Download Week 3 - Potentials and marbles of electrons

Week 3 - Potentials and marbles of electrons
Further, the dignity of the science itself
seems to require that every possible means
be explored for the solution of a problem so
elegant and so celebrated.
Carl Friedrich Gauss
Exercise 3.1: Potential from two point charges
Two positive point charges, each of magnitude q, are fixed on the y-axis at the point y = a and y = −a.
We set the potential to zero at an infinite distance from the charges.
a) Show that the potential at any point on the x-axis is
V =
1
2q
√
.
4πε0 a2 + x2
(1)
Solution:
The potential from a point charge at point r is
V =
1 q
4πε0 r
(2)
where r is √
the distance from the charge to the point. Along the x-axis the distance to either of the
charges is a2 + x2 . Adding together the potentials we therefore get
1
2.0
1.8
1.6
1.4
1.2
1.0
0.8
0.6
0.4 4
3
2
1
0
1
2
3
4
Figure 1: The potential from two point charges on the x-axis. The axes have wrong units, but x goes from
x = −4a to x = 4a.
V
=
=
1
q
1
q
√
√
+
4πε0 a2 + x2
4πε0 a2 + x2
1
2q
√
4πε0 a2 + x2
(3)
b) Graph the potential on the x-axis as a function of x over the range from x = −4a to x = 4a
Answer: See figure 1.
c) Find the x-component of electric field at the x-axis. What is it when x a? Is this result reasonable?
Answer: When x a this becomes
Ex ∼
1 2q
4πε0 x2
(4)
Solution:
At the the x-axis, the x-component of the electric field is given by
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Ex
dV
dx
2q
x
.
2
4πε0 (a + x2 ) 23
−
=
=
(5)
When x a this becomes
Ex ∼
1 2q
4πε0 x2
(6)
which is the field of a point charge of charge 2q placed at the origin. This makes sense because when
our distance from the origin is very large compared to the separation between the charges, we will
not be able to see this separation and therefore instead see a charge of 2q.
Exercise 3.2: Discussion Questions
a) Since a potential can have any value depending on the choice of reference level for zero potential, how
does a voltmeter know what to read when you connect it between two points?
Solution:
The keyword here is that we use two points. It is the potential difference which is read by the
voltmeter, and it’s also only this quantity that is physically relevant. The potential difference is the
only thing that is physically relevant because the electric field is related to the potential by
E(r) = −∇V (r)
(7)
so when we look at the x-component we see that
Ex = −
dV (x, y, z)
V (x + ∆x, y, z) − V (x, y, z)
= − lim
.
∆x→0
dx
∆x
(8)
In other words the electric field only cares about the potential difference between points and the
electric field is the quantity which produce measurable effects in the form of forces and accelerating
charges. Looking at the above equation we see that if we add a constant C to V it does not matter
because it vanishes when we take the difference. For example
Ex0 = −
∂ (V (x, y, z) + C)
∂x
=
=
(V (x + ∆x, y, z) + C) − (V (x, y, z) + C)
∆x
V (x + ∆x, y, z) − V (x, y, z)
− lim
= Ex
∆x→0
∆x
− lim
∆x→0
(9)
(10)
b) A student asked, ’Since electrical potential energy is always proportional to the potential, why bother
with the concept of potential at all?’ How would you respond?
Solution:
The electric potential is an important quantity in itself because of many reasons. When you talks
about electrical potential energy, you talk about the electrical potential energy of something. For
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example a charge q1 in the presence of a certain charge distribution. This energy might be totally
different from the energy that a charge q2 has in the precense of the same charge distribution. The
electric potenial however is a much more fruitfull concept because it does not depend on the specific
charge you choose to place in the presence of the particular charge distribution. Like the electric field,
it is a property of the space itself, so it’s much more usefull to state the potential in a region of space,
rather than the electric potential energy of a spefic charge being in that region. It is easy to find this
energy when we have the potential for any charge you would like to put there.
Another important and quite amazing property of the potential is revealed in equation (23.16) and
equation (23.20) in the book. It is important because it is much easier to find the potential, a scalar
quantity trough
V =
1
4πε0
Z
dq
r
(11)
dq
r̂
r2
(12)
than it is to find the electric field directly trough
E=
1
4πε0
Z
which is a vector quantity. The electric field can always be found after we’ve found the potential by
the relation E = −∇V which requires only differentiation and is easier than integration. Now the
electric field is determined from the potential trough E = −∇V and the electric field (3 numbers) at a
point determines a charge’s further motion from that point. But these 3 numbers (E) are determined
by 1 number (V ). Therefore the motion is really only determined by 1 number! If the charge knows
the potential V (r) at r it knows how to move! One single number is enough to describe the motion
of a charged particle trough 3-dimensional space. This is the amazing and quite profound part.
