Lecture Notes
8.022 ESG (Dourmashkin)
Jacob Hurwitz
September 20, 2010
Conservative and Non-Conservative Forces
RB
F~ · d~s = ∆K.
H
Recall that a conservative force is a force which performs no work on a closed loop: F~ · d~s = 0. Take
two paths from point A to point B. Because the work around the entire loop is zero, we know that the
work along one path from A to B equals minus the work along the other path from B to A. Switching
the limits of integration, the work along both paths from A to B is equal. That is, conservative forces are
path-independent.
Last class, we showed that Newton’s second law (in the integrated sense) gives W =
A
The electrostatic force is conservative not because of the inverse square law, but because the force is radially
symmetric. Draw a closed loop consisting of four segments: r = r1 , θ = θ1 , r = r2 , and θ = θ2 . The work is
zero along the arcs θ = θ1 and θ = θ2 , and the work is equal but opposite along the radial legs r = r1 and
r = r2 . Then it’s easy to see that the total work is zero.
Side note regarding polar coordinates: given any arbitrary path, you can decompose it into a radial change
of dr r̂ and a tangential piece (r + dr)dθ θ̂ = r dθ θ̂ to the first order. Thus, d~r = dr r̂ + r dθ θ̂.
Returning to our closed loop, you can decompose the two radial legs as d~r1 = dr r̂ and d~r2 = dr r̂. (Note:
you do not need a negative sign because dr can be positive or negative!) Thus:
Z r2
1
1 r2
1
kq1 q2
=
−kq
q
r̂
·
dr
r̂
=
kq
q
−
−
W1 =
1 2
1 2
r2
r r1
r2
r1
r1
On the other radial path, everything is the same except the limits of integration are flipped. This works
because the force is radially symmetric, that is, it does not depend on θ.
Given an arbitrary path, you can approximate it with many small tangential and radial paths. The same
argument applies to these pieces. This shows that the electrostatic force is conservative because of the
symmetry, not because of the inverse square.
Electric Potential Energy
Any force is made up of many other forces, which could be conservative or non-conservative: F~T = F~C + F~N C .
RB
RB
RB
Decomposing this, we see that A F~T · d~s = A F~C · d~s + A F~N C · d~s = ∆K. Conservative forces have
RB
potentials associated with them. Let’s define the potential as ∆U = U (B) − U (A) ≡ − A F~C · d~s, where
∆U is the change in potential energy of the system. Then ∆K = −∆U + WN C .
To recap: Last class, we used a math fact to introduce the work-energy theorem. Today, we introduced
conservative forces, which allow us to re-write the work-energy theorem as WN C = ∆U + ∆K ≡ ∆Emech .
If there are no non-conservative forces, then there is no change in mechanical energy. If there is a nonconservative force (like friction), then some of the mechnical energy will be converted into other types of
energy, like heat or sound.
To find the electric potential energy, we use the fundamental definition on an arbitrary path d~s = dr r̂ +
r dθ θ̂:
Z r2
Z r2
1
1
kq1 q2
dr
r̂ · d~s = −kq1 q2
= kq1 q2
−
U (r2 ) − U (r1 ) ≡ −
2
r2
r2
r1
r1 r
r1
We need to pick a zero-point reference. Let’s pick r1 = ∞ and set U (∞) = 0. Then for any arbitrary r2 = r,
we have U (r) = kq1 q2 /r. We could have picked any zero-point, but this works out the most conveniently.
This quantity represents negative the work done by the system force in moving a charge q2 originally infinitely
far away from q1 to only r away from q1 . (Do not say the work done by an external agent! This isn’t always
negative the work done by the system. For instance, if you lift a weight and increase its kinetic energy, then
you are performing more work than negative the change in potential energy.)
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Lecture Notes
8.022 ESG (Dourmashkin)
Jacob Hurwitz
September 20, 2010
Electric Potential
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~ s (P ) = F~q (P )/qt = kqs r̂sp /rsp
Recall that the electric field was defined as E
. Because we divided by the
t
test charge, electric field is a property of the source alone, independent of test charge. Let’s define a similar
quantity here. Denote the potential energy per test charge as ∆φ = φ(r2 ) − φ(r1 ) = [U (r2 ) − U (r1 )]/qt . This
represents the change in potential energy per test charge when moving a test charge from r1 to r2 away from
a point source charge qs .
Using the integral from above, we see that the change in electric potential energy per test charge is ∆φ =
kqs (1/r2 − 1/r1 ). Call this quantity the electric potential (not to be confused with electric potential energy).
Let’s once again set r1 = ∞, φ(∞) = 0, and r2 = r, so the electric potential at r is ∆φ = φ(r) = kqs /r.
In words: You take a test charge and start at infinity. You then bring it to a distance r away from the
source. You calculate how much the potential energy changed, and you divide the change by the test charge.
That’s what ∆φ is. Mathematically:
Rr
Z r2
− r12 F~ · d~s
U (r2 ) − U (r1 )
~ s · d~s
E
=
=−
∆φ = φ(r2 ) − φ(r1 ) =
qt
qt
r1
Thus, electric potential is associated only with the field!
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