The plates of a parallel-plate capacitor are 2.50 mm apart, and each carries a charge of magnitude 80.0 nC. The plates are in vacuum. The electric field between the plates has a magnitude of 4.00x10^6 V/m. What is (a) the potential difference between the plates; (b) the area of each plate; (c) the capacitance?

Video transcript

Hi, everyone. Let's take a look at this practice problem dealing with capacitors. So in this problem, we have a parallel plate capacitor made up of two metal plates. The gap between the two plates is exactly four millimeters apart and each holds an equal but opposite charge with a magnitude of 100 and 20 nano columns. The medium between them is vacuum and they experience an electric field of having an intensity of seven multiplied by 10 to the six volts per meter. There are three parts to this question. Part A, we need to determine the voltage difference across the plates. For part B, we need to determine the surface area of each plate. And for part C, we need to determine the capacitors capacitance. We're given four possible choices as our answers. For choice. A the potential difference is 17 kilovolts. The area is equal to 1.8 multiplied by 10th of the negative 3 m squared. And the capacitance is equal to 4.3 peak ofer adds for choice B, the potential difference is 28 klos. The area is 1.8 multiplied by 10th of the negative 3 m squared. And the capacitance is 2.2 PICO Ferres. For choice C, the potential difference is 28 kilovolts. The area is 0 to 1.9 multiplied by 10th of the negative 3 m squared. And the capacitance is equal to 4.3 PICO Ferres. And for choice D, the potential difference is 30 kilovolts. The area is 1.9 multiplied by 10th of the negative 3 m squared. And the capacitance is 4.3 PICO ferrets. So we're gonna start off by answering part a of this question and find the potential difference between the two plates. So for this, we just need to recall our relationship between the electric potential and the electric field field. So we call that formula is V is equal to E multiplied by D or V. Here is our potential difference. E is the electric field and D is the separation distance between the two plates. And so we'll go ahead and plug in values on the right hand side there. Since we know both our electric field and the separation distance, the V is going to be equal to the electric field will have seven multiplied by 10 to the six volts per meter. And for D will have the separate distance of four millimeters. However, I'm gonna need that in meters. So I need multiplied by the conversion factor of 1 m divided by 1000 millimeters. And so when I plug those values into my calculator, I get V is equal to 2.8 multiplied by 10 to the or volts. Now, if I look at my answer choices, they're all given in kill volts. So I'll need to write this um voltage in terms of kolts. So to do that, I'm gonna multiply by the conversion factor of one K volt divided by 1000 volts. So that means V is going to be equal to 28 kilts. So that's gonna be the answer to part A. Now, for part B, we're gonna use our definition for the electric field for a parallel play capacitor to recall that the electric field E is equal to Q divided by absolute not A where Q here is the charge on one of the plates. A is the area of a plate. And epsilon knot here is our constant. So I need to solve this for the area A. So I have A is equal to Q divided by epsilon knot. E and I was given the total chart and we have the electric field so I can plug in those values. And I know the value for A not as well. So have A is equal to or Q will have the charge which is the 120 nano cools. But I'm gonna need that in coolants. So I need to multiply by the conversion factor of one Coolum divided by one multiplied by 10 to the nine nano coons. And then this gets divided by epsilon, not that is constant. And so that's the 8.85 multiplied by tent the negative 12 that has use of Coulomb squared divided by Newton meter squared. And then this gets multiplied by our electro field, which was Z seven multiplied by 10 to the six volts per meter. So I can plug all those values into my calculator. I get my area A is equal to 1.9 multiplied by 10 to the negative 3 m squared. Here just kept two significant figures. That's my answer to part B. Now, for part C, we need to calculate the capacitance. And here we're just gonna use our definition of capacitance to recall their capacity C is equal to Q divided by VRQ is the charge and V is the voltage. So uh I have both of those quantities we have C is equal to our charge was the 20 nano Coolum. Again, we need to convert that into Coolum. So that's gonna be multiplied by the conversion factor of one Coolum divided by one multiplied by 10 to the nine nano columns. And then this gets divided by our potential difference, which is the 2.8 multiplied by 10 to the four volts. And so when we plug those values into my calculator, I get 4.3 multiplied by 10 to the negative 12 fair A. Now if I look at my answer choices, everything is given in PICO far ads. So I'm only 22 multiply this by conversion factor of one multiplied by 10 to the 12 Pica ferrets divided by one fair A and so that gives me my C is equal to 4.3 pika ferrets. And so that's my answer to part C. So if I look at uh my answer choices and compare them to the answers that I calculated the correct answer for this problem is choice C. So just a quick little recap of what we've done here. Part A, we used our relationship between the electric potential and the electric field to calculate our electric potential. For part B, we used our definition for the electric field of a parallel play capacitor to solve for the area. And for part C, we use our definition of capacitance to calculate the capacitance of this capacitor. So I hope that this has been useful and I'll see you in the next video.