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Discussion Question 4A
P212, Week 4
Electric Potential and Work
Electric Potential Energy can be considered unrealized kinetic energy for a particle in a fixed location.
It takes work to move a particle from a position of lower potential energy to a position of higher
potential energy. This increase in potential energy at rest may be converted to (or “realized as”) kinetic
energy when the particle is allowed to move from the position of higher potential energy back to the
position of lower potential energy.
Imagine that one of your hands is attached to a positive test charge q and the other hand holds a stake
that can attach you to the ground. You are first staked to the ground at a distance 10 m from a positive
charge Q. (Assume that gravity is zero for this problem, and that you and the test charge are small
enough to be treated as point charges). Let’s further assume, when you lift the stake, you (somehow)
float without friction… The total mass of you + the mass of charge q + the mass of the stake is M.
q
Q
10 m
(a) You lift the stake, so that suddenly you are free to move. As a consequence do you start to gain or
to lose kinetic energy? Gain
What does this imply about the sign of the work done by the field on you in the process? Positive
(b) You re-stake yourself at a distance 12 m from Q.
Has your potential energy increased or decreased? Decreased
How is the change in your potential energy, ΔU, related to the amount of work done by the field?
−ΔU = Wfield
This is consistent with (a) because the field does positive work on me, but my potential energy has
decreased.
q
(c) You lift the stake again, and now walk radially around Q from A
to B, as shown at right. What is the change in your potential energy?
Potential energy does not change
Would the change in potential energy be different if you had walked
from A to B in a straight line? Explain.
Potential energy would not change.
×B
×
A
Q
Discussion Question 4A
P212, Week 4
Electric Field Lines
(d) You are still a distance 12 m from Q. You lift the stake again and find that you are moving with
velocity v0 as you pass a point 14 m from Q. How is your velocity related to the change in the
potential energy?
Change in work = change in potential energy = change in kinetic energy:
ΔW= -ΔU = ½ m(v0)2
(e) You re-stake yourself at 14 m from Q. You now change q from a positive to a negative charge, and
un-stake yourself again.
What is the magnitude of your velocity as you pass a point at a distance 12 m from Q? v0
Does your potential energy increase or decrease as you move? Potential energy decreases as I move
towards Q
What is the sign of the work done by the field on you in this case? Positive
(c) Electric potential energy, U, is a property of an object, while electric potential, V = U/q (where q is
a test charge), is a property of space. Just as the electric field was defined as the force that would be
v v
exerted on a test charge ( E = F / q ), the change in electric potential can be defined as the work that
would be exerted on a test charge, ΔV = -WAB/q, where WAB is the work exerted by the field as particle
q moves from A to B. Both potential and potential energy can only be defined with respect to a
reference point.
Would your answer to (c) change if you considered potential, rather than potential energy? No.
For the case in part (e), if you doubled the value of q before you moved, how would ΔV change?
ΔV would not change
If you doubled the value of Q before you moved, how would ΔV change? ΔV would double.
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Discussion Question 4B
P212, Week 4
Electric Potential in a System with Cylindrical Symmetry
Consider a non-conducting cylinder of infinite length and radius a, which carries a volume
charge density . Surrounding this object is an uncharged conducting cylindrical shell. The
metal tube is also of infinite length, and its inner and outer radii are b1 and b2 respectively. In
this problem, we will define the potential to be zero at the outer surface of the conducting shell.
a = 3 cm
b1 = 6 cm
b2 = 8 cm
 = +7.5 C/m3
b1
b2
V=0
a

conducting shell,
uncharged
(a) What is the electric potential at a radius of 10 cm from the center of the cylinders?
(i) In all of these potential problems, we have to integrate the electric field from a point
where we know the potential to the point we are interested in. In this problem, we want
the potential at a point 10 cm from the axis of symmetry ... let's call this point 'A'. Now,
where do we know the potential? Write down an expression for VA (the quantity we
want) in terms of a potential we know and an integral over the electric field.
(ii) Now that our integral is set up, we need to know the electric field E, in all regions over
which we have to integrate. Write down an expression for E.
(iii)
Solve the integral!
(iv) This is important: check the sign of the integral you just solved using your physical
intuition: If we're moving 'downhill' (i.e. along the electric field), the electric potential
drops. If we've gone 'uphill', the electric potential should have increased.
The potential difference between V(r=10 cm) and V=0:
r

