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Magnitude at I or II:
If the charge on the plane is positive, then the direction of the electric field and the area vectors are as shown in [link] . Therefore, we find for the flux of electric field through the box
where the zeros are for the flux through the other sides of the box. Note that if the charge on the plane is negative, the directions of electric field and area vectors for planes I and II are opposite to each other, and we get a negative sign for the flux. According to Gauss’s law, the flux must equal . From [link] , we see that the charges inside the volume enclosed by the Gaussian box reside on an area A of the xy -plane. Hence,
Using the equations for the flux and enclosed charge in Gauss’s law, we can immediately determine the electric field at a point at height z from a uniformly charged plane in the xy -plane:
The direction of the field depends on the sign of the charge on the plane and the side of the plane where the field point P is located. Note that above the plane, , while below the plane, .
You may be surprised to note that the electric field does not actually depend on the distance from the plane; this is an effect of the assumption that the plane is infinite. In practical terms, the result given above is still a useful approximation for finite planes near the center.
Would Gauss’s law be helpful for determining the electric field of two equal but opposite charges a fixed distance apart?
yes, using superposition
Discuss the role that symmetry plays in the application of Gauss’s law. Give examples of continuous charge distributions in which Gauss’s law is useful and not useful in determining the electric field.
Discuss the restrictions on the Gaussian surface used to discuss planar symmetry. For example, is its length important? Does the cross-section have to be square? Must the end faces be on opposite sides of the sheet?
Any shape of the Gaussian surface can be used. The only restriction is that the Gaussian integral must be calculable; therefore, a box or a cylinder are the most convenient geometrical shapes for the Gaussian surface.
Recall that in the example of a uniform charged sphere, Rewrite the answers in terms of the total charge Q on the sphere.
Suppose that the charge density of the spherical charge distribution shown in [link] is for and zero for Obtain expressions for the electric field both inside and outside the distribution.
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