# Video: Finding the Electric Flux through a Surface given a Charge Distribution

The cross sections of the closed surfaces (a), (b), (c), and (d), are shown in the accompanying diagrams. Find the electric flux through the closed surface (a). Find the electric Flux through the closed surface (b). Find the electric flux through the closed surface (c). Find the electric flux through the closed surface (d).

04:56

### Video Transcript

The cross sections of the closed surfaces (a), (b), (c), and (d) are shown in the accompanying diagrams. Find the electric flux through the closed surface (a). Find the electric Flux through the closed surface (b). Find the electric flux through the closed surface (c). Find the electric flux through the closed surface (d).

If we represent electric flux in general with a Greek symbol capital π· with an π¬ subscript, then we want to solve for this value in the situations of (a), (b), (c), and (d). Weβll call the values of electric flux for each of these cases π· sub π¬ (a) π· sub π¬ (b) π· sub π¬ (c) and π· sub π¬ (d) respectively.

As we get started solving for π· sub π¬ (a), we can recall Gaussβs law. This law tells us that given a closed surface like we have in each of these four examples, if we integrate over the total area of that surface, multiplying each area element by the electric field that moves through that area element, then that surface integral will be equal to the total charge π enclosed by the surface divided by π naught, the permittivity of free space.

π naught is a constant value. And weβll treat it as exactly 8.85 times 10 to the negative 12 farads per meter. Now if we look back at the left-hand side of Gaussβs law, this is an expression where weβre multiplying an area element times the electric field. This can remind us of the relationship for electric flux. π· sub π¬, the electric flux, is equal to electric field multiplied by the area through which that field line passes.

Because weβre considering, in each of our four cases, a closed surface and weβre integrating over the entire thing, we can combine Gaussβs law with a relationship defining electric flux to write that the electric flux in any of these four scenarios is equal to the charge enclosed by the volume over π naught. Knowing that, letβs move ahead to solve for π· sub π¬ (a) considering scenario (a).

π· sub π¬ (a) equals the charge enclosed by our shape, weβll call it π sub π, over π naught. And as we look at situation (a), we see the amount of charge enclosed. Itβs 3.0 times 10 to the negative eighth coulombs. We see there is a charge outside our shape. But since itβs on the outside, we donβt include it in our calculation for π· sub π¬ (a). So we plug in our value for π naught and calculate this fraction. And we find that, to two significant figures, π· sub π¬ (a) is 3.4 times 10 to the third newtons-meter squared per coulomb. Thatβs the electric flux through the closed surface in (a).

Next, we move to scenario (b) to calculate π· sub π¬ (b). And as we look at the shape in scenario (b), we see that it encloses no charge. That means that π sub π, the name we can give to the charge enclosed by the shape in scenario (b), is equal to zero. And therefore, π· sub π¬ (b) is equal to zero as well. Now on to scenario (c) where this time our shape does enclose the charge. In this case π sub π, the name we can give to the charge enclosed by the shape in scenario (c), is equal to negative 2.0 times 10 to the negative sixth coulombs.

When we insert our value for π naught and calculate this fraction, we find that, to two significant figures, π· sub π¬ (c) is negative 2.3 times 10 to the fifth newtons-meter squared per coulomb. Thatβs the electric Flux through the surface in scenario (c). And finally, we turn to scenario (d) where our volume, in this case, encloses part of the conductor which extends infinitely far up and down. If π sub π is the charge enclosed in our volume in this case, then as we look at scenario (d), that charge π will equal the product of the surface charge density π and the area of the end cap of our volume.

In other words, π· sub π¬ (d) is equal to π, the surface charge density in the conductor, multiplied by the area, we can call it π΄, of the end cap. Weβre told the value for π as well as the value for the area of the end cap. And when we plug in those values along with our values for π naught and calculate this fraction, we find that, to two significant figures, it equals 90 newtons-meter squared per coulomb. Thatβs the electric flux through the volume in scenario (d).