Φ21 Fall 2006
1
HW6 Solutions
Problem 29.16
A proton with an initial speed of 800,000
m/s is brought to rest by an electric eld.
Part A. Did the proton
move into a region of higher potential or lower potential? Part B. What was the potential dierence that
stopped the proton?
Solution: Since the kinetic energy decreased, the potential energy must have increased. The proton has a
positive charge, so this is an increase in electric potential, i.e. moving to a region of
higher
potential.
Conservation of energy is used to say that the loss in kinetic energy equals the gain in potential energy. The
nal kinetic energy is zero while the initial kinetic energy is
K=
)(
)2
1
1(
mv 2 =
1.67 × 10−27 kg 8.0 × 105 m/s = 5.344 × 10−16 J
2
2
The change in electric potential is then found by
5.344 × 10−16 J = −∆K
∆V
2
= ∆U = q∆V
∆U
5.344 × 10−16 J
=
=
= 3340 V
q
1.6 × 10−19 C
Problem 29.46
What is the ratio
∆Vp /∆Ve
of the magnitudes of the potential dierences that will accelerate a proton and
an electron from rest to Part A. the same nal speed and Part B. the same nal kinetic energy?
Solution: To solve this problem, start with the requested expression and substitute for the variables until
only the given information is in the expression. The required formulas are
K = 12 mv 2 .
∆V = q∆U , −∆K = ∆U ,
−e 1 mp vp2
∆Vp
qp ∆Up
e∆Up
−e∆Kp
mp
=
=
=
= 12
= −1833
=−
2
∆Ve
qe ∆Ue
−e∆Ue
e∆Ke
me
e 2 me ve
(The book asks for the magnitude.) For part B, use the fourth fraction to say that
3
and
|∆Vp /∆Ve | = 1.
Equipotential Surfaces in a Capacitor
Part A. Is the electric potential energy of a particle with charge
q
the same at all points on an equipotential
surface?
Solution: Yes. That's what equipotential means.For a particle with charge
at potential
V,
the electric potential energy has a constant value
q
on an equipotential surface
qV .
Part B. What is the work required to move a charge around on an equipotential surface at potential
V
with
constant speed?
Solution:
W = 0.
Work changes the energy in a system. So
Ei + W = Ef
or
∆K + ∆U = W .
Since the
speed of the charge is described as constant and it moves along the equipotential surface, both the kinetic
and potential energy are constant and there is no work.
1
Part C. What is the work done by the electric eld on a charge as it moves along an equipotential surface
at potential V?
Solution: Work done by the electric eld = 0. The work done by the electric eld is equal to the change in
the electric potential energy, which is zero as the charge moves along the equipotential surface.
Part D. The work
WE
done by the electric eld
~
E
in displacing a particle with charge
q
along the path
d~
is given by
~ cos θ
~ · d~ = q|E||
~ d|
WE = q E
where theta is the angle between
equipotential surface, what must
Solution:
θ = π/2rad.
θ
~
E
d~.
WE
and
be for
Since in general,
~
E
is not equal to zero, for points on an
to equal 0? Express your answer in radians.
This shows equipotential surfaces are always perpendicular to the electric eld.
Now assume that a parallel-plate capacitor is attached across the terminals of a battery as shown in the gure. The electric eld
~
E
in the
region between the plates points in the negative z direction, from higher
to lower voltage.
Part E. Find the electric potential
V (x, y, z)
inside the capacitor if the origin of the coordinate
is at potential 0.
variables
E , x, y ,
Solution:
~ = (x, y, z)
X
system O = (0, 0, 0)
at a point
Express your answer in terms of some or all of the
and
z.
V (x, y, z) = E · z
is constant as
x
and
the proper derivative to correspond to the given
~
E
y
change an also has
eld.
The equation of an equipotential surface at a potenial
z=
V0
is given by
Figure 1:
and its
~
E
Sketch of a capacitor
eld direction.
V0
E
This is the equation of a plane that is parallel to the plates of the capacitor and perpendicular to the electric
eld. In particular, the lower plate, which is at zero potential, corresponds to the surface z = 0.
Part F. What is the distance
your answer in terms of
Solution:
∆z = ∆V /E
E
and
∆z between
∆V .
two surfaces separated by a potential dierence
derives directly from the solution to the previous part.
2
∆V ?
Express
4
Stopping the Proton
(Note: This problem has randomized numbers for each student.)
An innitely long line of charge has a linear charge density of
6.00 × 10−12
C/m. A proton is at distance 12.5
cm from the line and is moving directly toward the line with speed 1200 m/s. How close does the proton
get to the line of charge?
Solution: The electric eld near an innite line charge is directed away from the positive charge with a
magnitude calculated by Gauss's Law to be
E=
λ
2π²0 r
Since the force on the proton is non-uniform, the distance travelled is most easily calculated by conservation
of energy.
∆K + ∆U
(
)
1
2
0 − mP vi + e∆V
2
The electric potential
∆V
= 0
= 0
is calculated by integrating the electric eld.
∫
∆V
~ · d~l
E
= −
∫
=
rf
λ
dr
2π²
0r
ri
−λ
rf
ln
2π²0
ri
= −
The distance of closest approach is
1
mv 2
2
−mv 2 π²0
λ
rf
−λ
rf
ln
2π²0
ri
rf
= ln
ri
−mv 2 π²0
= ri exp
eλ
= 0.117 m
= e
3