Download Electric Fields and Forces (Chapter 20

Brock University
Physics 1P22/1P92
Winter 2014
Dr. D’Agostino
Solutions for Tutorial 5: Electric Fields and Forces (Chapter 20)
1. When I take clothes out of a clothes-dryer right after it stops, some of the clothes stick
to me. Is my body charged? If so, how did it get a charge? If not, then why does this
happen? [2 points]
Solution: My body is almost certainly neutral. However, all the rubbing of the clothes
with each other and with the metal drum of the dryer transfers charge so that some of
the clothes end up being charged. When one of the charged pieces of clothing touches
my uncharged body, my body becomes polarized, and then there is an attraction
between the charged clothing and my uncharged body.
The same effect is operative when you rub a balloon on your hair and then it sticks to a
wall. The same effect also allows pollen to stick to bees, as explained in the textbook.
2. A light-weight neutral metal ball hangs by a thread. When a charged metal rod is held
near, the ball moves towards the rod, touches the rod, and then quickly “flies away”
from the rod. Explain each step in this behaviour. [2 points]
Solution: When the charged rod is brought near the ball, charge polarization takes
place in the ball, resulting in attraction between the ball and the rod. When the ball
touches the rod, charge is transferred between the rod and the ball so that they each
end up with a net charge of the same sign. Because the rod and ball have net charges
of the same sign, they repell each other.
3. A typical commercial aircraft is struck by lightning about once per year. When this
happens, the external metal skin of the airplane might be burned, but the people
and equipment inside the aircraft experience no ill effects. Explain why this is so. [2
points]
Solution: When lightning strikes, there is a tremendous transfer of charge from the
cloud to the object struck. When lightning strikes the metal plane, this charge is
distributed over the surface of the plane. There will initially be movement of charges
over the surface of the plane (over a very short time interval) but a situation of static
equilibrium will quickly be established.
Since we have established that there is no electric field inside a conductor, the passengers and equipment inside the plane will experience no effect.
Similarly, if you are inside a car struck by lightning, you will be safe. The tires will
most likely be damaged as this tremendous amount of charge moves through them to
ground.
4. An electric dipole is formed from ±1 nC point charges spaced 2.0 mm apart. The
dipole is centred at the origin, oriented along the y-axis. Determine the strength and
direction of the electric field at the point (10 mm, 0 mm). [4 points]
Solution: Assume that the positive charge is at (0, 1.0 mm) and the negative charge is
at (0, −1.0 mm). Of course, you could also make the opposite assumption, because the
statement of the problem doesn’t specify this; if you make the opposite assumption,
the electric field in the result will have the same magnitude but opposite direction.
Draw a diagram!
In your diagram, you’ll notice a key angle,
1
−1
= 5.71◦
θ = tan
10
The distance r from each of the charges to the point (10 mm, 0 mm) at which we wish
to calculate the field satisfies
2
2
r2 = 1 × 10−3 + 10 × 10−3 = 1.01 × 10−4 m2
The electric field created by the positive charge at the point (10 mm, 0 mm) is
KQ
KQ
KQ
~
E1 =
cos θ, − 2 sin θ = 2 (cos θ, − sin θ)
2
r
r
r
The electric field created by the negative charge at the point (10 mm, 0 mm) is
KQ
KQ
KQ
~2 = −
E
cos
θ,
−
sin
θ
=
(− cos θ, − sin θ)
r2
r2
r2
~1 + E
~ 2:
By the principle of superposition, the total electric field is the vector sum E
~ =E
~1 + E
~2
E
~ = KQ (cos θ, − sin θ) + KQ (− cos θ, − sin θ)
E
r2
r2
~ = KQ (cos θ − cos θ, − sin θ − sin θ)
E
r2
~ = KQ (0, −2 sin θ)
E
r2
The magnitude of the electric field is
KQ
(2 sin θ)
r2
(8.99 × 109 ) (1 × 10−9 )
E=
(0.199)
1.01 × 10−4
E = 18 kV/m
E=
The electric field is in the −y direction.
Remember that if you placed the dipole in the opposite direction to the one chosen
here then the magnitude of the electric field will be the same as calculated here, but
the direction of the electric field will be in the +y-direction.