In most cases, the source of flux will be described as a vector field: Given a point x,y,z , there's a formula giving the flux vector at that point. From our intuition, it should look something like this:. However, this formula only works if the vector field is the same at every point. Make sense so far? Now, we need to figure out how much orientation actually matters.
Like we said before, if the field and the surface are parallel, then there is zero flux. If they are perpendicular, there is full flux. In this diagram, the flux is parallel with the top surface, and nothing enters from that direction. Mathematically, we represent surfaces by their normal vector, which sticks out of the surface. How much, exactly? Well, this is a job for the dot product , which is the projection of the field onto the surface.
The dot product gives us a number from 0 to 1 that tells us what percent of the field is passing through the surface. So, the equation becomes:. Good question. For a surface like a plane, the normal vector is the same in every direction.
More complicated shapes may have a normal vector that varies quite a bit. In this case, try to break the shape into smaller regions like spheres, cylinders and planes and find the flux in each part. Then, add up the flux in each region to get the total flux keeping in mind positive and negative flux. If the shape is more complicated than that, you may need a computer model or more advanced theorems; but at least you know what is happening behind the scenes.
Imagine a tube, that lets water pass right through it. We hold the tube under a waterfall, wait a few seconds, then ask what the flux is. I want a numeric answer — what is the flux? You might think we need to know the speed of the waterfall, the size of the tube, the orientation, etc. Remember our convention for flux orientation: positive means flux is leaving, negative means flux is entering.
In this example, water is falling downward, or entering the tube. This means the top surface has negative flux it appears to be siphoning up water. The water passed through the top and is now leaving the bottom, which is positive flux:.
Ah, this beautiful diagram shows what is going on. Assuming the same amount of water is leaving and entering the rate of water falling is a constant , the net flux would be zero. My possibly incorrect answer: If we increased the rate, it means more water would enter than leaves, for a brief moment. Even though net flux is zero, this is different from having zero flux pass through each surface.
If you are in an empty field, no shape will generate any flux. But if you are in a field where flux is canceling, changing your shape or orientation could create a non-zero flux. Recognize the difference between having zero flux because the field is zero, vs. Measuring flux is about drawing imaginary boundaries, not having a physical shape.
Flux is what is passing through the sides of a bucket at a moment in time. And one more point. Ab aap Whatsapp pe solutions paa saktey h, hum aapko message karenge. Ab aap Whatsapp pe solutions paa saktey h, hum aapko ping karenge. Study Materials. Why use Doubtnut? Instant Video Solutions.
Request OTP. Updated On: Share This Video Whatsapp. Text Solution. 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.
If the planar surface is perpendicular to the electric field vector, the maximum flux would be obtained. If the planar surface were parallel to the electric field vector, the minimum flux would be obtained. What are the maximum and minimum values of the flux in the preceding question? The net electric flux crossing a closed surface is always zero. True or false? The net electric flux crossing a closed surface is always zero if and only if the net charge enclosed is zero.
The net electric flux crossing an open surface is never zero. A uniform electric field of magnitude is perpendicular to a square sheet with sides 2.
What is the electric flux through the sheet? Calculate the flux through the sheet of the previous problem if the plane of the sheet is at an angle of to the field. Find the flux for both directions of the unit normal to the sheet.
Note that this angle can also be given as. The electric flux through a square-shaped area of side 5 cm near a large charged sheet is found to be when the area is parallel to the plate. Find the charge density on the sheet. Two large rectangular aluminum plates of area face each other with a separation of 3 mm between them. The plates are charged with equal amount of opposite charges,.
The charges on the plates face each other. Find the flux through a circle of radius 3 cm between the plates when the normal to the circle makes an angle of with a line perpendicular to the plates.
A square surface of area is in a space of uniform electric field of magnitude. The amount of flux through it depends on how the square is oriented relative to the direction of the electric field. Find the electric flux through the square, when the normal to it makes the following angles with electric field: a , b , and c.
Note that these angles can also be given as. A vector field is pointed along the z -axis, a Find the flux of the vector field through a rectangle in the xy -plane between and.
Leave your answer as an integral. Consider the uniform electric field What is its electric flux through a circular area of radius 2. Repeat the previous problem, given that the circular area is a in the yz -plane and b above the xy- plane. An infinite charged wire with charge per unit length lies along the central axis of a cylindrical surface of radius r and length l.
What is the flux through the surface due to the electric field of the charged wire? Learning Objectives By the end of this section, you will be able to: Define the concept of flux Describe electric flux Calculate electric flux for a given situation. The numerical value of the electric flux depends on the magnitudes of the electric field and the area, as well as the relative orientation of the area with respect to the direction of the electric field.
N field lines cross surface. The same number of field lines cross each surface. Area Vector 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.
The area vector of a flat surface of area A has the following magnitude and direction: Magnitude is equal to area A Direction is along the normal to the surface ; that is, perpendicular to the surface. The direction of the area vector of an open surface needs to be chosen; it could be either of the two cases displayed here.
The area vector of a part of a closed surface is defined to point from the inside of the closed space to the outside. This rule gives a unique direction. Electric Flux 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 :.
Electric flux through a cube, placed between two charged plates. Electric flux through the bottom face ABCD is negative, because is in the opposite direction to the normal to the surface. The electric flux through the top face FGHK is positive, because the electric field and the normal are in the same direction.
The electric flux through the other faces is zero, since the electric field is perpendicular to the normal vectors of those faces. The net electric flux through the cube is the sum of fluxes through the six faces.
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