MASSACHUSETTS INSTITUTE OF TECHNOLOGY
Department of Physics
8.02
Spring 2013
Problem Set 3 Solutions
Problem 1: Electric Potential and Electric Potential Energy
If you place a negatively charged particle in an electric field, the charge will move
1.
from higher to lower electric potential and from lower to higher potential energy.
2.
from higher to lower electric potential and from higher to lower potential energy.
3.
from lower to higher electric potential and from lower to higher potential energy.
4.
from lower to higher electric potential and from higher to lower potential energy.
Explain your reasoning.
Answer: 4. From lower to higher potential and higher to lower potential energy.
Objects always move to reduce their potential energy, DU º U f - U i < 0 . The electric potential
difference is defined as the potential energy difference per charge.
DV = DU / q
Since DU < 0 (always) and q < 0 (negative charge), DV º V f - Vi > 0 .
negative charges move towards a higher electric potential.
So V f > Vi . Thus
Problem 2 Electric Potential Difference
(a) What are the x- and the y-components of the electric field E at point P ?
(b) What is the electric potential V at point P , assuming that V = 0 at infinity?
(c) A fifth point-like object with charge +9Q is slowly moved in from infinity to point P.
An external agent in moving the charged object in this process must do how much
work?
Answers
(a) What are the x- and the y-components of the electric field E at point P?
x component: Ex = -
Q
4pe 0 (a / 2)2
y component: E y = -
-
Q
4pe 0 (a / 2)2
2Q
4pe 0 (a / 2)2
+
=-
2Q
4pe 0 (a / 2)2
=
6Q
4pe 0 a 2
2Q
4pe 0 a 2
(c) What is the electric potential V at point P, assuming that V
V=
Q
4 2Q
éë -2Q - 2Q + Q - Q ùû = 4pe 0 a
4pe 0 (a / 2)
(c) A fifth charge with charge +9Q is slowly moved in from infinity to point P. How much
work must be done by an external agent moving the charge in this process?
é 4 2Q ù
36 2Q 2
DW = qDV = (9Q) ê ú=4pe 0 a
ë 4pe 0 a û
Problem 3 Energy Stored in a Uniformly Charged Spherical Shell
Consider a spherical shell of radius R that has charge Q distributed uniformly on the surface.
We want to know the potential energy U of this sphere of charge. You can make use of the fact
that outside a spherical distribution of charge, the electric field and electric potential is the same
as if all the charge were at the center. We shall calculate the stored potential energy by
determining the work done in assembling the charge distribution.
(a) If during this charging process, the shell has charge q , how much work does an external
agent do in moving an amount of charge dq from infinity and adding it to the shell?
(b) Determine the potential energy stored in the shell when the charge on the shell is Q . (You
will need to set up an calculate a simple integral.)
(c) An interstellar dust grain, roughly spherical with a radius of R = 3´10-7 m , has acquired a
negative charge such that the potential difference V (R) -V (¥) = -0.15 V . (i) How many extra
electrons has it picked up? (ii) What is the magnitude of the electric field on its surface?
Answer:
(a) Because the electric field outside a shell of charge q is the same as if the charge were
located at the origin (Gauss’s Law).
The change in potential energy in moving a charge dq from infinity and adding it to the shell is
given by
(b) We now find the total energy stored in the shell by integrating
Q
(dq)keq 1 keQ 2
.
=
R
2
R
0
U=ò
(c) (i) Again we can assume that for regions outside the sphere, the spherical charge distribution
can be treated as a point charge located at the origin, as soon the electric potential difference
kQ
V (R) -V (¥)= e = -0.15 V .
R
Therefore the charge on the dust grain is
Q=
RDV (3´10-7 m)(-0.15 V)
=
= 5´10-18 C 31 electrons .
9
2
-2
ke
(9.0 ´10 N × m ×C )
(ii) The magnitude of the electric field on the surface of the dust grain is
E=
keQ ( -DV )
(0.15 V)
=
=
= 5.0 ´105 V × m -1 .
2
-7
R
R
(3´10 m)
Problem 4: Energy Minimization
Consider two metal shells of radius R1 and R2 , quite far apart from one another compared with
these radii. (a) Given a total amount of charge Q , which we have to divide between the shells,
how should it be divided to make the potential energy of the resulting charge distribution as
small as possible? Hint: Use your results from problem 3. First calculate the potential energy
when charge Q - q is placed on one shell and charge q is placed on the other shell. Then
minimize the energy has a function of q . You may assume that any charge put on these shells
distributes itself uniformly. (b) When you have found the optimal distribution of charge, what is
the potential difference between the shells?
Answer:
Suppose we divide up the charge and put Q - q on shell 1 and q on shell 2.
If we assume the shells are very far apart so that we can ignore any interaction between the two
shells then the stored energy (see Problem 3 part (a)), is
1 ke (Q - q) 1 keq
.
U=
+
2
R1
2 R2
2
2
In order to find the charge q that maximizes the stored energy we set the derivative of the stored
energy with respect to q equal to zero and solve for the charge q on the shell 2,
0=
k (Q - q) ke q
dU
=- e
+
Þ
dq
R1
R2
æ 1
1ö Q
qç + ÷ =
Þ
è R1 R2 ø R1
q=
R2
Q
R1 + R2
Therefore the charge Q - q on shell 1 is
Q-q=
R1
Q.
R1 + R2
The potential difference between infinity and the surface of shell 1 with V (¥) = 0 is
V (R1 ) =
ke (Q - q)
Q
.
=
R1
R1 + R2
The potential difference between infinity and the surface of shell 2 with V (¥) = 0 is
kq
Q
.
V (R2 ) = e =
R2 R1 + R2
Therefore
V (R1 ) -V (R2 ) = 0 .
The shells are at the same potential when the stored in the system is minimized. If we had
connected a wire between the two shells then the charge would redistribute itself until the
potential difference between the two shells is zero. So the condition of equal potentials and
minimization of stored energy are equivalent.
Problem 5: Fusion and Fission One way of thinking about fusion and fission is simply thinking
about electrical effects and not considering nuclear effects.
(a)
If you model a proton as a uniformly charged sphere of radius about a fermi
( 1 fm = 1´10-15 m ), how fast would two protons have approach each other in order to fuse
(come into contact)?
(b)
If you had 1 kilogram of 236
92 U (approximate radius 7 fm ) and it underwent fission, in
which each parent nucleus split into two adjacent nuclei of half the charge and half the volume,
how much energy would be released?
Convert this to units of kilotons of TNT
12
( 1 kiloton TNT = 4.18 ´10 J ). It is estimated that the recent meteor (mass approximately
107 kg ) that exploded over Siberia released 500 kilotons TNT.. How does this compare to the
energy released from 1 kilogram of
236
92
U?
Solution:
(a) The question is most easily handled by thinking about energy. When they are in contact,
there is a potential energy


