Review of Electrical Potential (A makeup for yesterday)

Note on equations 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)

Ball on a hill analogy for potential energy U 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 = U final – U initial = -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) h U = mgh U =0

Ball on a hill analogy for potential energy U Now we choose U = 0 at the top of the hill So ∆U = U final – U initial = -mgh And again the change in kinetic energy ∆KE = -∆U = mgh h U = 0 U = -mgh

Ball on a hill analogy for potential energy U 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?)

Relating the ball on a hill to E&M 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 )

Relating the ball on a hill to E&M 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 )

One last clarification Going back to the first case with two positive charges +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

One last clarification So the potential difference is then And the change in potential energy is So if the charge is repelled, r final > r initial 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 +q r