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BROCK UNIVERSITY
Mid-term Test 1: June 2013
Course: PHYS 1P22/1P92
Examination date: 26 June 2014
Time of Examination: 18:00–18:50
Number of
Number of
Number of
Instructor:
pages: 4 (+ formula sheet)
students: 38
hours: 1
S. D’Agostino
A formula sheet is attached at the end of the test paper. No other aids are permitted except
for a non-programmable, non-graphing calculator.
Solve all problems in the space provided.
Total number of marks: 20
SOLUTIONS
1. Two 2.0-cm-diameter disks spaced 2.0 mm apart form a parallel-plate capacitor. The
electric field between the disks is 5.0 × 105 V/m.
(a) [2 marks] Determine the voltage across the capacitor.
(b) [2 marks] Determine the charge on each disk.
(c) [2 marks] An electron is launched from the negative plate. It strikes the positive
plate at a speed of 2.0 × 107 m/s. Determine the electron’s speed as it left the
negative plate.
Solution: (a) Because the electric field is constant,
V = Ed = 5 × 105 2 × 10−3 = 1000 V
(b) The capacitance of the capacitor is
2
C=
(8.85 × 10−12 ) π (1 × 10−2 )
ε0 A
=
= 1.39 × 10−12 F = 1.39 pF
d
2 × 10−3
The magnitude of the charge on each disk is therefore
Q = C∆V = 1.39 × 10−12 (1000) = 1.39 × 10−9 C = 1.39 nC
(c) By the principle of conservation of mechanical energy,
∆K + ∆U = 0
∆K = −∆U
1 2 1 2
mv − mvi = q∆V
2 f
2
1
m vf2 − vi2 = q∆V
2
2q∆V
vf2 − vi2 =
m
2q∆V
vi2 = vf2 −
m
2 (1.6 × 10−19 ) (1000)
−7 2
2
−
vi = 2 × 10
9.11 × 10−31
13
2
vi = 4.87 × 10
vi = 6.98 × 106 m/s
vi = 7.0 × 106 m/s
2. At a distance r from a point charge, the electric potential is 3000 V and the magnitude
of the electric field is 2.0 × 105 V/m.
(a) [2 marks] Determine the distance r.
(b) [3 marks] Determine the electric potential and the magnitude of the electric field
a distance r/2 from the point charge.
Solution: (a) Using the expressions for the potential near a point charge and the
magnitude of the electric field near a point charge, we can solve for the distance r:
V =
KQ
= 3000 V
r
and
E=
KQ
= 2 × 105 V/m
r2
Dividing the two quantities in the previous line, we obtain
KQr−1
KQr−2
r−1
r−2
r
r
=
V
E
3000
2.0 × 105
= 1.5 × 10−2 m
= 1.5 cm
=
(b) Using the same sort of proportional reasoning that was used in Part (a),
Vr/2
KQ(r/2)−1
2/r
=
=
=2
−1
Vr
KQr
1/r
Thus,
Vr/2 = 2Vr = 2(3000) = 6000 V
Similarly,
Er/2
KQ(r/2)−2
(2/r)2
=
=
= 22 = 4
Er
KQr−2
(1/r)2
Thus,
Er/2 = 4Er = 4 2.0 × 105 = 8.0 × 105 V/m
3. [4 marks] 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).
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
~1 =
cos θ, − 2 sin θ = 2 (cos θ, − sin θ)
E
2
r
r
r
The electric field created by the negative charge at the point (10 mm, 0 mm) is
KQ
KQ
KQ
~
E2 = − 2 cos θ, − 2 sin θ = 2 (− cos θ, − sin θ)
r
r
r
~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 )
(0.199)
E=
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.
4. [5 marks] Circle the best answer in each case.
(a) A lightweight neutral metal ball hangs from a string. When a charged metal rod
is brought near the ball, the ball moves towards the rod, touches the rod, and
then
i.
ii.
iii.
iv.
v.
sticks to the rod.
moves away from the rod.
moves back and forth, touching the rod repeatedly.
[It depends on whether the charge on the rod is positive or negative.]
[Not enough information is provided.]
Answer: (ii)
(b) A metal rod A and a metal sphere B touch each other while each is on an insulating
stand. A positively charged rod is brought near (but not touching) the end of
rod A that is farther from B. While the charged rod is still close, A and B are
separated. The charged rod is then moved far away. Sphere B is then
i.
ii.
iii.
iv.
v.
positively charged.
negatively charged.
neutral and unpolarized.
neutral and polarized.
[Not enough information is provided.]
Answer: (i)
(c) The electric field has zero magnitude throughout some region of space. In the
same region of space,
i.
ii.
iii.
iv.
v.
the electric potential is constant and must have the value 0.
the electric potential is constant, but could have any value.
the electric potential varies linearly with position.
the electric potential varies inversely as 1/r.
[Not enough information is provided.]
Answer: (ii)
(d) The electric potential is zero throughout some region of space. In the same region
of space,
i.
ii.
iii.
iv.
v.
the electric field is constant and must have magnitude 0.
the electric field is constant, but could have any magnitude.
the electric field varies linearly with position.
the electric field varies inversely as 1/r.
[Not enough information is provided.]
Answer: (i)
(e) The electric potential is 300 V at x = 0 cm, and −100 V at x = 5 cm, and varies
linearly with x. If a positive charge is released from rest at x = 2.5 cm, and is
subject only the force described by the electric potential, the charge will
i.
ii.
iii.
iv.
move to the right.
move to the left.
remain at rest at x = 2.5 cm.
[Not enough information is provided.]
Answer: (i)