Electric Flux | Brilliant Math & Science Wiki

A flux through a given surface can be “inward” or “outward” depending on which way counts as “in” or “out”—that is, flux has a definite orientation.

For a closed surface (a surface with no holes), the orientation of the surface is generally defined such that flux flowing from inside to outside counts as positive, outward flux, while flux from the outside to the inside counts as negative, inward flux. To remember this choice of orientation, we divide the closed surface into many small patches of surface and assign a vector \( \mathbf{a}_i \) to each small patch of surface that indicates the normal (perpendicular) to the surface. In addition, the magnitude of each \( \mathbf{a}_i \) is defined to be the area of the corresponding patch.

If the surface is partitioned into patches that are sufficiently small, then the electric field \( \mathbf{E}_i \) at all points on each patch essentially becomes constant. In that case, the electric flux \( \Phi_\text{patch} \) through the patch is given by the dot product, which calculates the component of \( \mathbf{E}_i \) parallel to \( \mathbf{a}_i \):

\[ \Phi_\text{patch} = \mathbf{E}_i \cdot \mathbf{a}_i \]

The total electric flux through the entire surface, meanwhile, is the sum over all patches:

\[ \Phi = \sum_i \mathbf{E}_i \cdot \mathbf{a}_i.\]

As the \( \mathbf{a}_i \) become vanishingly small, as in the case of a continuous surface, the sum is replaced with a surface integral:

\[ \Phi = \int_S \mathbf{E} \cdot d\mathbf{a}. \]

The “\( S \)” in the limits of integration indicates that the integral is to be taken across the entire surface for all infinitesimal surface elements \( d \mathbf{a} \). Fortunately, the electric flux can often be computed without resorting to computing the integral explicitly.

Compute the electric flux across a spherical surface of radius \( R \) that contains a charge \( q \) at its center.

In this case, the electric field is the same at all points on the surface. Furthermore, the field is always perpendicular to the surface. Therefore, the perpendicular component of the electric field summed across the entire surface is simply the electric field at a distance \( R \) from the charge multiplied by the area of the surface. Thus

\[ \Phi = E \cdot 4 \pi R^2 = \frac{1}{4\pi\epsilon_0} \frac{q}{R^2} \cdot 4 \pi R^2 = \frac{q}{\epsilon_0}. \]

Compute the electric flux across a cube of side length \( s \) placed (a) perpendicular to and (b) \( 45^\circ \) with respect to a uniform electric field of magnitude \( E \) as shown.

If the cube is placed parallel to the field, then the four faces parallel to the field have zero flux. Of the two faces with nonzero flux, one face contains flux \( -E s^2 \) (“inward” flux) while the other contains flux \( E s^2 \) (“outward” flux). Therefore, the total flux is zero.

If the cube is placed \( 45^\circ \) with respect to the field, then two faces each contain flux \( -E s^2 \cos{45^\circ} = -E s^2 / \sqrt{2} \). Similarly, the other two faces with nonzero flux contain \( E s^2 \cos{45^\circ} = E s^2 / \sqrt{2} \). Therefore, the total flux is again zero.