Electrical Potential Presentation
Review of Electrical Potential
(A makeup for yesterday)
Note on equations
General equation (always works as long as you do the calculus right)
V b
V a
b
E
d
a
Specific equation for a point charge with infinity as your reference point (i.e. where V = 0)
V = kq r
Specific equation for finding the potential difference in a uniform, constant electric field (i.e. the field from an infinite plane of charge)
V
Ed
(d is the distance between your reference point and the point you are interested in)
Ball on a hill analogy for potential energy
U
U = mgh h
U =0
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)
Ball on a hill analogy for potential energy
U
U = 0 h
U = -mgh
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
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 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
V
kQ r
One last clarification
r
+q
+Q
So the potential difference is then
V
kQ r final
kQ r initial
And the change in potential energy is
U
q kQ r final
kQ r initial
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)