 a2
 a2 r dr
 a2 r
Er 
; V (r )  0    Er dr V (r )  
ln



b
  r 2  r
2  b r
2  b2
2
2
V(r=10 cm) = -( a2 / 2 0) ln(r/b2) = -8.5 x 107 V
(b) What is the electric potential at a radius of 7 cm?
(v) Here we go again. We'll call our
new point of interest 'B'. Write
down an expression for VB in
terms of a known potential and
an integral over the electric field.
a = 3 cm
b1 = 6 cm
b2 = 8 cm
 = +7.5 C/m3
b1
b2
a

(vi) Now you need the electric field
E. What could it be ...?
V=0
(vii) Solve the integral ... (this is too
easy)
(viii) By the way: what general statement can you make about the electric potential inside a
conductor?
V(r=7 cm) = 0, potential always constant within any conductor
(c) What is the electric potential at the center of the non-conducting cylinder?
Same procedure, just a bit more work than before …
b1
0-V (a )    Er dr ; V (a ) 
a
 a2 b1 dr  a2 b1

ln
2  a r
2  a
a
a
0
0
V (a )  V (0)    Er dr ;V (0)  V (a )   Er dr
 r
r

=
; V ( 0 )  V (a ) 
2  r 2 
2 
2
Er =
V (0)  V (a ) 