There are more reasons, but by now the student is probably satisfied.
c) It is easy to produce a potential difference of several thousand volts between your body and the floor
by scuffing your shoes across a nylon carpet. When ou touch a metal doorknob, you get a mild
shock. Yet contact with a power line of comparable voltage would probably be fatal. Why is there a
difference?
Exercise 3.3: Curvature, charge density and lightning rods
Figure 2: Two spheres separated by a large distance.
Let’s study the effect of curvature by making a simple model in which we imagine two spherical conductors
of radius R1 and R2 connected by a conducting wire (figure 2). Assume that the spheres are far away
from each other, so that to a good approximation, the potential of one does not effect the potential of the
other. Further assume that the wire is so thin that all the charge is stored on the spheres, respectively
Q1 and Q2 . Hint: Remember what you know about potential difference in conductors.
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a) Set the potential to zero at infinity and find the ratio Q1 /Q2 .
Answer:
Q1
R1
=
Q2
R2
(13)
b) If R1 > R2 , which sphere has the largest charge?
Answer: if R1 > R2 , Q1 > Q2 .
Solution:
From the assumption that the presence of the second sphere does not effect the potential of the first
their respective potentials must be
Q1
4πε0 R1
and
Q2
.
4πε0 R2
(14)
Now since they are connected by a conducting wire, the whole surface consisting of the two spheres
and the wire constitute an equipotential. Therefore the potentials of the two spheres most be equal,
implying
Q1
Q2
Q1
R1
=
⇒
=
4πε0 R1
4πε0 R2
Q2
R2
(15)
Or equivalently
Q1 = Q2
R1
R2
(16)
which means that if R1 > R2 , Q1 > Q2 .
c) What sphere has the greatest charge density?
Answer: The charge density of the smaller sphere with radius R2 is higher than the density of the
larger sphere.
Solution:
Since the surface area of the spheres goes as 4πR2 the charge density is
σ=
Q
4πR2
(17)
and therefore
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σ2
Q2
=
σ1
Q1
R1
R2
2
=
R1
.
R2
(18)
So the charge density of the of the smaller sphere with radius R2 is higher than the density of the
larger sphere!
d) There has been a controversy between British and American scientist of whether a lightning rod
should have a pointy end or a rounded end in the form of a ball. The English claim it should be a
ball and the Americans claim it should be a pointy end. Who do you support and why?
Answer: There is no clear cut answer to this exercise. If you want to read up on the topic, check out
the section “Should a lightning rod have a point?” on http://en.wikipedia.org/wiki/Lightning_rod
Exercise 3.4: Manipulating protons
a) How much work would it take to push two protons very slowly from a separation of 20.0 nm1 (a
typical atomic distance) to 3.0 fm (a typical nuclear distance)?
Solution:
You can either memorize and use the formulas or remember the method and reasoning. I would highly
recommend the second option. To find the work needed to push the second proton neared the first
(from point a to point b) we have to oppose the electrical force. I.e. the work that we have to do is
opposite the electical force, F = −Fe . So that we get
Z
Wab
b
−
=
Fe · dl
(19)
a
2
Z b
dr
e
−
4πε0 a r2
e2
1 1
−
4πε0 b a
=
=
(20)
(21)
where we have used that the electrical field is conservative so that we might just as well go the easy
road along the radial distance. Plugging in the numbers we get
1
1
−
= 0.49 MeV
(22)
Wab = 1.46 MeVfm
3.0 fm 20 × 106 fm
One might wonder how we can ever push a proton along a path against the electric force with a force
equal in magnitude to the electric force. Wouldn’t the proton just stand still? The anwer is yes, so
let the force get a little tiny additon of F + f . Repeating the argument we get
Z
Wab
b
= −
(Fe + f ) · dl
(23)
a
2
11
Z b
Z b
e
dr
−
f · dl
4πε0 a r2
a
Z b
e2
1 1
−
−
f · dl
4πε0 b a
a
= −
(24)
=
(25)
nm = 10−9 m.
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and now take the limit in which f → 0. The second integral vanishes and the answer remains the
same.
b) How much work is needed to assemble an atomic nucleus containing three protons such as beryllium
(Be) if we model it as an equilateral (equal sided) triangle of sides 2.0 fm (1 fm = 10−15 m) with a
proton at each corner? Assume the protons started from very far away (infinity).
