1.a.
At x = 1.5m, V = 0 =
𝑘(𝑞1 )
2.5
+
𝑘(2 𝑥 10−9 )
1.0
q1 = -5 x 10-9C
1.b.
1.c.
𝛥𝑉
4𝑉
|E| = 𝛥𝑅 = 0.2𝑚 = 20V/m
1.d.
𝑞
2 𝑥 10−9
ΦE = 𝜀 = 8.85 𝑥 10−12 = 226
0
𝑁·𝑚2
𝐶
1.e.i.
W = -ΔU = -q·ΔV = -(1.6 x 10-19)(-8V) = 1.28 x 10-18J
1.e.ii.
1
K = 1.28 x 10-18 = 2(1.67 x 10-27)v2
v = 39153 m/s
1.f.
Left. The electron accelerates antiparallel to the electric field and the electric field points
rightward, from higher to lower electric potential.
2.a.
It is the same through both. The circuit forms a single loop and the current is constant
throughout that loop.
2.b.
It depends upon the resistance of the sample. By Ohm’s law for a common current, the potential
difference will be proportional to the resistance. But the relative resistance is unknown for the
resistor and wire sample.
2.c.
Graph
ε - VR on the vertical axis and IR on the horizontal axis.
2.d.
0.35
y = 0.4845x - 0.0028
0.3
Voltage
0.25
0.2
0.15
0.1
0.05
0
0
0.1
0.2
0.3
0.4
Current
2.e.
R = 0.48Ω
2.f.
3
0.48 = ρ·𝜋·0.0012
ρ = 5.03 x 10-7 Ω·m
0.5
0.6
0.7
0.8
2.g.i.
Less than. If the ammeter is not ideal, it will have some resistance which will cause a decrease
𝜀 − 𝑉𝑅
in both I and VR. If r =
, then r will be greater than the actual value.
𝐼
2.g.ii.
Greater than. A non-ideal voltmeter has some less-than-infinite resistance, which decreases the
total resistance of the circuit. This allows a greater current to flow through the ammeter.
3.a.
Right.
3.b.i.
Less than. The counter-clockwise current induced in the loop is out of the page and opposes the
initial magnetic field.
3.b.ii.
Greater than. The current in the bar to the right creates a magnetic field below the bar which is
into the page and aligns with the initial magnetic field.
3.c.
M·a = -M·g 𝑑𝑣
𝑑𝑡
+
𝐵2 𝐿2 𝑣
𝑀𝑅
3.d.
vT =
−𝑀𝑔𝑅
𝐵2 𝐿2
𝐵2 𝐿2 𝑣
𝑅
+g=0
3.e.
P=
P=
𝜀2
𝑅
=
(𝐵𝐿𝑣)2
𝑅
𝑀 2 𝑔2 𝑅
𝐵2 𝐿2
3.f.
𝑑𝑣
𝑑𝑡
+
𝐵2 𝐿2 𝑣
𝑀𝑅
+g=0
Integrate with substitution
v=
−𝑀𝑔𝑅
𝐵2 𝐿2
𝐵2 𝐿2
(1- 𝑒 − 𝑅𝑀 𝑡 )