Question

Why is the work-energy theorem relevant to Electric Potential Difference?

Answer 1

I have an Electric Potential Difference problem $$\Delta V$$
in which it asks to solve for velocity.

The problem solution says to use $$(KE_i + PE_i) = (KE_f + PE_f)$$ and use the fact that $$ KE = \frac{1}{2}mv^{2} $$

I never would have thought to use that.

Why is that relevant to this subject?

Answer 2

Electric potential is an example of what is called a "conservative field." Gravity is another example.

This has a lot of implications, but for this question, it means that energy is conserved, as the work needed to get from one place to another doesn't depend on the path taken, in an isolated system involving a charged particle and an electric potential difference.

What this means is that we can treat it a bit like you would gravity, and the work-energy theorem can be used.

If we have a stationary charge q, and start another charge p with an initial velocity, we can tell it's velocity at any other point by using the work-energy theorem. We simply find it's total energy (Kinetic + Potential) at some point (usually the start, but it doesn't have to be. Do it wherever it's most convenient!), and then find it's potential energy at whatever point we're looking at.

The particles kinetic energy will be the difference between the total energy and it's potential energy at that point.

Answer 3

Small correction: Electric field is the conservative field.

Electric potential is actually a consequence of that fact. But all the rest holds.

Answer 4

So does the work-energy theorm apply to any system where energy is conserved?

And will it always be $$ KE = \frac{1}{2}mv^{2}$$?

Answer 5

I understand that $$PE$$ can have different applications like $$PE = mgh$$ and $$ PE = k_e \frac{Q}{r}$$

Answer 6

Hmmm, good question. I'm hesitant to say that it works for ANY system where energy is conserved, but I can't really think of an example where that isn't true.

It'll hold for conservative fields.

The equations for potential energy will depend on the field you are in.

Answer 7

I guess to limit the scope of my question, I could ask "Does it always apply to any conserved energy system in a problem I would see on an exam?"

And I know PE will differ for each field type, but does Kinetic Energy remain the same at 1/2mv^2 ?

Answer 8

For ANY field?

Answer 9

I'm not sure what level you are at.

At this point, you probably know two conservative fields where this is true: gravity and electric field

It holds for those.

In physics, whether you conserve energy or not depends on what you are looking at, and needs to be evaluated on a case by case basis.

Answer 10

For instance, ice melting in water, in the real world.

If we look at just the water and just the ice, energy isn't actually conserved. Heat is being exchanged with the surrounding environment (we usually say that we "isolate" the system, meaning we keep outside interactions from happening, so we can then conserve energy).

However, if we look at the same situation, but choose the entire universe as our system, then energy is conserved, as all heat exchanges are then accounted for.

Energy will be conserved in an isolated system. But that doesn't necessarily mean that the work-energy theorem works. It requires conservative fields/forces.

For instance, a block sliding down an incline, from rest, but in the presence of friction does NOT follow the work-energy theorem. You can modify some things and still use it, though.

Answer 11

Makes sense to me.
Thank you for you for your help!