You have two identical capacitors and an external potential sourc...

Question

You have two identical capacitors and an external potential source. (a) Compare the total energy stored in the capacitors when they are connected to the applied potential in series and in parallel. (b) Compare the maximum amount of charge stored in each case. (c) Energy storage in a capacitor can be limited by the maximum electric field between the plates. What is the ratio of the electric field for the series and parallel combinations?

Accepted Answer

Welcome back everybody. We are taking a look at two capacitors that are placed in a loop with a battery. We are told that they are identical and they each have a capacitance of four times 10 to the negative six fair ads or for micro fare adds. And we are told that it is a voltage difference of volts. We're also told that the distance between them is one millimeter or 10.1 m. Now here's the thing. The pastors can be placed in either series or parallel. So this is what we are tasked with finding. Were tasked with finding a what the ratio of the stored energy is for the two different arrangements, be what the ratio of the maximum charge is for the two different arrangements and see what the ratio is of the electric fields for the two different arrangements. Now, before diving into all of these ratios, we need to calculate the individual values. So we're gonna use these formulas. Now these formulas are only per capacitor. So we're gonna kind of have to multiply them when it is appropriate here. So the let's see here the ratio of the energy stored for a capacitor, a single capacitor is gonna be one half times the capacitance times the voltage per capacitor. The maximum charge is just going to be the capacitance times the voltage and then the electric field which is going to be for both capacitors is going to be the voltage per capacitor divided by the distance between them. So let's go ahead and calculate all of these values for the capacitors when they are in the series arrangement. Now when they are in the series arrangement, it is important to note that the vultures per capacitor is going to be the total voltage difference divided by the number of capacitors. So this will just be five volts. Let's see here. Let's go ahead and start with the stored energy here. This is going to be one half times four times and to the negative six Times. Let's see here, five squared. And all of this is going to be multiplied by two since there are two capacitors, this gives us a stored energy of 10 to the negative fourth jewels. Now let's look at the maximum charge in series arrangements. This is going to be two for two. Capacitors times capacitance of four times 10 to the negative six times five. This gives us four times 10 to the negative fifth columns. Now let's go ahead and find the electric field in a series arrangement. This is just going to be the voltage per capacitor which is five divided by our distance of 50. which gives us an electric field in series of five times 10 to the third volts per meter. Great. So now I'm just going to scroll down just a little bit here so that we can go ahead and now calculate all of these values when the capacitors are arranged in parallel or the voltage. This time the voltage difference is just going to be equivalent to the individual voltage per capaci. So now for the stored energy we have one half times four times 10 to the negative six times squared times two. Since there are two capacitors, this gives us four times 10 to the negative fourth jewels for the maximum charge, apologies have been putting S as the subscript, this should be P for the maximum charge in a parallel arrangement, we have two for two capacitors times four times 10 to the negative six times 10, which is equal to eight times 10 to the negative fifth columns. And then finally for the electric field in a parallel arrangement we have 10 divided by 100.1 giving us 10 to the fourth volts per meter. Now that we have calculated the individual values here, we are ready to calculate our ratios. So the ratio of the stored energy of series two parallel is going to be 10 to the negative fourth, divided by four times 10 to the negative fourth, simply giving us a ratio of 100.25 for our maximum stored charge here of Qs compared to Q P. We have four times 10 to the negative fifth divided by eight times to the negative fifth giving us a ratio of 0.5. And finally for the ratio of the electric fields we have five times 10 to the third divided by 10 to the fourth, which gives us a ratio of 0.5. So now we have our ratio for the stored energy for the maximum charge and the electric fields corresponding to our final answer choice of B. Thank you all so much for watching hope this video helped. We will see you all in the next one.

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