5.9 Energy Stored in Capacitors from Office of Academic Technologies on Vimeo.

**5.09 Energy Stored in Capacitors**

All right. Let’s now try to calculate the energy stored in the electric field of the capacitor. As you recall, we said capacitors are the devices which provide small electric field packages in the electric circuits so that we can store energy into these field lines. If we consider a parallel plate capacitor, let’s say the upper plate is charged positively and the lower plate is charged negatively to some q Coulombs and in doing so, we generate a potential difference of, let’s say, V volts between the plates and generate an electric field, which originates from the positive plate and enters into the negative plate, filling the whole regions between the plates.

Now, once we fully charge this capacitor with a capacitance of, let’s say, C farads, if we move an incremental amount of charge, dq, from positive plate to negative plate, then a certain work will be done by the electric field. That work will be equal to-directly from the definition of potential–the incremental work done will be equal to, in moving that through the potential difference of V volts, V times dq.

Since the capacitance is equal to amount of charge stored in the plates of the capacitor divided by the potential difference between the plates of the capacitor, we can express the potential difference in terms of the charge and the capacitance of the capacitor as q over C. Therefore we can express the incremental work done in terms of the charge and the capacitance as q over C times dq, simply substituting this quantity for the potential difference, V, in this equation.

Now, if we discharge the whole capacitor, in other words, if we just completely move all this positive charge to here, so that they neutralize, and look at the total work done, that will be in that case the sum of all these incremental works done by moving these incremental charges, which eventually makes the whole charge along the plate. The summation will be integration, and that will be integral of dw, which is going to be equal to integral of q over C times q.

The limits of integration is such that we start from zero and go up to the total charge originally stored in the capacitor. If we take this integral, total work done is going to be equal to–since the capacitance is constant, we take it outside of the integral–we will have 1 over C times integral of qdq integrated from zero to q and the total work done is going to be equal to q squared over 2 times C evaluated at zero and q. Of course, we substitute the boundaries, then we will end up with q squared over 2C.

If you recall from work energy theorem, this is equal to negative of the change in potential energy, in other words u sub f minus u sub i. Here, the final potential energy is the potential energy associated with the negative charge. And as you remember, when we were talking about the potential, we said that as long as we deal with the potential difference, the origin is irrelevant.

And if we call the potential of the positively charged plate as V and potential of the negatively charged plate as zero, then the potential difference between these two plates becomes V minus zero, which is V volts. By applying the same approach over here, we will say let u sub f, which is equal to the potential energy associated with the negatively charged plate is equal to zero, and u sub i is equal to potential energy associated with the positively charged plate; let’s denote that as u.

Then q squared over 2C becomes equal to minus zero minus u. From here, minus minus will make positive. The potential energy stored in the electric field of this capacitor becomes equal to q squared over 2C. Using the definition of capacitance, which is C is equal to q over V, we can express this relationship. Let me use subscript E here to indicate that this is the potential energy stored in the electric field of the capacitor, is equal to q squared over 2C.

If we solve this for q, that will be equal to CV. If we substitute over here, we will have C squared V squared over V. One of these “C”s will cancel. This can also be expressed. Therefore, in terms of the capacitance of the capacitor and the potential difference between the plates as one-half CV squared. Or, we can also express the same equation or expression in terms of the charge and the potential difference by substituting for C. We will have q squared over 2. For C, we will write down q over V. One of these charges will cancel. Therefore that will be equal to one-half QV.

So all these three expressions will give us the energy stored in the electric field of a capacitor in three different forms, namely q squared over 2C, or one-half CV squared, or one-half qV.