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BROCK UNIVERSITY
Test 1: January 2015
Course: PHYS 1P22/1P92
Examination date: 24 January 2015
Time of Examination: 11:00–11:50
Number of pages: 5 + formula sheet
Number of students: 107
Instructor: S. D’Agostino
No aids are permitted except for a non-programmable, non-graphics calculator.
Solve all problems in the space provided.
Total number of marks: 20
SOLUTIONS
1. [4 points] A parallel-plate capacitor has a capacitance of 7.0 µF when filled with a
dielectric. The area of each plate is 1.5 m2 and the separation between the plates is
1.0 × 10−5 m. Determine the dielectric constant of the dielectric in the capacitor.
Solution:
κε0 A
d
Cd
κ=
ε0 A
(7.0 × 10−6 )(1.0 × 10−5 )
κ=
(8.85 × 10−12 )(1.5)
κ = 5.3
C=
2. [4 points] The drawing shows an electron entering the lower left side of a parallel
plate capacitor and exiting at the upper right side. The initial speed of the electron is
7.00×106 m/s. The capacitor is 2.00 cm long, and its plates are separated by 0.150 cm.
Assume that the electric field between the plates is uniform everywhere and determine
its magnitude.
Solution: A full solution is contained at the end of the Chapter 18 lecture notes. The
result is that the magnitude of the electric field is 2100 N/C.
Be aware of an important rounding issue that came up in some students’
work on this problem. Read the full solution in the lecture notes for an
explanation of this issue.
C
~v1
6.
3
m
m
A
3.5 mm
3. [4 marks] Particle A has positive charge 3.2
nC and is fixed in place. Particle B has
positive charge 8.3 nC and is also fixed in
place. Particle C has positive charge 5.9 nC,
is initially located as indicated in the figure, has initial speed 5.3 km/s, and has mass
9.4 × 10−12 kg.
m
4.3 m
B
(a) Determine the electric potential at the initial position of Particle C.
(b) Determine the initial electric potential energy of Particle C.
(c) Determine the speed of Particle C when it is far from Particles A and B.
Solution: A full solution is contained at the end of the Chapter 19 lecture notes. The
results are:
(a) 1.9 × 104 V
(b) 1.1 × 10−4 J
(c) 7.2 km/s
4. Answer each question briefly and clearly, in at most a few sentences. Your explanation
may include formulas or diagrams, if you wish. Remember, brevity and clarity are
courtesy.
(a) [2 points] You have three metal rods with identical size, shape, and composition.
Rod A has a charge of +6 units, and Rods B and C are neutral. Explain how to
get a negative charge on one of the rods using only the three rods present.
Solution: Place Rods B and C in contact. Bring Rod A close to, but not touching,
Rod B. This causes negative charge to flow from Rod C to Rod B. While Rod
A is still close, separate Rods B and C, leaving a negative charge on Rod B and
positive charges on Rods A and C.
This works provided that all the rods are insulated from the environment during
the process.
(b) [2 points] Some long-distance transport trucks have metal strips attached to
them that drag along the road. Explain their purpose.
Solution: Rubbing between tires and road may allow large charges to build up
on the truck after a long journey. Then when the driver steps out of the truck,
while holding on to the metal frame, a large spark may occur, which could set
flammable contents of the truck afire. The metal strip prevents charge buildup
on the truck by discharging continuously.
(c) [2 points] Rank the positions according to their potentials, in order from highest
to lowest. The electric field is uniform.
−
−
+
A
+
B
−
C
−
−
+
D
+
E
−
F
−
+
+
+
−
+
Solution: The electric field points from the positive plate to the negative plate
and points in the direction of decreasing potential. Thus, the closer a point is to
the positive plate, the higher its potential. Thus, in order of decreasing potential,
VB > VE > VC = VF > VD > VA
(d) [2 points] Rank the positions according to their potentials, in order from highest
to lowest. The electric field is caused by a positively charged particle indicated
by the “⊕”.
A
B
⊕
C
D
F
E
Solution: The electric field points radially outwards, away from the central point
charge, and points in the direction of decreasing potential. Thus, the closer a point
is to the centre, the higher its potential. Thus, in order of decreasing potential,
the positions are
VC = VD > VF > VE > VB > VA