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Review of Electrical PotentialPowerPoint Presentation

Review of Electrical Potential

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Review of Electrical Potential

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(A makeup for yesterday)

General equation (always works as long as you do the calculus right)

Specific equation for a point charge with infinity as your reference point (i.e. where V = 0)

Specific equation for finding the potential difference in a uniform, constant electric field (i.e. the field from an infinite plane of charge)

(d is the distance between your reference point and the point you are interested in)

Remember that you can choose your zero of potential (U) to be anywhere

Here we choose U = 0 at the bottom of the hill

So ∆U = Ufinal – Uinitial = -mgh

And the change in kinetic energy ∆ KE = -∆U = mgh (i.e. the ball has a real speed at the bottom of the hill as the potential energy is converted into kinetic energy)

U = mgh

h

U =0

Now we choose U = 0 at the top of the hill

So ∆U = Ufinal – Uinitial = -mgh

And again the change in kinetic energy

∆KE = -∆U = mgh

U = 0

h

U = -mgh

The important point is that nature always wants to be in the lowest potential energy configuration

In this case that means the ball wants to roll down hill (amazing how much physics can complicate the obvious, eh?)

- Consider a positive charge +Q fixed at the origin with a small test charge +q close by

+q

+Q

- Intuitively, we know +q will be repelled from +Q, just like a ball is repelled from the top of a hill
- In physics-speak, +q is forced towards a lower potential energy (U) or the charge wants ∆U to be negative
- Finally, to relate this to the electrical potential (V), U = qV, so a positive test charge +q will move towards lower V (in other words it wants ∆V to be negative)

- Now consider the same thing but with a negative test charge -q

-q

+Q

- Intuitively, we know -q will be attracted to +Q
- In physics-speak, -q is forced towards a lower potential energy (U) or the charge wants ∆U to be negative
- Finally, to relate this to the electrical potential (V), U = qV, so a negative test charge -q will move towards higher V (in other words it wants ∆V to be positive)

- Going back to the first case with two positive charges

r

+q

+Q

- To flesh this out a little more, if we wanted to find the change in electrical potential (a.k.a. if we wanted to find the potential difference ∆V), we first find the potential from +Q because this creates the field through which we move the test charge +q

r

+q

- So the potential difference is then
- And the change in potential energy is
- So if the charge is repelled, rfinal > rinitial and ∆U is then negative like we expected from the ball on a hill analogy
- Think about this on your own for the case of a negative test charge (you should still get a negative change in potential energy for the –q being attracted to +Q)

+Q