Solution: Generally the work one has to do to assemble a collection of point charges is
W
=
=
1 X qi qj
4πε0 i<j |ri − rj |
1
q1 q2
q1 q3
q2 q3
+
+
+ ...
4πε0 |r1 − r2 | |r1 − r3 | |r2 − r3 |
(26)
(27)
where the first term comes about because of the work needed to push the charge q2 to r2 in the
presence of electric field set up by q1 at r1 , and the second and third terms comes about because of
the work needed to push the charge q3 in the presence of the electric field set up by q2 and q3 at r1
and r2 .
Now for the three protons in the equilateral of sides d = 2 fm triangle q1 = q2 = q3 = e and
|r1 − r2 | = |r1 − r3 | = |r2 − r3 | = d. So we get
2
e2
e2
e
1
+
+
(28)
W =
4πε0 d
d
d
3e2
=
= 2.2 MeV = 3.5 × 10−13 J
(29)
4πε0 d
where 1.0 MeV = 1.0 × 106 eV. This is a much more convenient (and used) unit at the atomic scale.
The mass energy of a proton is forexample just about 1 MeV trough Einstein’s E = mc2 .
Answer:
W = 2.2 MeV = 3.5 × 10−13 J
Exercise 3.5: Plotting the potential from an uniformly charged
sphere
In the next exercise you will be asked to find the field from an uniformly charged solid sphere, but first;
let’s have a look at the potential and play around with its properties.
To find the field, it is often easier to find the potential first. The potential inside an uniformly charged
sphere is
Q
r2
V (r) =
3− 2
2 · 4πε0 R
R
while outside the sphere, it is
V (r) =
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Q 1
.
4πε0 r
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a) Use the contour3d() function in Mayavi to visualize the field in 3D. Set the number of contours to
20 and the opacity to 0.5 to be able to see through the contours (see the code in the hint below on
how to do this).
Hint: Use the following code to find the distance to each point in space. You need to use the distance
to calculate the potential.
from numpy i m p o r t ∗
try :
from m a y a v i . mlab i m p o r t ∗
except I m p o r t E r r o r :
from e n t h o u g h t . m a y a v i . mlab i m p o r t ∗
x , y , z = mgrid [ − 1 0 0 . : 1 0 1 . : 5 . , − 1 0 0 . : 1 0 1 . : 5 . , − 1 0 0 . : 1 0 1 . : 5 . ]
r = s q r t ( x ∗∗2 + y ∗∗2 + z ∗ ∗ 2 )
V = 0∗ x
for i in range ( len ( r ) ) :
for j in range ( len ( r ) ) :
for k in range ( len ( r ) ) :
# This p r i n t s out the d i s t a n c e to the p o i n t i n q u e s t i o n .
# Replace i t with your c a l c u l a t i o n of the p o t e n t i a l .
print r [ i ] [ j ] [ k ]
V[ i ] [ j ] [ k ] = . . . # i n s e r t p o t e n t i a l as f u n c t i o n of r here
...
c o n t o u r 3 d ( x , y , z , V , c o n t o u r s =20 , o p a c i t y =0.5)
b) Use the Numpy function gradient() to find the electric field as E = −∇V . Plot the electric field
using the quiver3d() function.
Solution: An example script:
from numpy i m p o r t ∗
try :
from m a y a v i . mlab i m p o r t ∗
except I m p o r t E r r o r :
from e n t h o u g h t . m a y a v i . mlab i m p o r t ∗
x , y , z = mgrid [ − 1 0 0 : 1 0 1 : 5 . ,
R = 40
Q = 1.0
−100:101:5. ,
−100:101:5.]
r = s q r t ( x ∗∗2 + y ∗∗2 + z ∗ ∗ 2 )
V = 0∗ x
for i in range ( len ( r ) ) :
for j in range ( len ( r ) ) :
for k in range ( len ( r ) ) :
i f r [ i ] [ j ] [ k ] < R:
V [ i ] [ j ] [ k ] = Q / ( 8 ∗ p i ∗ R) ∗ ( 3 − r [ i ] [ j ] [ k ] ∗ ∗ 2 / R∗ ∗ 2 )
else :
V[ i ] [ j ] [ k ] = Q / (4 ∗ p i ∗ r [ i ] [ j ] [ k ] )
c o n t o u r 3 d ( x , y , z , V , c o n t o u r s =20 , o p a c i t y =0.5)
Ex , Ey , Ez = g r a d i e n t (V)
Ex = − Ex
Ey = − Ey
Ez = − Ez
#q u i v e r 3 d ( x , y , z , Ex , Ey , Ez )
c) Are you able to see where in space the sphere is just by looking at the plot of your potential?