9 109 V m C-1 1.6 1019 C
 kq 
U  qV  q   
2 1015 m
 r 

2
 1013 J
Now we can equate this with kinetic energy to get the velocity.
K  mv 2  U  1013 J  v 
1013 J
 60 1012 m s  8 106 m s
27
1.7 10 kg
This is a decent fraction of the speed of light. These protons are moving!
(b) The daughter nuclei must have a radius such that their volume are half that of the parent
nucleus, so
2V f = Vi Þ 2rf3 = ri3 Þ rf = (2)-1/3 ri = (2)-1/3 (7 fm) = 5.6 fm 6 fm
So now we have two nuclei each of charge 46e and r = 6 fm touching each other, so they have
(
)(
9 ´109 V m C-1 46 ×1.6 ´10-19 C
kq 2
U»
»
r
11´10-15 m
)
2
» 4 ´10-11 J (per U nuclei)
If we had a kg of the stuff how many such nuclei do we get? The atomic mass is 236 so:
 1000g  1 mole   6.02 1023 nuclei 
24
  2 10 nuclei


kg
236
g
mole




1kg  
Multiplying we find a total energy of ~ 1014 J ~ 27 kilotons of TNT! The meteor released 20
times as much energy and it was seven orders of magnitude more massive.
Notes: We are not making a mass-energy conversion calculation for fission in which we measure
the mass-energy of the initial and final states of the uranium nucleus decaying into other states.
238
for
example
The
U ® 23490Th + 42 He .
mU = 395.292542 ´10-27 kg ,
92
mTH = 388.638445 ´10-27 kg ,
and
mU c2 - mTH c2 - mHec2 = DE = 6.848514 ´10-13 J .
mHe = 6.646477 ´10-27 kg .
In
one
kilogram
Then
there
are
N = (1kg)/(395.292542 ´10-27 kg) = 2.52977199 ´1024 atoms . So if the entire sample decayed
into thorium and helium, then the energy released in just this first decay is
DE = N DEsingle decay = (2.52977199 ´1024 atoms)(6.848514 ´10-13 J) = 1.732518 ´1012 J
= 0.414 kilotons of TNT
You would then calculate the energy released in all the decays shown belwo in the decay table of
uranium 238
U.
92
Problem 6: Energy Stored in a Spherical Distribution of Charge
(a) A spherical volume of radius R is filled with charge of uniform density r . Determine the
potential energy U of this sphere of charge by building up the sphere layer by layer, making use
of the fact that outside a spherical distribution of charge, the electric field and electric potential is
the same as if all the charge were at the center. In order to understand the differential
contribution to the stored energy, consider a uniformly charged sphere of radius r and charge
density charge of uniform density r . How much charge is contained in the sphere? Now add a
uniformly charged spherical shell of radius r and thickness dr . How much charge is contained
in that shell? How much work does an external agent do in moving that amount of charge dq
from infinity and adding it to the sphere? Is this the same as the increase in potential energy dU
of the charge distribution? Now integrate until you have built up the entire sphere to find the
potential energy U of this sphere of charge.
(b) At the beginning of the twentieth century the idea that the rest mass of the electron might
have a purely electrical origin was very attractive, especially when the equivalence of energy and
mass was revealed by special relativity. One defect of this model is that nothing holds the charge
together. Imagine the electron as a ball of charge of constant volume density r out to some
maximum radius r0 . Set the potential energy of this system equal to mec 2 , where
me = 9.11´10-31 kg is the mass of the electron, and c = 2.998 ´108 m ×s-1 is the speed of light in
vacuum. (i) Determine r0 . (ii) What happens to the mass if the charge density is keep constant
but the radius is doubled?
Answer
(i) Consider a uniformly charged sphere of radius r and charge density charge of uniform
density r . The charge contained in the sphere is q = r (4 / 3)p r 3 . When we add a
uniformly charged spherical shell of radius r and thickness dr , we are adding a
charge dq = r 4p r 2 dr .
When the charge dq is moved from infinity and added to the sphere, the change is potential
energy is
dU =
keqdq
= -dW = dWext .
r
The change in potential energy is equal the negative of the work done by the electrostatic force
