How does flux change with distance

The flux through is therefore Designating as a unit vector normal to see Figure b , we obtain. 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 Figure.

Since is a unit normal to a surface, it has two possible directions at every point on that surface Figure 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 Figure 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 :.

Figure 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? The reason is that the sources of the electric field are outside the box. Therefore, if any electric field line enters the volume of the box, it must also exit somewhere on the surface because there is no charge inside for the lines to land on.

Therefore, quite generally, electric flux through a closed surface is zero if there are no sources of electric field, whether positive or negative charges, inside the enclosed volume. Any smooth, non-flat surface can be replaced by a collection of tiny, approximately flat surfaces, as shown in Figure. If we divide a surface S into small patches, then we notice that, as the patches become smaller, they can be approximated by flat surfaces.

This is similar to the way we treat the surface of Earth as locally flat, even though we know that globally, it is approximately spherical. To keep track of the patches, we can number them from 1 through N. Now, we define the area vector for each patch as the area of the patch pointed in the direction of the normal.

Let us denote the area vector for the i th patch by We have used the symbol to remind us that the area is of an arbitrarily small patch. With sufficiently small patches, we may approximate the electric field over any given patch as uniform. Let us denote the average electric field at the location of the i th patch by. Therefore, we can write the electric flux through the area of the i th patch as. The flux through each of the individual patches can be constructed in this manner and then added to give us an estimate of the net flux through the entire surface S , which we denote simply as.

This estimate of the flux gets better as we decrease the size of the patches. However, when you use smaller patches, you need more of them to cover the same surface. In the limit of infinitesimally small patches, they may be considered to have area dA and unit normal.

Since the elements are infinitesimal, they may be assumed to be planar, and may be taken as constant over any element. Then the flux through an area dA is given by It is positive when the angle between and is less than and negative when the angle is greater than.

The net flux is the sum of the infinitesimal flux elements over the entire surface. With infinitesimally small patches, you need infinitely many patches, and the limit of the sum becomes a surface integral. With representing the integral over S ,. In practical terms, surface integrals are computed by taking the antiderivatives of both dimensions defining the area, with the edges of the surface in question being the bounds of the integral.

To distinguish between the flux through an open surface like that of Figure and the flux through a closed surface one that completely bounds some volume , we represent flux through a closed surface by. If you only integrate over a portion of a closed surface, that means you are treating a subset of it as an open surface. Flux of a Uniform Electric Field A constant electric field of magnitude points in the direction of the positive z -axis Figure.

What is the electric flux through a rectangle with sides a and b in the a xy -plane and in the b xz -plane? Strategy Apply the definition of flux: , where the definition of dot product is crucial.

Significance The relative directions of the electric field and area can cause the flux through the area to be zero. Flux of a Uniform Electric Field through a Closed Surface A constant electric field of magnitude points in the direction of the positive z -axis Figure. What is the net electric flux through a cube? Strategy Apply the definition of flux: , noting that a closed surface eliminates the ambiguity in the direction of the area vector.

Solution Through the top face of the cube,. Through the bottom face of the cube, because the area vector here points downward. Along the other four sides, the direction of the area vector is perpendicular to the direction of the electric field. Therefore, the scalar product of the electric field with the area vector is zero, giving zero flux. The net flux is. Significance The net flux of a uniform electric field through a closed surface is zero.

What is the electric flux through the plane surface of area located in the xz -plane? Assume that points in the positive y -direction. Strategy Apply , where the direction and magnitude of the electric field are constant. Solution The angle between the uniform electric field and the unit normal to the planar surface is. Since both the direction and magnitude are constant, E comes outside the integral. All that is left is a surface integral over dA , which is A.

Therefore, using the open-surface equation, we find that the electric flux through the surface is. Significance Again, the relative directions of the field and the area matter, and the general equation with the integral will simplify to the simple dot product of area and electric field.

Check Your Understanding What angle should there be between the electric field and the surface shown in Figure in the previous example so that no electric flux passes through the surface?

Place it so that its unit normal is perpendicular to. Inhomogeneous Electric Field What is the total flux of the electric field through the rectangular surface shown in Figure? Strategy Apply. We assume that the unit normal to the given surface points in the positive z -direction, so Since the electric field is not uniform over the surface, it is necessary to divide the surface into infinitesimal strips along which is essentially constant.

As shown in Figure , these strips are parallel to the x -axis, and each strip has an area. Solution From the open surface integral, we find that the net flux through the rectangular surface is. Significance For a non-constant electric field, the integral method is required. Check Your Understanding If the electric field in Figure is what is the flux through the rectangular area? Conceptual Questions Discuss how to orient a planar surface of area A in a uniform electric field of magnitude to obtain a the maximum flux and b the minimum flux through the area.

We can get around this by comparing the luminosities of two objects, either two different objects, or the same object before or after some great change in temperature, radius, or both:. Often we will use the Sun as object 2, comparing its luminosity to that of another star. But even luminosity is not the true measure of how bright an object appears.

A flashlight at 2 feet is blinding, but a searchlight at 10 miles is not. How can this be if the searchlight has a higher luminosity? As light leaves a source, it spreads out in a spherical pattern.

As the photons get farther away from the light source, they spread out and become less and less concentrated there are only so many photons to go around, after all. Thus the light source appears dimmer the farther away it is. This is expressed mathematically using the inverse-square relation:.

Remember that R is the actual radius of the light source, and D is the distance of the light source. The units for R and D don't really matter, as long as they are the same both R and D in km, for example. We can get around using Sigma here the same way we did above, by comparing the brightnesses of two objects. Apparent brightness decreases as the square of the distance.