Electricity - Kinetic/Potential Energy Quick Question

[sciencebooks]

So in normal electrical circuits, we have KE=qV=-PE

And in capacitors, we have PE=(1/2)QV

Can anybody explain to me (1) where the 1/2 comes from and (2) why there is no longer the - sign?

Thanks!

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[milski]

The first formula looks a bit better if you write it as KE=-PE=q.ΔV It describes what happens to a particle which moves to a new position where the difference in potential between these two positions is ΔV. KE=-PE comes from conservation of energy - another way to write it is KE+PE=0. Since the particle did not exchange any energy with other objects, it's total energy stays the same, only PE turns to KE or vise versa, depending on the sign of the change in potential.

The second formula tells you what the energy is to charge a capacitor to charge Q. Let's consider what this energy is. When you start, the capacitor is fully discharged. That is, the potential between the two plates is 0. To add a charge to it at that point requires 0 work. As you start increasing the charge on its plates, you also increase the potential between them and that starts to require more and more work to add the next discrete amount of charge. There are different ways to calculate what the sum of all these separate amounts of energy is but it all leads to a total of QV/2.

Just a quick hint about one way: the first charge needs 0 work. At that point the potential between the two plates is q/C, so the second one needs q*q/C amount of work. The third one has a potential of 2q/C and needs 2q*q/C of work and so on... the last one has a charge of(Q-q)/C and needs (Q-q)*q/C of work. When you do the sum of q/C(q+2q+3q+...+(Q-q)) you'll get Q^2/(2C) or QV/2.

 

[sciencebooks]

The first formula looks a bit better if you write it as KE=-PE=q.ΔV It describes what happens to a particle which moves to a new position where the difference in potential between these two positions is ΔV. KE=-PE comes from conservation of energy - another way to write it is KE+PE=0. Since the particle did not exchange any energy with other objects, it's total energy stays the same, only PE turns to KE or vise versa, depending on the sign of the change in potential.

The second formula tells you what the energy is to charge a capacitor to charge Q. Let's consider what this energy is. When you start, the capacitor is fully discharged. That is, the potential between the two plates is 0. To add a charge to it at that point requires 0 work. As you start increasing the charge on its plates, you also increase the potential between them and that starts to require more and more work to add the next discrete amount of charge. There are different ways to calculate what the sum of all these separate amounts of energy is but it all leads to a total of QV/2.

Just a quick hint about one way: the first charge needs 0 work. At that point the potential between the two plates is q/C, so the second one needs q*q/C amount of work. The third one has a potential of 2q/C and needs 2q*q/C of work and so on... the last one has a charge of(Q-q)/C and needs (Q-q)*q/C of work. When you do the sum of q/C(q+2q+3q+...+(Q-q)) you'll get Q^2/(2C) or QV/2.

Thank you! It's one of those facts I'd just remembered but this actually helps. It all comes down to a transfer of energy. Could you sum it up as "creating charge requires energy" or is that a gross oversimplification? (I've been had questions testing it conceptually, but I've noticed that circuits are quite popular so it's better to be safe than sorry.)

 

[milski]

Thank you! It's one of those facts I'd just remembered but this actually helps. It all comes down to a transfer of energy. Could you sum it up as "creating charge requires energy" or is that a gross oversimplification? (I've been had questions testing it conceptually, but I've noticed that circuits are quite popular so it's better to be safe than sorry.)

Sort of. You cannot really create charge. "Moving charge requires energy" would be a somewhat better generalization but then you need to remember that if there is no change in potential between the initial and final position of that charge, the required energy is zero.

 

[sciencebooks]

Sort of. You cannot really create charge. "Moving charge requires energy" would be a somewhat better generalization but then you need to remember that if there is no change in potential between the initial and final position of that charge, the required energy is zero.

Oh yeah! You're right, it'd actually probably be bad to think of "creating charge" because that's something that could easily be a false answer. No change in potential = no delta V = zero work/energy by the equation. Makes sense. I think I forget to consider equations with conceptual facts which actually helps me a lot. Thank you again, .