Polystyrene has dielectric constant 2.6 and dielectric strength 2.0x10^7 V/m. A piece of polystyrene is used as a dielectric in a parallel-plate capacitor, filling the volume between the plates. (a) When the electric field between the plates is 80% of the dielectric strength, what is the energy density of the stored? (b) When the capacitor is connected to a battery with voltage 500.0 V, the electric field between the plates is 80% of the dielectric strength. What is the area of each plate if the capacitor stores 0.200 mJ of energy under these conditions?

Video transcript

Hi everyone today we are going to find the energy density stored in the capacitor. When the paper slap is inserted between the plates of an airfield parallel play capacitor. And also next we are going to find the area of the plates when the capacity itself is attached to a hundredfold power source, maintaining an electric field constant of a quarter. The dialectic strength and an energy store of 12.5 micro jewel. Okay so first what we want to do is to actually identify all the necessary information given in this particular problem. So we have the dielectric constant K which is 3.7. And we also have the dialectic strength which is going to be E of 1.6 times 10 to the power of seven fold per meter. And then we also have the potential difference or the power source which is a hundredfold. And lastly we have the energy store which is a big queue of 12.5 micro jewel. Remember that we have to make everything in as a unit. So we want to multiply it by 10 to the power of minus six and that will be the energy stored in chill. Like So so these are pretty much all the information given. So now we can actually start recalling the equation that we're gonna use. So recall that the energy density you in this case a small U. Which is being asked in the first part of this problem, in an electric field for the case in which a dielectric is present is going to be represented by this formula right here which is how how k epsilon? Not e squared. So DK is gonna be the dielectric constant. The epsilon is gonna be the electric constant and e is going to be the magnitude of the electric field or the dialectic strength. And in this case the U. S as I have previously mentioned is the electric energy density. So we actually have all the necessary information here to calculate for you. So we can just plug all of this because the epsilon naught is actually a constant. So we can just start plugging the values to calculate our first problem. So first you is going to be half que Absolutely not E squared which is going to be half multiplied by 3.7 multiplied by epsilon is going to be 8.854 times standard, apart of minus 12 column square per newton meter. And now we will have the e square which is given here 1.6 times 10 to the power of seven fault per meter squared, just like so okay, so this will actually come up to be uh 2. times 10 to the border of two shows per meters cube. Like so this is gonna be the answer for our first part which is 2.62 joules per meter cube. So we can essentially cross out option A A and also option C. Okay, so now we want to move on to the second part so to find the area. We will also need to find the plate separation. So recall that the plate separation is going to be represented by this formula de Which is going to be close to the power source potential difference divided by E. Which is going to be the biotic strength. So we can start by plugging this values that is known already four per meter. So we know that the separation distance is gonna be 2.5, multiplied by 10 to the power of minus five m. And that is not the area. So what we want to do next is to find what the area is. So so we want to recall that the formula connecting the electric energy density small you in terms of the energy store large you and the area. And the separation distance is going to be represented by this formula right here. So the energy density is gonna be equal to the energy stored divided by the area, multiplied by the separation distance which is essentially just the normal density formula. And we can rearrange this so that we can find what the area is. So we just move the A. To the other side and the other you. So it's gonna be you over small. You multiplied by D. And we actually know all these values because we just calculated the small you from part one so we can just block everything in directly. The large shoe is going to be 12.5 times 10 to the power of -6 jewels. And then the small you that we just calculated 2.62 times 10 to the part of two shows per meter cube. And then the D. is gonna be 2.5 times down to the part of -5 m like. So and this will give us an A of 1.9 times 10 to the power of minus three m squared. Which essentially the A. Is then going to be um 19 centimeters squared in centimeters. So uh that will actually be our answer for the area of the plate. So with energy density of 2.62 times standard, part of two jewels per meter cube and area of plate of centimeters squared. The answer is then gonna be option B. Like so Okay, so um let me know if you guys have any problem with this and if you do there are other videos along the same topic that you guys should check out because we have a lot of other videos on our website regarding this particular topic and that will be all. Thank you