When a 360-nF air capacitor (1 nF = 10^-9 F) is connected to a power supply, the energy stored in the capacitor is 1.85x10^-5 J. While the capacitor is kept connected to the power supply, a slab of dielectric is inserted that completely fills the space between the plates. This increases the stored energy by 2.32x10^-5 J. (a) What is the potential difference between the capacitor plates? (b) What is the dielectric constant of the slab?

Video transcript

Hi everyone today we are going to determine the potential difference fee and also the dielectric constant K. Off the sheet. So first we are going to start with creating a list of what is given in our problem first we have the foreign nano ferret parallel plate capacitor which means that it will have an initial capacity C. Not for an F. Or recall that nano will be multiplied by 10 to the power of minus nine to become the as a unit of ferret. And then it is mentioned that it has nothing between the plane and is charged by the potential difference fee which is going to be determined in our problem. So the capacity itself stores an initial energy you not of 1.02 Multiplied by 10 to the power of -5 jewels. And the volume between the plates is totally filled with a sheet of dialectic while the power source is still going to be connected so it's still gonna be connected to the same potential difference fee. And because of that the energy of the capacitor is going to be enhanced by this value. So that will be the U. Value which is going to be the same. You not plus the enhancement which is 2. times 10 to the power of minus five and that will be 1.2 plus 2. multiplied by 10 to the power of minus five jewels which is going to actually be 3.8 times 10 to the power of minus five jules like. So okay so now that we have created a list of everything that is given in our problem, what we want to do next is to actually start solving So recall that the energy that is actually going to be stored in our capacitor is going to be given by this formula right here so you not is going to equal to how C not multiplied by fi squared. That is going to be the energy formula before the insertion of the dielectric. So we just want to rearrange this formula right here to solve for fee because we know that you know as we know what is so you not to, you not oversee not will equals two free square Sophie will just equals to the square root of two, you not over see not like so and this will then Can be solved. So this will be two multiplied by our unit here which is 1.2 times 10 to the power of -5 jewels and then defied it, defied it By Our C. Not which is four times 10 to the -9. The feed will then be 71. 1 fault. So now we have the potential difference fee across the plates of the capacitor. Now what we have to look at is the after insertion of the dielectric while the power source is still connected since the capacity remains connected to the power supply, the potential difference doesn't change when the dielectric electric is inserted so essentially defeat is still going to be the same. The dielectric constant K. We want to recall is going to be given by the C. Oversee not and replacing the C. We can try replacing the C by using this formula right here. So rearranging the energy store formula then see not will actually be to you not over free squared. Right? So in this case uh this will also correlate with C equals to two you over free squared with this to being the same free so we can pluck this to formula into this K. Dielectric constant formula here too then obtain to you over free squared and this is gonna be to you not overfeed squared. So we can cross out the fee squared and the tooth. So the K will essentially be equals to you over you not. Right, okay. So defining you with you not will actually gives us the dielectric constant and we can just actually block in the U. Value that we have calculated here. And do you not value here to find K. So K will then be 3.8 times 10 to the power of minus five tools over. Do you not? 1.2 times 10 to the power of minus five tool like so so Okay well then equals to 3. and with a potential difference of 71.41 fall and a K. Dielectric constant of 3.72. The answer isn't going to equals two option D. Just like so. Alright, that will be all for this particular problem. If you guys are still confused about this, there are other lesson plans on this particular topic, so please check it out. And that will be all.