when moving the charge. Because this increases the potential energy, and external energy must
supply the work. If the added charge begins and ends at rest, then the work done by the external
agent is equal to the negative of the work done by the electrostatic force and is therefore equal to
the change in potential energy. Using our results above for q = r (4 / 3)p r 3 and dq = r 4p r 2 dr ,
the change in potential energy is
dU =
ke r (4 / 3)p r 3 r 4p r 2 dr
= 3ke ( r (4 / 3)p )2 r 4 dr .
r
We now add up the contributions shell by shell, which requires integrating r from r = 0 to
r = R,
3
3 ( r (4 / 3)p R3 )2
2 5
.
U = 3ke ( r (4 / 3)p ) ò r dr = ke ( r (4 / 3)p ) R = ke
5
5
R
r=0
r=R
2
4
Now we use the fact that the charge Q contained in the sphere is equal to Q = r (4 / 3)p R3 , and
therefore the stored potential energy in the uniformly charged sphere is
U = ke
3 Q2
.
5 R
b) From our results from part a)
ke
3 (-e)2
= mec 2 .
5 r0
We can solve for the radius
r0 = ke
3 (-e)2 3 (9.0 ´109 N × m 2 ×C-2 )(1.6 ´10-19 C)2
=
= 1.7 ´10-15 m .
2
-31
8
-1
5 mec
5 (9.1´10 kg)(3.0 ´10 m ×s )
(ii) What happens to the mass if the charge density is keep constant but the radius is doubled?
We begin by using our energy relation above to determine that
me = ke
3 (-e)2
.
5 r0c 2
Note that -e = re (4 / 3)p r03 . If we keep the charge density constant then there is more charge,
and so
3 ( re (4 / 3)p r0 )
3 ( re (4 / 3)p ) 5
= ke
r0 .
2
5
5
r0c
c2
3 2
me = ke
If we double the radius then the mass is 25 = 32 times as large.
2
Problem 7 Charged Rod
A thin rod extends along the x -axis from x = -d to x = d . The rod carries a positive charge Q
uniformly distributed along its length.
a) Find the electric potential at a point x > d along the x -axis. Indicate clearly where you
have chosen your zero reference point for your potential.
b) A positively charged particle of mass m and charge q is released from rest from the
point x = 4d . What is the speed of the particle when it reaches the point x = 7d ?
Solution
a) We choose our electric potential to be zero at infinity. Our expression for the electric
potential is
dq¢
V (P) -V (¥) = ke ò
source r - r ¢
Choose as an integration element a small element of length dx¢ with charge dq¢ = l dx¢ where
l = Q / 2d . The vector form the origin to the field point P is r = x î and the vector from the
origin to the any point on the ring is r¢ = x¢ î .
Therefore in the above expression r - r¢ = x î - x¢ î is the vector from the integration element to
the field point P and the distance is r - r¢ = x - x¢ . So the integral becomes
V (x) = kel
x¢ =d
x¢ =d
kQ
k Q x+d
dx¢
= - e ln(x - x¢ ) x¢=- d = e ln
(x - x¢ )
2d
2d x - d
x¢ =- d
ò
(b) The change in potential energy of the charged particle is
DU = q(V (7d) -V (4d)) =
=
keQq æ 7d + d
4d + d ö keQq æ 8 æ 5 ö ö keQq 4
ln
- ln
=
ln - ln
=
ln
ç
2d è 7d - d
4d - d ÷ø
2d çè 6 çè 3÷ø ÷ø
2d
5
The particle starts from rest so the change in kinetic energy is DK = (1/ 2)mv 2f . We now use the
energy law, 0 = DK + DU , and find that
0=
keQq 4 1 2
ln + mv f .
2d
5 2
Therefore the speed of the particle when it reaches the point x = 7d is
vf =
keQq 5
ln .
md
4
Problem 8: Charged Disk
A thin disk, radius 3 cm , has a circular hole of radius 1cm in the middle. There is a uniform
surface charge density of
on the disk.
(a) What is the electric potential at the center of the hole? (Assume zero potential at infinite
distance form the center.)
(b) An electron, starting from rest at the center of the hole, moves out along the axis,
experiencing no forces except repulsion by the charges on the disk. What is the speed of
the electron that it ultimately gains? The mass of the electron is 9.11´10-31 kg .
(c) A non-conducting disk of radius R lies in the xy-plane and carries a non-uniform charge
density
. Determine the electric potential on the axis of the disk a distance z
from its center. Check the limit where z >> R to see that you recover the point-charge
limit. Hint: the following integral may be helpful:

