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Check out this video to observe what happens to the flux as the area changes in size and angle, or the electric field changes in strength.
For discussing the flux of a vector field, it is helpful to introduce an area vector This allows us to write the last equation in a more compact form. What should the magnitude of the area vector be? What should the direction of the area vector be? What are the implications of how you answer the previous question?
The area vector of a flat surface of area A has the following magnitude and direction:
Since the normal to a flat surface can point in either direction from the surface, the direction of the area vector of an open surface needs to be chosen, as shown in [link] .
Since is a unit normal to a surface, it has two possible directions at every point on that surface ( [link] (a)). For an open surface, we can use either direction, as long as we are consistent over the entire surface. Part (c) of the figure shows several cases.
However, if a surface is closed, then the surface encloses a volume. In that case, the direction of the normal vector at any point on the surface points from the inside to the outside. On a closed surface such as that of [link] (b), is chosen to be the outward normal at every point, to be consistent with the sign convention for electric charge.
Now that we have defined the area vector of a surface, we can define the electric flux of a uniform electric field through a flat area as the scalar product of the electric field and the area vector, as defined in Products of Vectors :
[link] shows the electric field of an oppositely charged, parallel-plate system and an imaginary box between the plates. The electric field between the plates is uniform and points from the positive plate toward the negative plate. A calculation of the flux of this field through various faces of the box shows that the net flux through the box is zero. Why does the flux cancel out here?
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