Application of Gauss's Law to Charge Distribution

In summary, the conversation discusses how to find the charge per unit area on a large, horizontal sheet of plastic that has a uniform charge density on its surface, given a 10.0 g piece of Styrofoam with a net charge of -0.700 \muC suspended in equilibrium above the center of the sheet. The conversation covers using the equations F = Eq and σ = \frac{Q}{A} to find the electric field and charge per unit area on the sheet, and suggests treating the sheet as being infinitely big and choosing a suitable Gaussian surface to simplify the calculations.
  • #1
zachfoltz
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Homework Statement


A 10.0 g piece of Styrofoam carries a net charge of -0.700 [itex]\mu[/itex]C and is suspended in equilibrium above the center of a large, horizontal sheet of plastic that has a uniform charge density on its surface. What is the charge per unit area on the plastic sheet?


Homework Equations



Fe = [itex]\frac{KqQ}{r^2}[/itex] (q is the charge of the styrofoam- given, and Q is the total charge of the plastic sheet. K is Coulumbs Constant = 8.99x10^9)
Fg = mg
σ = [itex]\frac{Q}{A}[/itex] (Q is the total charge of the sheet, A is its area)
[itex]\Phi[/itex] = [itex]\oint[/itex] E[itex]\cdot[/itex]dA = qin0 (E is the electric field. dA is an infinitesimal area VECTOR hence the dot product. qin is the total charge inside a Gausian surface. ε0 is the permittivity of free space which is equal to 1/(K4[itex]\pi[/itex]) which means 1/ε0 = K4[itex]\pi[/itex]


The Attempt at a Solution


Obviously since the Styrofoam is floating at equilibrium Fg = Fe and I set those equal to each other, but since I don't know "r" - the height that the Styrofoam is suspended at, or "A" - the area of the plastic sheet, or "Q" (the total charge of the sheet) I don't know how to solve for σ. I couldn't come up with a Gaussian surface that would make E and A vectors parallel to simplify the surface integral, so I don't know if I can do anything with that equation. Thank you for taking the time to read and hopefully help.
 
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  • #2
zachfoltz said:
Fe = [itex]\frac{KqQ}{r^2}[/itex] (q is the charge of the styrofoam- given, and Q is the total charge of the plastic sheet. K is Coulumbs Constant = 8.99x10^9)
That gives the force between two point charges. Not relevant here.

What's the field from a charged sheet? What field is required to put that Styrofoam piece in equilibrium? (Treat it as a particle.)
 
  • #3
Oh, I didn't know that only applied to particles. The formula for a force from a uniformly distributed charge on a charge q would be: F= q[itex]\oint[/itex] E[itex]\cdot[/itex]dA. Unfortunately I don't know how to solve that integral because I can't think of a Gaussian surface that would make E[itex]\cdot[/itex]da = E*dA (because the vectors are parallel), or a surface on which E is always constant (the same distance away at all points) therefore able to take out of the integral.
 
  • #4
zachfoltz said:
The formula for a force from a uniformly distributed charge on a charge q would be: F= q[itex]\oint[/itex] E[itex]\cdot[/itex]dA.
The force would be F = Eq. The first job is to find the field from a uniform sheet of charge.

Unfortunately I don't know how to solve that integral because I can't think of a Gaussian surface that would make E[itex]\cdot[/itex]da = E*dA (because the vectors are parallel), or a surface on which E is always constant (the same distance away at all points) therefore able to take out of the integral.
Treat the charged sheet as being infinitely big. How would the field lines look? Choose a Gaussian surface that has its surfaces parallel and perpendicular to the charged surface. (Many shapes will do.)

This is a standard exercise, so your textbook may have a discussion worth reviewing.
 
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  • #5
THANK YOU. It just clicked. Much gratitude brother!
 

FAQ: Application of Gauss's Law to Charge Distribution

How does Gauss's Law explain the distribution of electric charge?

Gauss's Law states that the electric flux through a closed surface is equal to the total amount of charge inside that surface divided by the permittivity of free space. This means that the electric field created by a distribution of charge is directly related to the amount of charge contained within a given surface.

What is the significance of Gauss's Law in understanding charge distribution?

Gauss's Law is a fundamental principle in the study of electrostatics and plays a crucial role in understanding the behavior and distribution of electric charges. It allows us to calculate the electric field at any point in space and provides a relationship between electric flux and charge distribution.

How is Gauss's Law applied to different types of charge distributions?

Gauss's Law can be applied to various charge distributions, such as point charges, line charges, and surface charges. The mathematical approach to applying the law may vary depending on the type of charge distribution, but the underlying principle remains the same.

Can Gauss's Law be used to calculate the electric field inside a conductor?

Yes, Gauss's Law can be used to calculate the electric field inside a conductor. Since the electric field inside a conductor is zero, the flux through any closed surface within the conductor must also be zero. This allows us to use Gauss's Law to determine the charge distribution on the surface of the conductor.

What are the limitations of Gauss's Law in analyzing charge distribution?

Gauss's Law is only applicable to static electric fields, and it does not take into account the effects of magnetic fields or time-varying fields. It also assumes that the medium surrounding the charge distribution is homogenous and isotropic, which may not always be the case. Additionally, the calculation of electric field using Gauss's Law may become complex for more complicated charge distributions.

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