This work is licensed by OpenStax University Physics under a [ "article:topic", "flux", "authorname:openstax", "area vector", "electric flux", "license:ccby", "showtoc:no" ][ "article:topic", "flux", "authorname:openstax", "area vector", "electric flux", "license:ccby", "showtoc:no" ]Flux of a Uniform Electric Field through a Closed Surface The figure below for questions 2 and 3 shows four Gaussian surfaces surrounding a … What should the magnitude of the area vector be? 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 \(30^o\). As you change the angle of the hoop relative to the direction of the current, more or less of the flow will go through the hoop.
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. Again, flux is a general concept; we can also use it to describe the amount of sunlight hitting a solar panel or the amount of energy a telescope receives from a distant star, for example.To quantify this idea, Figure \(\PageIndex{1a}\) shows a planar surface \(S_1\) of area \(A_1\) that is perpendicular to the uniform electric field \(\vec{E} = E\hat{y}\). In electromagnetism, electric flux is the measure of the electric field through a given surface, although an electric field in itself cannot flow.
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. In the limit of infinitesimally small patches, they may be considered to have area \[\Phi = \int_S \vec{E} \cdot \hat{n}dA = \int_S \vec{E} \cdot d\vec{A} \, (open \, surface).\]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 \(\PageIndex{2}\) and the flux through a closed surface (one that completely bounds some volume), we represent flux through a closed surface by\[\Phi = \oint_S \vec{E} \cdot \hat{n} dA = \oint_S \vec{E} \cdot d\vec{A} \, (closed \, surface)\]where the circle through the integral symbol simply means that the surface is closed, and we are integrating over the entire thing. 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. The net electric flux through the cube is the sum of fluxes through the six faces. Electric flux becomes zero, when electric field is parallel to the surface area ΔS, as the angle becomes 90 ° and value of cos 90 ° is zero The direction of ΔS is determined by area vector, as it … Why does the flux cancel out here?The reason is that the sources of the electric field are outside the box.
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. 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 \(\vec{A}\).
The factors of The same, it turns out, holds for other surfaces as well. Since both the direction and magnitude are constant, \[\Phi = \int_S \vec{E} \cdot \hat{n} dA = EA \, cos \, \theta\]\[= (10 \, N/C)(6.0 \, m^2)(cos \, 30^o) = 52 \, N \cdot m^2/C.\]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.What angle should there be between the electric field and the surface shown in Figure \(\PageIndex{9}\) in the previous example so that no electric flux passes through the surface?Place it so that its unit normal is perpendicular to \(\vec{E}\).Example \(\PageIndex{4}\) : Inhomogeneous Electric FieldWhat is the total flux of the electric field \(\vec{E} = cy^2\hat{k}\) through the rectangular surface shown in Figure \(\PageIndex{10}\)?Apply \(\Phi = \int_S \vec{E} \cdot \hat{n}dA\).
The electric flux through the top face (FGHK) is positive, because the electric field and the normal are in the same direction.