Download Electric Potential Problem Set: Solutions! 1. A charged particle

Electric Potential Problem Set: Solutions!
1. A charged particle moves in a region of space that contains an electric field. When the particle moves towards
the east, the particle gains 125 kV of electric potential. (a) What is the direction of the electric field?
The electric field points from high electric potential to low electric potential. In this case, motion must be
opposite to the direction of the electric field is electric potential increases.
(b) The potential energy stored in the interaction of this charge with the electric field is reduced by 1.00 × 10 –13 J as
a result of the motion. How much charge is on the object, and is it positively or negatively charged?
PE  qV Solve for charge:
PE
1.00 1013 J
q 

 8.00 1019 C
V
125, 000V
If the electric potential is increasing, yet the potential energy of the charge is decreasing, then the charge must
be negative (due to the opposite behavior).
(c) How strong is the electric field if the object moved 2.50 m?
V  Ex  E 
V
x

125, 000V
V
 5.00 104
2.50 m
m
2. An electric field is used to accelerate an alpha particle (with charge of +2e and mass of 6.64 × 10−27 kg) from rest
to a speed of 4.00 × 10+6 m/s in a distance of 25.0 cm.
(a) Should the electric field point in the direction of motion or opposite to the motion?
The electric field should point in the direction of intended motion of the alpha particle so that the particle will
speed up.
(b) Determine the change of potential energy for the alpha particle due to motion through the electric field. Does the
potential energy increase or decrease?
By energy conservation, if the kinetic energy of the particle is increasing, then the potential energy should be
decreasing.
PE  12 mv 2  12  6.64 1027 kg  4.00 106
m
s

2
 5.311014 J
(c) What is the change of electric potential over this region of space? Does the electric potential increase or
decrease?
V 
PE
q

5.312 1014 J
 1.66 105V
19
2 1.6022 10 C 
Since the motion is in the direction of the electric field, this will be in the direction of decreasing electric
potential as well. Also, for a positive charge, a decreasing potential energy corresponds to a decreasing
electric potential.
(d) What is the strength of the electric field?
E
V
x

1.66 105 V
V
 6.63 105
0.250 m
m
(e) How would the answers be different if the alpha particle had a negative charge instead?
To make any particle accelerate to a greater speed, the potential energy must decrease by energy
conservation. Since the charge is negative, the electric field will need to point in the opposite direction. The
electric potential will thus be increasing in the direction of motion of the particle.
3. {Review Problem! Possible bonus question…} The unstable nucleus uranium-236 can be regarded as a
uniformly charged sphere of charge Q = +92e and radius R = 7.4 × 10 –15 m. In nuclear fission, this can divide into
two smaller nuclei, each with ½ the volume and ½ the charge of the original uranium-236 atom. This is one of the
reactions that occurred in the nuclear weapon that exploded over Hiroshima, Japan in 1945.
(a) Find the radii of the two “daughter” nuclei of charge +46e.
Hint: Assume the nuclei have the same density and the daughter nuclei receive ½ the total mass.
Let r be the radius of the daughter nuclei and R the radius of the unranium-236.
Volumedaughter 
r3 
4 3 1
1 4 3
r  Volumeuranium 
R
3
2
2 3
1 3
R
R  r  3  5.87 1015 m
2
2
(b) In a simple model for the fission process, immediately after the uranium-236 nucleus has undergone fission, the
daughter nuclei are at rest and just touching, as shown in the figure above. How much potential energy is stored in
the two daughter nuclei due to the electrostatic repulsion? Treat the two daughter nuclei as point charges placed two
nuclear radii apart.


8.988 109 Nm
 46 1.6022 1019 C 
kQ1Q2
C2
PEo 

 4.16 1011 J
15
2rdaughter
2  5.87 10 m 
2
2
2
(c) Use momentum conservation to justify the daughter nuclei will move apart at equal speeds, and find the kinetic
energy of each daughter atom after they have moved very far apart.
The two nuclei start initially at rest, so the total momentum is zero. The Coulomb repulsion between the
charges pushes equally on each of the identical masses (Newton’s 3 rd law), so each mass accelerates the same
amount. The final speeds later on must be equal to that the total momentum stays zero.
KE f , each  12 PEo  2.08 1011 J
(d) In this model the sum of the kinetic energies of the two daughter nuclei, calculated in part (b), is the energy
released in the fission of one uranium nucleus. Calculate the energy released by fission of 10.0 kg of uranium-236.
The mass of one uranium-236 atom is 3.92 × 10–25 kg. Express your answer in joules and in kilotons of TNT.
1 kiloton TNT releases 4.18 × 1012 J when it explodes.

  4.16 1011 J 
1atom
15
Energy  10.0 kg 236
U

  1.06 10 J
92


25
3.92

10
kg
atom



 1 kilotonTNT
Energy  1.06 1015 J  
12
 4.18 10 J

  254 kilotonTNT
