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Homework
on Electrical Storage (solutions)

Feel free to work on the homework in groups.
The work you hand in, however, should reflect your understanding of the
material and be in your own words.Students
who turn in identical (or close to identical) homework assignments will
be asked to explain their answers orally to the TA or prof. A student
who cannot explain how he or she arrived at a given answer will be charged
with academic dishonesty.

You should show all of your calculations (neatly)
and justify all of your answers for full credit.

Capacitors and Electric
Fields

Two metal plates are separated by a
distance d=3.0 mm as shown below. The left plate has 8.0 mC (8 millicoulombs)
of extra positive charge, and the right plate has 8.0 mC of extra negative
charge. This charge distribution produces a uniform electric field
between the

plates having a magnitude of 600
N/C.

1.
In what direction does the electric
field between the plates point?

The electric field points from the positive
to the negative plate- left to right.

2.
If I place an electron at a point
halfway between the plates, which way will it move?

The electric field points in the direction of the
force that would be on a positive charge. An electron will move in the
opposite direction of the electric field because of its negative charge.
Therefore it will move toward the left. One could also think in
terms of the electron being attracted to the positively charged plate.

3.
What is the force on the electron
due to the electric field between the plates?

The magnitude of the force on a charge in an electric
field is given by the expression

F = qE = (1.6 x 10-19
C)(600 N/C) = 9.6 x 10-17 N.

4.
What is the potential difference
(voltage) between the plates?
V = E d;

(600 N/C) * (0.0030 m) = 1.8
V

5.
What is the capacitance of the
arrangement?

Q = C V, where Q is the charge
stored on one plate of the capacitor; (0.080 C) / (1.8 V) = 140
mF

6.
Through the magic of age-old technology
known as imagination, the spacing between the plates of the capacitor is
doubled while the capacitor is still hooked up to the power supply (so
the voltage across the plates does not change). What is the new (a)
capacitance, (b) charge on the capacitor, and (c) electric field between
the plates?

(a) Capacitance varies inversely
with distance, so the capacitance will be halved. 70 mF

(b) Q = C V; with voltage constant,
the charge will be halved because of our result for (a). 4.0 mC

(c) V = E x; since voltage is
constant, the field will be halved. 300 N/C

7.
Sketch the electric field lines
for the charge arrangement shown below. Be sure to appropriately
indicate both the direction and the magnitude of the electric field.
The negative charge on the left has a charge of -12.0 mC,
while the positive charge on the right has a charge of +6.0 mC