Answer: Yes. Note that the distance between any two equipotential surfaces decreases as you move
out from the center of the sphere. As soon as you go past the surface of the sphere, the distances
start to increase. Do you see this? If not, try to play around with the value of R in the script.
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See the course pages for information on how to set up and use Python and Mayavi.
Exercise 3.6: The electron marble
We want to construct a marble solely out of electrons. Let’s say we have a glass marble with R = 1 cm
in radius and weighing m = 10 g. Consider that we remove all the protons and are left with only the
electrons, summing up to a charge of Q = 48 180 C.
a) Let us model the marble as a sphere with uniform charge density ρ. Find the electric field inside and
outside the sphere.
Hint: When you’re inside the sphere, the total charge enclosed is no longer Q.
Answer: For r < R:
E=
For r > R:
ρ
r
3ε0
Q
4πε0 r2
E=
Solution:
We use Gauss’ law to determine the electric field for r < R and r > R. Since the charges are placed
uniformly over the volume, we can use a constant ρ as the charge density. For r < R we have
I
Qenc
E · dA =
ε0
S
which gives
2
=
E4πr2
=
E
=
E4πr
Z
1
ρdV
ε0 V
ρ 4 3
πr
ε0 3
ρ
r
3ε0
Furthermore, for r > R we get
E4πr2
=
Q
ε0
E
=
Q
4πε0 r2
b) To construct this marble, we assemble the electrons slowly one by one, until the charge is Q, and
placed uniformly over the marble’s volume. This might remind you of what you did in exercise 4, but
this time, we’ll use the energy density to calculate the total energy. The energy density is a topic of
the next chapter in the book, but we’ll skip ahead a bit and introduce the concept here:
Due to the conservation of energy, the work done will be stored as energy in the the electric field,
with the energy density as
1
uE = ε0 E 2
2
Find the total required work by integrating the energy density over the space inside and outside the
sphere.
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Answer:
3 Q2
5 4πε0 R
Solution:
The work needed is equal to the total energy stored in the electric field after the charges are assembled.
Now we may use that the energy density in the electric field is given as
uE =
1
ε0 E 2
2
We can integrate over all space where we have an electric field. To do this, we make up a small
infinitesimal volume element dV = 4πr2 dr and see that this results in an infinitesimal energy element
dU = uE dV =
1
ε0 E 2 4πr2 dr
2
For the field where r < R, we get
R
Z
U1
=
0
ρ 2
1
ε0 (
r) 4πr2 dr
2
3ε0
R
4πρ2 r4
dr
2
0 2 · 3 ε0
4πρ2 R5
5 · 2 · 3 2 ε0
Z
=
=
Since the charge density is uniform, we have that the total charge must be the density times the
volume, Q = ρ4/3πr3 , so that
3 Q
ρ=
4 πR3
Inserting for ρ we get the energy as
U1 =
Q2
5 · 2 · 4πε0 R
For r > R we get
Z
∞
U2 =
R
1 Q2
1 Q2
dr =
2
2 4πε0 r
2 4πε0 R
In total, this is
U = U1 + U2 =
3 Q2
5 4πε0 R
This can also be solved by using
U=
ρ
2
R
Z
4πr2 drV (r)
0
or by modeling the total charge by bringing shells with charge ρ4πr2 dr from r → ∞ for a sphere of
radius r and then integrate r from 0 to R.
By insertion of Q = 48 180 C, we find
U = 1.253 · 1021 J
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c) If we were actually able to assemble such a marble of electrons, the electrons would obviously want to
immediately repel each other. The number of electrons where found last week to be ne = 3.011 · 1023 .
If we assume that all the electrons are thrown out into space with an equal velocity for each electron,
what would the speed of each electron be? (Is your answer reasonable or should we consider the
effects of relativity?)
Answer: 1.87 · 1023 m/s. No it is not reasonable, and relativity should be taken into account.
Solution:
The speed would be determined by the potential energy going over to kinetic energy. If each electron
gets the same amount of the potential energy, each would gain
Ue =
U
1.253 · 1021 J
=
ne
3.011 · 1023
If all this potential energy was to be turned into kinetic energy, the velocity of each electron would
be given by Ue = 1/2me ve2 .
r
2Ue
ve =
= 9.56 · 1013 m/s
me
which obviously is out of proportions compared to the speed of light. This means that the problem
should be solved by proper use of relativity.
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