Download 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  V   E  d 
b
b
a
a
Specific
equation for a point charge with infinity as
kq
your reference point (i.e. where V = 0)
V=
Specific
r
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
Remember
U =0
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)
Ball on a hill analogy for potential energy
U
U=0
h
Now
U = -mgh
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
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
Intuitively, we
+q
+Q
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
Intuitively, we
-q
+Q
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
And
kQ
kQ
V 

rfinal rinitial
the change in potential energy is
 kQ

kQ

U  q

r

r
 final initial 
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)