Electrical PE Stored in a Capacitor

[collegelife101]

I just had a question on how TPR derived the equation: PE= 1/2mv^2

It begins with the formula change in PE = qV

It then says that the if the final voltage is V, then the average voltage during the charging process is 1/2V.

I was wondering why we're interested in the average voltage here? Does this mean that we are deriving the average PE? I would think that we would take the difference in voltage, so Vf-Vi (with the initial being zero)?

Any help is appreciated. Thanks!

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[NextStepTutor_1]

So I'm really baffled by why they would start with PE = 1/2*mv^2 because that seems like it might be trying to relate the potential energy to the kinetic energy. Anyway, I'll try to derive the PE equation from scratch since that will probably be easier.

So, a parallel plate capacitor has 2 plates with opposite charges on them. Lets say that the charge stored in the capacitor is Q. Furthermore, if the capacitor is fully charged, and connected to a voltage source of voltage V, then we can say that the voltage between the plates is also V. Thus, the potential energy (energy of position) is the energy that is stored by holding the positive charge away from the negative charge. So, the PE = q * dV. Here, the q = Q, and the dV = V. Thus, PE = QV.

However, you'll immediately correct me by saying that the PE of a capacitor is actually 1/2*Q*V. And you're right, and that's where the idea that the average voltage is 1/2 comes in. So, if the average voltage is 1/2V, we get that dv = 1/2 * V, so PE = Q * 1/2*V = 1/2*Q*V!

The reason the average is 1/2 is because of how a capacitor functions. A capacitor "charges up" because charges are deposited to the 2 plates. As the capacitor charges up, the voltage it contains increases. Thus, we need to look at the average voltage when we consider potential energy because the potential changes every time an charge is added or removed from the plates. In reality, the derivation for this requires calculus, but you don't need to know that on the MCAT!

 

[Chrisz]

I just had a question on how TPR derived the equation: PE= 1/2mv^2

It begins with the formula change in PE = qV

It then says that the if the final voltage is V, then the average voltage during the charging process is 1/2V.

I was wondering why we're interested in the average voltage here? Does this mean that we are deriving the average PE? I would think that we would take the difference in voltage, so Vf-Vi (with the initial being zero)?

Any help is appreciated. Thanks!

The first equation is derived from the fact that the capacitor is made of two plates. One is positively charged while the other is negatively charged. So there is an electric field between the space between the two plates. If you place a positive charge in between, there is attraction and repulsion, so potential energy of the positive charge will eventially transformed into kinetic.

If you know calculus, you can actually derive the above formula quite easily. But it is not necessary, because with the above reasoning Delta PE=-Delta KE. You can get exactly the same equation.

Next thing is to derive PE=1/2(CV^2)=1/2(qV)

We know PE=qV at each specific moment

so, dPE=Vdq
dPE=(q/C)dq
PE=1/2(q^2/C) =1/2(CV^2)=1/2(qV) (These are equivalent terms, because we know the fact q=CV)
For the above derivation, we integrated the differential equation