dr
 ln  r  r 2  z 2 


r z
2
2
Answers
(a) Consider a circular ring of radius r  and width dr . The charge on the ring is
dq¢ = s dA¢ = s (2p r ¢ dr ¢)
The field point P is located at the center of the disk, at z = 0 .
From the figure, we also see that the distance from a point on the ring to P is r ¢ . Therefore, the
contribution to the electric potential at P is
dV = ke
d q¢
s 2p r ¢d r ¢
= ke
= kes 2p d r ¢ .
r¢
r¢
By summing over all the rings that make up the disk, we have
R2
V (0) = ò kes 2p d r ¢ = kes 2p (R2 - R1 )
R1
= (9.0 ´109 N × m 2 ×C-2 )(-10-5 C × m -2 )2p (2.0 ´10-2 m) = -1.1´104 V
The change in the potential energy of the electron is
DU = (-e)(V (¥) -V (0)) = eV (0) .
By conservation of energy
0 = DK + DU Þ DK = -eV (0)
Because the electron started from rest, DK = (1/ 2)mev 2 . Therefore the speed of the electron is
v= = -
2eV (0)
me
2(1.6 ´ 10-19 C)(-1.1´ 104 V)
= 6.3´ 107 m ×s -1
-31
(9.1´ 10 kg)
This result gives a speed for the electron that is 0.2 c . At these speeds we can no longer use
our classical definition of kinetic energy and mass but must use relativistic versions and so our
result is only an approximation, not exact.
(c) Consider a circular ring of radius r  and width dr . The charge on the ring is
dq   dA 
0R
r
 2 r  dr    2 0 Rdr 
The field point P is located along the z -axis a distance z from the plane of the disk.
From the figure, we also see that the distance from a point on the ring to P is r  (r 2  z 2 )1/ 2 .
Therefore, the contribution to the electric potential at P is
dV = ke
2ps 0 Rd r ¢
d q¢
.
= ke
r
r ¢2 + z 2
By summing over all the rings that make up the disk, we have
V ( P) = ke 2ps 0 R ò
R
0
dr¢
= ke 2ps 0 R ln éê r + r 2 + z 2 ùú
2
2
ë
û
r¢ + z
é R + R2 + z 2 ù
ú
= ke 2ps 0 R ln ê
z
êë
úû
In the limit z >> R ,
 R  R2  z 2 
R
 Rz 

ln 

ln
1

  ln 


z
z
 z 



R
0
 R

 z
and the potential simplifies to the point-charge limit:
(2ps 0 R2 )
R
Q
V (P) » ke 2ps 0 R × = ke
= ke
z
z
z
As expected, at large distance, the potential due to a non-conducting charged disk is the same as
that of a point charge Q.