General Physics
Homework, 2nd set, solutions
1) If two points are at the same potential, then no NET work was done in moving a test charge
from one point to the other. Along some segments of the path, some positive work might have
been done, but along other segments of the path, negative work would then have been done.
And if the object was moved along an equipotential line, then no work would have been done
along any segment of the path.
Along any segment of the path where positive or negative work was done, a force would
have to be exerted. If the object was moved along an equipotential line, then no force would
have been exerted along any segment of the path.
This is analogous to climbing up and then back down a flight of stairs to get from one
point to another point on the same floor of a building. Gravitational potential increased while
going up the stairs, and decreased while going down the stairs. A force was required both to
go up the stairs and down the stairs. If instead you walked on the level from one point to
another, then the gravitational potential was constant, and no force was needed.
2) The electric field is never zero, except at infinity. Between the charges their fields have the
same sign, so they will always add up and can never compensate each other. On the other
parts of the line, i.e. not between the charges, the electric fields have opposite sign, so they
may cancel, where qi /ri2 has the same magnitude for the two charges, which can be fulfilled
only once on the side of the weaker charge.
The electric potential caused by the two charges have always an opposite sign. The
potential will therefore be zero at one point between the charges, where qi /ri has the same
magnitude for the two charges. There will be one other point beyond the weaker charge, where
this condition is fulfilled and the potential is zero.
3) Two equipotential lines cannot cross. That would indicate that a region in space had two
different values for the potential. For example, if a 40-V line and a 50-V line crossed, then the
potential at the point of crossing would be both 40 V and 50 V, which is impossible. Likewise,
the electric field is perpendicular to the equipotential lines. If two lines crossed, the electric
field at that point would point in two different directions simultaneously, which is not possible.
4) A positive charge must be placed in the center to compensate the effect of the negative
charges in the corners.
Equilibrium is achieved, if the electric field in the corners vanishes. The problem is symmetric, so we need to calculate the field only for one corner and also only one component of
it, here the horizontal or x component. Use the Pythagoras’ rule to calculate the length of the
diagonal of the square (∼ 17 cm) and to break up a diagonal vector in its x and y components,
√
each having 1/ 2 of the length of the diagonal vector.
We find with Q as the unknown charge in the center
Ex = −
14 µC
14 µC
2Q
1
1
−√
+√
=0
2
2
(12 cm)
2 2 (12 cm)
2 (12 cm)2
⇒
Q = (14 µC)
1
1
+√
4
2
!
' 13.4 µC
This is an unstable equilibrium, because a small displacement of the central charge makes it
feel a force that pulls it further away from the center, as some charges would then be closer.
5) The kinetic energy gained is equal to the work done on the electron by the electric field.
The potential difference must be positive for the electron to gain potential energy. Use Eq.
17-2b.
W = −q V = e V = 21, 000 eV = 3.36 · 10−15 J
6) The magnitude of the electric field can be found from Eq. 17-4b.
E=
V
120 V
=
' 3, 430 V/m
d
0.035 m
7) By the work energy theorem, the total work done, by the external force and the electric
field together, is the change in kinetic energy. The work done by the electric field is given by
Eq. 17-2b.
Wext + Welec = Ekin.,final − Ekin.,ini. = 0.5 mJ
Welec = −q (Vb − Va ) = 0.5 mJ − Wext = −1.5 mJ
8)
⇒ Vb − Va = −50 V
9) The electric potential is given by Eq. 17-5.
V =
1 Q
= 7.2 · 105 V
4π 0 r
The potential energy equals the work done to bring the charges together.
U=
1 Q1 Q2
' 1.2 · 10−13 J
4π 0 r
10) Just touching implies the distance between the two atoms equals the sum of their radii, i.e.
3.5 · 10−15 m. We assume that all of the energy the proton gains in being accelerated by the
voltage is changed to potential energy just as the protons outer edge reaches the outer radius
of the oxygen nucleus, so the required voltage is minus the difference in electric potential a)
touching and b) at infinity.
V =−
1 Qoxygen
= −3.3 · 106 V = −3.3 MV
4π 0
r