If the walls of a box containing a charge are expanded, the total flux through the box remains the same, as the increased number of field lines captured by the expanded walls is exactly canceled by a decrease in field strength farther from the walls.But what if the box contains no charge?
This allows us to write the last equation in a more compact form. Adopted or used LibreTexts for your course?
Therefore, the perpendicular component of the electric field summed across the entire surface is simply the electric field at a distance Compute the electric flux across a cube of side length If the cube is placed parallel to the field, then the four faces parallel to the field have zero flux. The net electric flux through the cube is the sum of fluxes through the six faces.
Therefore, the scalar product of the electric field with the area vector is zero, giving zero flux.The net flux is \(\Phi_{net} = E_0A - E_0 A + 0 + 0 + 0 + 0 = 0\).The net flux of a uniform electric field through a closed surface is zero.Example \(\PageIndex{3}\): Electric Flux through a Plane, Integral MethodA uniform electric field \(\vec{E}\) of magnitude 10 N/C is directed parallel to the Apply \(\Phi = \int_S \vec{E} \cdot \hat{n} dA\), where the direction and magnitude of the electric field are constant.The angle between the uniform electric field \(\vec{E}\) and the unit normal \(\hat{n}\) to the planar surface is \(30^o\). In that case, the direction of the 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:\[\Phi = \vec{E} \cdot \vec{A} \, (uniform \, \hat{E}, \, flat \, surface).\]Figure \(\PageIndex{5}\) shows the electric field of an oppositely charged, parallel-plate system and an imaginary box between the plates. In electromagnetism, electric flux is the measure of the electric field through a given surface, although an electric field in itself cannot flow. (a) The electric flux through a closed surface due to a charge outside that surface is zero.
We want to hear from you.A macroscopic analogy that might help you imagine this is to put a hula hoop in a flowing river.
However, when you use smaller patches, you need more of them to cover the same surface.
The figure below for questions 2 and 3 shows four Gaussian surfaces surrounding a … If you only integrate over a portion of a closed surface, that means you are treating a subset of it as an open surface.Example \(\PageIndex{1}\): Flux of a Uniform Electric FieldA constant electric field of magnitude \(E_0\) points in the direction of the positive Apply the definition of flux: \(\Phi = \vec{E} \cdot \vec{A} \, (uniform \, \vec{E})\), where the definition of dot product is crucial.The relative directions of the electric field and area can cause the flux through the area to be zero.A constant electric field of magnitude \(E_0\) points in the direction of the positive Apply the definition of flux: \(\Phi = \vec{E} \cdot \vec{A} \, (uniform \, \vec{E})\), noting that a closed surface eliminates the ambiguity in the direction of the area vector.Through the top face of the cube \(\Phi = \vec{E}_0 \cdot \vec{A} = E_0 A\).Through the bottom face of the cube, \(\Phi = \vec{E}_0 \cdot \vec{A} = - E_0 A\), 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. What should the direction of the area vector be? We assume that the unit normal \(\hat{n}\) to the given surface points in the positive From the open surface integral, we find that the net flux through the rectangular surface is\[\begin{align*} \Phi &= \int_S \vec{E} \cdot \hat{n} dA = \int_0^a (cy^2 \hat{k}) \cdot \hat{k}(b \, dy) \\[4pt] &= cb \int_0^a y^2 dy = \frac{1}{3} a^3 bc. The electric flux through the top face (FGHK) is positive, because the electric field and the normal are in the same direction. Excel in math and science. Furthermore, the field is always perpendicular to the surface. Existing user? Figure \(\PageIndex{2b}\) shows a surface \(S_2\) of area \(A_2\) that is inclined at an angle \(\theta\) to the For discussing the flux of a vector field, it is helpful to introduce an area vector \(\vec{A}\). If Now consider a planar surface that is not perpendicular to the field. What are the implications of how you answer the previous question?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 \(\PageIndex{3}\).Since \(\hat{n}\) is a unit normal to a surface, it has two possible directions at every point on that surface (Figure \(\PageIndex{1a}\)).
The factors of The same, it turns out, holds for other surfaces as well. Already have an account?
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