A 3.0-cm-diameter parallel-plate capacitor has a 2.0 mm spacing. ...

Question

A 3.0-cm-diameter parallel-plate capacitor has a 2.0 mm spacing. The electric field strength inside the capacitor is 1.0×10^5 V/m.b. How much charge is on each plate?

Accepted Answer

Hello, fellow physicists today we solve the following practice problem together. So first off, let's read the problem and highlight all the key pieces of information that we need to use. In order to solve this problem. The overlap area between two parallel plates is 50 centimeters squared. The separation of the plates is 2.0 millimeters. The plates are charged to create an electric field strength of 8.0 multiplied by 10 to the power of four volts per meter between them determine the total charge on the plates. So that's our end goal is we're trying to determine the total charge on the plates. Awesome. We're also given some multiple choice answers. Let's read them off to see what our final answer might be. A is 350 nano Couls B is 3.5 nano Coulombs. C is 180 micro coulombs and D is 0.071 coulombs. Awesome. So first off, let us assume that the electric field between the plates is uniform. Now, we must recall and use the equation to describe how a charge on plate Q is related to the electric field. Capital E So we're gonna represent the electric field as capital E. So our equation for the electric field is, is E is equal to A divided by epsilon subscript zero, which is also equal to Q divided by capital A all divided by epsilon subscript zero. So we need to note that A A is the charge density area and epsilon subscript zero is the permittivity of free space constant and capital A is the area between the parallel plates. Awesome. And then don't forget that Q is the charge on the plate. OK. So let us use the electric field E is equal to Q divided by A all divided by epsilon subscript zero to solve for our final answer. But first, let's make a couple of notes here that one Newton per Coolum is equal to one volt per meter. And we need to note that the electric field is equal to 8.0 multiplied by 10 to the power of four Newton per Coolum using this logic. OK. So now we must rearrange the energy equation to solve for Q. So we're trying to isolate and solve for Q to find our final answer. So if we use a little bit of algebra to rearrange our energy equation and let's call it fall mar market in read here, we'll call it equation one for energy equation. So rearranging equation one to solve Q for Q. So when we isolate Q, we get that Q is equal to epsilon subscript zero is multiplied by the area between the plates multiplied by the electric field. So at this stage, we could plug in our known variables and solve for Q. So let's do that. So Q is equal to epsilon subscript zero, the numerical value for that. So this is the permittivity of free space constant which is 8.8 five multiplied by 10 to the power of negative 12 Coolum squared per Newton multiplied by meters squared. Awesome. And then multiplied by the area between the plates which is 50 centimeters squared. And that number is provided to us in the problem. But we need to make a quick little conversion here because it's in centimeters and we need it in meters meters squared. So to do that, all we need to do is we need to remember that in 1 m there is 100 centimeters. So 1 m divided by 100 centimeters and then we need to square that value and then we'll get the centimeters squared to cancel out leaving us with meters squared, which is exactly what we need multiplied by the electric field which is provided to us as eight point zero multiplied by 10 to the power of four volts per meter. Awesome. So when we plug all of that into a calculator, we shall get that Q is equal to 3.5 multiplied by 10 to the power of negative nine cool lumps. But as you've noticed, none of our multiple choice answers match our Coolum value. So let's convert our Coolum value to Nano Coolum to see if our final answer will match any of the multiple choice answers. So all we have to do is take 3.5 multiplied by 10 to the power of negative nine. And we need to multiply it by 10 to the power of nine, which is our conversion factor to convert Coombs to nano Coolum. So when we multiply 3.5 multiplied by 10 to the power of negative nine by 10 to the power of positive nine, we will get 3.5 Nano Coolum, which is our final answer. And I'm gonna box that in green. So looking at our multiple choice answers, that means the correct answer has to be the letter B 3.5 Nano Kums. Thank you so much for watching. Hopefully that helped and I can't wait to see you in the next video. Bye.

Struggling with Physics?

A 3.0-cm-diameter parallel-plate capacitor has a 2.0 mm spacing. The electric field strength inside the capacitor is 1.0×10^5 V/m.b. How much charge is on each plate?

Watch next

Relationships Between Force, Field, Energy, Potential practice set