Video: Calculating the Electric Flux through a Closed Surface due to a Point Charge

A point charge of 12 𝜇C is located at the center of a cube. If there are no other charges present, what is the electric flux through one face of the cube?

02:44

Video Transcript

A point charge of 12 microcoulombs is located at the center of a cube. If there are no other charges present, what is the electric flux through one face of the cube?

We can call the value of our point charge, 12 microcoulombs, 𝑄. We want to solve for the electric flux through one face of the cube, what we’ll call 𝛷 sub 𝐸 one. Let’s start off by drawing a sketch of our charge and our cube.

At the geometric center of a cube sits a charge we’ve called 𝑄. The charge is positive. And so we know the electric field lines from the charge will point radially outward in all directions. We want to solve for the electric flux through one of the six sides of the cube.

We recall from Gauss’s law that if we integrate over an entire closed surface, adding together the electric field components through each area element, then that total surface integral will equal the charge enclosed by the surface over the permittivity of free space, 𝜖 naught. The constant 𝜖 naught we’ll assume is exactly 8.85 times 10 to the negative 12 farads per meter.

Along with Gauss’s law, we can recall that electric flux, 𝛷 sub 𝐸, is equal to electric field multiplied by the area through which the field lines move. Now since in our integral in Gauss’s law we’re integrating over an entire closed surface, that means that this integral will be equal to an expression for 𝛷 sub 𝐸.

That is, in the case of our charge and our cube, we can write that the total electric flux through the entire cube is equal to the charge enclosed by the cube 𝑄 over 𝜖 naught. We don’t want the total electric flux over the entire cube. But rather we want the electric flux through one of the six faces of the cube.

However, since the charge 𝑄 is at the geometric center of the cube and the area of each of the six faces is the same, we can write that 𝛷 sub 𝐸 one, the flux through one of the cube’s faces, is equal to 𝑄 over 𝜖 naught times one-sixth.

We’ve been given the value of 𝑄 in the statement. And 𝜖 naught is a known constant. So we’re ready to plug in and solve. When we do, being careful to write our charge 𝑄 in units of coulombs, when we enter these values on our calculator, we find, to two significant figures, that 𝛷 sub 𝐸 one is 2.3 times 10 to the fifth newtons meter squared per coulomb. That’s the electric flux moving through one of the six faces of this cube.

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