# Video: Finding the Electric Flux through a Cubic Surface That Intersects a Uniformly Charged Cylinder

The electric field at 2.0 cm from the center of a long copper rod of radius 1.0 cm has a magnitude of 3.0 N/C and is directed outward from the axis of the rod. What is the charge per unit length of the copper rod? What would be the electric flux through a cube of side 5.0 cm situated such that the rod passes through opposite sides of the cube perpendicularly?

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### Video Transcript

The electric field at 2.0 centimeters from the center of a long copper rod of radius 1.0 centimeters has a magnitude of 3.0 newtons per coulomb and is directed outward from the axis of the rod. What is the charge per unit length of the copper rod? What would be the electric flux through a cube of side 5.0 centimeters situated such that the rod passes through opposite sides of the cube perpendicularly?

We can call the distance at which the electric field is measured from the center of the long copper rod, 2.0 centimeters, 𝑑 and the magnitude of the electric field at that point, 3.0 newtons per coulomb, 𝐸. We want to solve first for the charge per unit length of the copper rod, which we’ll call 𝜆. And then we want to solve for the electric flux passing through the sides of a cube. We’ll call that flux 𝜙 sub 𝐸.

We can start off by drawing a diagram. We have a charged copper rod with a radius. And we’re told it’s a long rod. So we can approximate it as having an infinite length. At a point at distance 𝑑 from the axis of the rod, we measure the electric field. And we find at that point that the electric field has a value of 3.0 newtons per coulomb. Based on this, we wanna figure out what is the charge per unit length of the rod, 𝜆.

We can recall that the electric field created by an infinitely long charged rod is equal to the linear charge density of the rod, 𝜆, over two 𝜋 times 𝜖, naught the permittivity of free space, multiplied by 𝑟, the distance from the center of the rod to the point where the field is being measured.

𝜖 naught is a constant, which we’ll treat as exactly 8.85 times 10 to the negative 12th farads per meter. Because the point at which we’ve measured the electric field is completely outside our spatially extended charged rod, we don’t need to account for the radius of that rod given as 1.0 centimeter. We only need to account for the distance from the center of the rod 𝑑.

Since we want to solve for 𝜆, let’s rearrange this equation to do so. 𝜆 equals two 𝜋 𝜖 naught times 𝑑 times the electric field 𝐸. We know each one of these values and can plug in for them now. When we do, we make sure to use units of meters when we plug in for distance 𝑑. When we multiply all these numbers together, we find that, to two significant figures, 𝜆 is 3.3 times 10 to the negative 12th coulombs per meter. That’s the charge per unit length of the copper rod.

Now that we’ve solved for 𝜆. We want to find out what the electric flux would be if we put a box around our charged rod where the box was a cube with sides, each of length 5.0 centimeters, and arranged so that the rod went through opposite sides of the cube perpendicularly.

We know that because the rod is charged, it will create flux through the cube. And we can recall Gauss’s law to tell us what that is. This law says that the electric flux through a closed surface is equal to the charge the surface encloses over 𝜖 naught. In our case, our charge 𝑄 is spread out over the rod. It’s distributed across it.

The charge enclosed by the cube will be equal to the linear charge density 𝜆 times the length of one of the cube’s edges, 5.0 centimeters. We’ve solved for 𝜆 in part one. And 𝜖 naught in the denominator is a constant of known value. So we can plug in and solve for the flux.

When we do plug in, we’re careful to convert the cube side length into units of meters to agree with the units in the rest of our expression. Plugging these values into our calculator, we find that 𝜙 sub 𝐸 is 0.019 newtons meter squared per coulomb. This is the total electric flux passing through the walls of the cube.