Two large rectangular aluminum plates of area 200 centimeters squared face each other with a space between them. The facing areas of the plates are charged with equal amounts of opposite charges, totaling 40.0 microcoulombs. Find the flux through a circle of radius 2.00 of centimeters between the plates when the normal to the circle makes an angle of 5.00 degrees with a line perpendicular to the faces of the plates.
We’re told in this statement the area of each plate, 200 centimeters squared, which we’ll call capital 𝐴. The magnitude of the charge on each plate, 40.0 microcoulombs, we’ll call 𝑄. The radius of the circle that exists between the plates, 2.00 centimeters, we’ll call 𝑟. And the angle between the normal of that circle and the faces of the plates, 5.00 degrees, we’ll call 𝜃.
We want to solve for the flux; that is, the electric Flux that moves through the circle. We’ll call that 𝛷 sub 𝐸. Let’s start our solution by drawing a diagram of this scenario. In this situation, we have two plates of equal sizes, 𝐴, set a distance apart from one another. And they’re equally and oppositely charged. We’ll say that the plate on the left has charge plus, 𝑄 where 𝑄 is 40.0 microcoulombs, and the plate on the right has a charge minus 𝑄.
In between these oppositely charged plates, there is a ring of radius lowercase 𝑟. If we draw normal vectors for each of the two plates as well as the circular disc. Then the angle between the normal vector of the circle and the normal vector of either one of the plates is equal to 𝜃, which is given as 5.00 degrees. Because these two parallel plates are equally and oppositely charged, an electric field we can call 𝐸 is set up between them. The field lines of this electric field move through the ring. And it’s that electric flux, 𝛷 sub 𝐸, we wanna calculate.
We can recall that electric flux, 𝛷 sub 𝐸, is equal to the electric field magnitude times the area that the field lines move through multiplied by the cosine of the angle between these two vectors, 𝐸 and 𝐴. Now in our case, we know the angle 𝜃 between these two vectors. The area 𝐴 isn’t given, but the radius of the loop is. And we can use that along with the fact that the area of a circle is equal to its radius squared times 𝜋 to calculate 𝐴.
What remains then is to calculate 𝐸, the electric field magnitude between these equally and oppositely charged plates. To solve for 𝐸, we can recall that, in this case, it’s equal to the service charge, 𝜎, of the plates it’s between divided by the permittivity of free space, 𝜖 naught. 𝜖 naught is a constant whose value we’ll assume is exactly equal to 8.85 times 10 to the negative 12 farads per meter. The 𝐸 in our equation then is equal to 𝜎 over 𝜖 naught. And that is equal to the charge on each plate multiplied by the area of each plate, given by capital 𝐴, over 𝜖 naught.
Since we know both of those values, as well as 𝑟 and 𝜃, we’re ready to plug in to solve for 𝛷 sub 𝐸. 𝛷 sub 𝐸 is 𝑄 times 𝐴 over 𝜖 naught times 𝜋𝑟 squared cosine of 𝜃. Or, this expression where we’ve been careful to write our area in units of meters squared and our distance in units of meters, as well as our charge in units of coulombs. We’re now ready to plug in these values on our calculator and solve for 𝛷 sub 𝐸. To three significant figures, it equals 28.3 times 10 to the fourth newtons-meter squared per coulomb. That’s the electric flux through this circular ring.