PHYS 2421
Final Exam
Fall 2009
NAME___________________________________
Problem 1 (10 points). Select True or false
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In an electric field, a positive charge will move perpendicularly to any equipotential surface.
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A charged conductor has a hole inside with a point charge in it, the E field inside the conductor is zero. .
A charged conductor has a hole inside, the E field in the hole’s surface is zero. .
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Capacitors connected in series always yield a smaller total capacitance. .
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An 80 W light bulb has a smaller resistor than a 40 W light bulb. .
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Capacitors connected in parallel store different amounts of charge.
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Three current-carrying parallel wires attract each other, thus all currents are in the same direction.
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If two parallel wires have equal currents in opposite directions the B field is never zero between the wires.
Velocity selectors (E and B fields crossed) also work with neutral particles.
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A Weber/m2 can be used as a unit of magnetic field.
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Problem 2 (10 points).
A) Determine the potential produced by the charges at the
point A. The triangle is equilateral.
B) Determine the energy needed to bring a charge q3 = 1.5 nC
from far away to point A.
6
6 the charge
C)Solution
The force
C) (magnitude
E=qV = 1.5and
×10−direction)
C −1.21felt
× 10by
V = −1.82 J
q3 = 1.5 nC placed at A due to the other charges.
(
)(
)
q1
q  8.988 ×109 Nm 2 /C2 
−6
+k 2 =
 ( 2 − 2 ) ×10 C = 0 V
d
d 
0.01 m

−6
B) E=qV = (1.5 ×10 C ) ( 0 V ) = 0 J
Solution A) d = 0.01 m, V = k
qq
qq
C) F = k 1 23 − cos 300 iˆ + sin 300 ˆj + k 2 2 3 cos 300 iˆ + sin 300 ˆj
d
d
k
k
k
= 2 − q1q3 + q2 q3 cos 300 iˆ + 2 ( q1q3 + q2 q3 ) sin 300 ˆj = 2 2 ( q1q3 ) sin 300 ˆj
d
d
d
9
2
2
8.988 ×10 Nm /C
=2j
2 ×10−9 ×1.5 ×10−9 C 2 sin 300 = 1.685 × 10−5 N j
2
( 0.04 m )
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(
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)
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Problem 3 (10 points). Determine the value of the EMF
ε.
Solution. Calling IL to the current in left branch, and traversing
the outer loop clockwise yields:
-3 IL + 24 - 1.8x7 = 0 IL = (24-1.8x7)/3 = 3.8 A.
Then, the current through middle branch is
IM = 3.8 – 1.8 = 2.0 A, downward.
Traversing the right loop clockwise yields:
+2x2 + e -1.8x7 = 0 e = 1.8x7 – 4 = 8.6 V
Problem 4 (10 points) Determine the magnitude and direction of the
current I2 that will produce a zero net magnetic field at the center of
the ring.
Solution: The field produced by the straight wire is B = µ0I1/2πD
and the field produced by the ring at its center is B = µ0I2/2R. To
have zero field at the center, I2 should be counterclockwise, then
BNet = µ0I1/2πD- µ0I2/2R = 0
I2 = RI1/πD = 0.6 x 3.5 / (3.14150 x 1.2) = 0.56 A
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Problem 5 (10 points) In the circuit shown both
capacitors are charged initially to 45 V.
A) How long after closing the switch will the potential
across each capacitor be dropped to 10 V?
B) What will be the current a that time?
Solution:
The equivalent capacitance is CEq = 15 µF + 20 µF = 35 µF, the equivalent resistance is
REq = 30 Ω + 50 Ω = 80 Ω, and the total charge accumulated is q0 = 35 µF x 45 V = 1575 µC.
When V = 10 V the total charge should be q = 35 µF x 10 V = 350 µC.
The charge will decrease as: q = q0 Exp[ - t / RC ] fl t = — RC Ln[ q/q0 ]
Then the time for when V =10 V will be
t = — (80 Ω)(35 µF) Ln[350 µC / 1575 µC] = = — (80 Ω)(35 µF) (—1.5) = 4200 µS = 0.0042 s
The current will be I = I0 Exp[- t / RC], where the initial current is I0 = 45 V / 80 Ω = 0.56 A.
Then I = (0.56 A) Exp[- (0.0042 s) / (80 Ω)(35 µF)] = 0.125 A