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Gauss's law: plane of charge

  • #1
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Homework Statement


Find the electric field near a large nonconducting plane of charge density sigma.


Homework Equations




The Attempt at a Solution



I am working through chapter 24 of Giancoli and am having trouble understanding the concepts. This is example 24-4, and the magnitude of the electric field is found to be
sigma / (2*epsilon_0). But shouldn't the total electric field be zero?

The Gaussian surface would be a cylinder with tops parallel to the plane. The field points normally outward from the tops, but in opposite directions. Wouldn't this mean the total field is 0?
 

Answers and Replies

  • #2
Doc Al
Mentor
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The Gaussian surface would be a cylinder with tops parallel to the plane. The field points normally outward from the tops, but in opposite directions. Wouldn't this mean the total field is 0?
No. When adding up the flux through each segment of the Gaussian surface, think of the surface area as pointing outward. So at both ends of this Gaussian cylinder the field points in the direction of the surface area vector, thus they add and do not cancel.
 
  • #3
104
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But don't the surface normals point in opposite directions? Suppose the plane of charge is in the xy-plane. Then the gaussian surface would have its top surface normal pointing in the +z direction and the bottom surface normal pointing in the -z direction.

Mathematically, I can see why the magnitude of the field is non-zero:

[tex]
\begin{align*}
\oint \mathbf{E} \cdot \mathbf{dA} &= +E\hat{z} \cdot +A\hat{z} + (-E\hat{z}) \cdot (-A\hat{z}) \\
&= 2EA = \frac{\sigma A}{\epsilon_0} \, .
\end{align*}
[/tex]

But if the field points in the direction of the surface normal at each end of the cylinder, and each end of the cylinder is the same distance from the plane, why wouldn't they cancel?

[tex]
E\hat{z} \text{ (top) } + (-E \hat{z}) \text{ (bottom) } = 0 \, .
[/tex]
 
  • #4
Doc Al
Mentor
44,867
1,114
But don't the surface normals point in opposite directions? Suppose the plane of charge is in the xy-plane. Then the gaussian surface would have its top surface normal pointing in the +z direction and the bottom surface normal pointing in the -z direction.
Exactly right. And since the field at those surfaces points in the same direction as the surface normal, each surface makes a positive contribution to the total flux.

Mathematically, I can see why the magnitude of the field is non-zero:

[tex]
\begin{align*}
\oint \mathbf{E} \cdot \mathbf{dA} &= +E\hat{z} \cdot +A\hat{z} + (-E\hat{z}) \cdot (-A\hat{z}) \\
&= 2EA = \frac{\sigma A}{\epsilon_0} \, .
\end{align*}
[/tex]
Exactly.

But if the field points in the direction of the surface normal at each end of the cylinder, and each end of the cylinder is the same distance from the plane, why wouldn't they cancel?

[tex]
E\hat{z} \text{ (top) } + (-E \hat{z}) \text{ (bottom) } = 0 \, .
[/tex]
Why would the field at the top surface cancel the field at the bottom surface? They are at different locations! (You cannot add the fields at different locations.)

The field at the bottom surface is already the total field at the bottom; similarly for the top surface.
 
  • #5
104
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Why would the field at the top surface cancel the field at the bottom surface? They are at different locations! (You cannot add the fields at different locations.)
Yikes! That's what I was trying to do. Now it makes a lot more sense. Thanks!
 

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