a
0
r dr 
a2
4 
 a 2  a2 b1  a 2

ln 
4 
2  a 4 
V(r=0) = ( a2 / 4 0) [ 2 ln(b1/a) + 1 ] = 4.55 x 108 V
2
(d) It is customary to use infinity as a reference point, and set the potential to zero there.
Why could we not choose infinity as the zero of potential in this particular problem?
From the solution to part (a), we can see that the voltage drop between infinity and any finite
point is infinite, which makes infinity a non-useful reference point in this problem.
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Discussion Question 4C
P212, Week 4
Electric Potential due to a System of Point Charges
Three point charges are fixed at the positions shown
in the figure. We are interested in the electric
potential surrounding this system of charges ... we'll
set the potential at infinity to be zero in all of these
problems. Also, we'll work with algebraic answers in
this problem, so you can put your calculator away.
(a) What is the electric potential at the origin?
y
-q
d
-q
2d
2d
q
x
(i)
It is always a good idea to anticipate as much of the answer as possible before you
begin your calculations. In potential problems, the sign is particularly important. So,
using only your physical intuition, what will be the sign of the electric potential at the
origin? (Ask yourself: if I bring in a positive charge from infinity, where the potential
is set to zero, will I be pushing it ‘uphill’ or letting it roll ‘downhill’?)
(ii)
Now we’re ready to calculate. You could do this problem like the last one, using
integrals over the electric field. However, we already know an expression for the
potential due to a point charge. What is this expression? Remember, any formula for
potential is only good to within an additive constant ... check your formula: does it set
the potential to zero at the location we want?
(iii)
Superposition to the rescue! The total potential at the origin is just: the potential due
to charge q1, plus the potential due to charge q2, plus ... you get the idea. Write down
the potential V due to each of the three charges.
(iv)
Finally, calculate the net potential at the origin. Is the sign correct?
V=
−kq −kq kq −kq
+
+
=
d
2d 2d
d
y
-q
d
-q
2d
2d
q
x
(b) If you placed a negative charge –Q at the origin, what potential energy would it have?
What is the maximum speed it would achieve if it was initially at rest and you let it go?
" −kq % kQq
U = −QV = −Q $
= U(0)
'=
d
# d &
()
(
)
(
U 0 + KE(0) = U x = 2d + KE x = 2d
mv
2
2
2d
=
)
kQq
→ v 2d = ∞
0
(c) What is the total potential energy of the system of 3 charges?
Let's think for a second ... what does this question mean? The potential energy of the system
of 3 charges is equal to the work it took to assemble this configuration. When all 3 charges
are at infinity, we define them to have zero potential energy. So ... bring in one charge at a
time from infinity, and calculate how much work was required at each step …
Zero work to place left charge. Work required for right charge WR =
2
Work required to add top charge: WT =
U = total work = WR + WT = -
K ( −q )
( 2d )2 + d 2
+
K ( −q )( +q )
( 2d )2 + d 2
K ( q)( -q)
4d
Kq 2
=−
.
4d
=0
Kq 2
4d
Did you check the sign of your result using your physical intuition?
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Discussion Question 4D
P212, Week 4
Superposition and Conductors
As we saw in the last problem, superposition is an extremely useful tool in potential problems.
However, we have to be very careful with superposition when conductors are around … let’s
explore this!
Two thin plates of infinite area and made of insulating material are on either side of the origin
and a distance d = 5 cm away from it. They carry uniformly-distributed surface charges with
the values given below. In this problem, we will be concerned with the electric potential
difference between the two plates. For convenience, we’ll use the left-hand plate as our
reference and set the electric potential to zero there. (This is indicated by the ground symbol in
the figure.)
y
a
= -4 C/m2
2
b = +2 C/m
d = 5 cm
a
b
d
x
d
(a) What is the electric potential Vb of the right-hand plate?
E
V x
a
2
d
b
xˆ ; V x
d
V x
2d
d
0
o
0
Vb = +| a– b| d /
a
b
2
0
o
2d
0
dx
a
b
E x dx
d
o
= 3.39 x 1010 V
(b) Now suppose a positive point charge Q = +5 C is placed at the location
(x,y) = (-5 cm, +2 cm). What is the electric potential difference V between the points
(x,y) = (d,0) and (-d,0) on the x axis? (Remember, superposition …)
First calculate V due to point charge
KQ
KQ
Vc (d ) Vc ( d )
2
2
(d 5) 2
(d 5)2
KQ
2
2
( 2d )
2
a
Now include V due to sheets VS (d ) VS ( d )
2
b
KQ
2
2
d
o
total V
d
V
d
a
b
d
o
KQ
( 2d )2
22
KQ
2
Now get rid of the point charge. Instead, suppose that an uncharged metal slab of thickness 2 cm
is placed parallel to the two plates and centered on the origin.
Does superposition still work?
YES … we just have to be a bit careful.
uncharged
metal slab y
a
b
d
x
d
2 cm
(c) If we considered the uncharged metal slab all on its own, in isolation, what potential
difference Vb – Va would it cause between the two plates?
Vb
Va
0 (produces no fields)
d
d
(d) If we considered the charged insulating sheets all on their own, in isolation, what potential
difference Vb – Va would they cause between the two plates?
Same as in part (a)
Va
Vb
d
d
2
Pure superposition would suggest that adding these two contributions together gives the correct
net result. But not quite …
(e) Calculate the full potential difference Vb – Va between the plates: find the electric field
everywhere between the plates, and integrate it to find Vb – Va.
Width of metal plate is w and is located in region -w/2 < x < w/2.
In all regions except -w/2 < x < w/2, E
a
V (d ) V ( d )
V (d ) V ( d )
b
2
a
b
2
w /2
d
( w /2
dx
w /2
w /2
a
b
2
xˆ
0 dx
d ) 0 (d w / 2)
a
b
2
d
w /2
a
b
2
dx
2d w
2.71 1010V
(f) Can you explain why adding the results of (c) and (d) together did not work?
Conductors respond to the presence of charged objects by internally rearranging their own
charges. Thus, when we apply superposition to a collection of objects that includes one or more
conductors, we must remember that the conductors are affected by the other objects.
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