinite sheet?

The kinetic energy associated with a charge on a Gaussian infinite sheet is the energy that the charge possesses due to its motion. This energy is given by the equation KE = 1/2mv^2, where m is the mass of the charge and v is its velocity.

5. How is the kinetic energy of a charge on a Gaussian infinite sheet related to the electric field?

The kinetic energy of a charge on a Gaussian infinite sheet is related to the electric field through the work-energy theorem. This theorem states that the work done by a force (in this case, the electric field) is equal to the change in kinetic energy of the object. So, as the electric field does work on the charge, its kinetic energy will change accordingly.

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