Download Chapter 5 Problem Set

Chapter 5 Problem Set
2. A charge of +2 X 10-6 C is 20 cm from a charge of -3 X 10-6 C.
Find the magnitude and
direction of the force on each charge.
We are asked to find the magnitude and direction of the electric force on each charge.
According to Coulomb’s Law, the force on each charge will be the same (the force exerted by
charge 1 on charge 2 is the same as the force exerted by charge 2 on charge 1).
We are given: the charges [(charge 1(Q1) = +2 X 10-6 C)(charge 2(Q2) = -3 X 10-6 C)], and the
distance between them [distance (R) = 20 cm = 0.20 m]
Relevant formula: F = K*(Q1Q2)/R2 where K is the electric force constant and
K = 9 X109 N*m2/C2
plugging in to this formula we get:
F = [(9 X 109 N*m2/C2) X (2 X 10-6 C * 3 X 10-6 C)]/(0.2 m)2
= [(9 X 109 N*m2/C2) X (6 X 10-12 C2)]/0.04 m2
= [5.4 X 10-2 N*m2]/4 X 10-2 m2 = 1.35 N towards each other
8. A hydrogen molecule consists of two hydrogen atoms whose nuclei are single protons.
Find the force between the two protons in a hydrogen molecule whose distance apart is 7.42
X 10-11 m. (The two electrons in the molecule spend more time between the protons than
outside them, which leads to attractive forces that balance the repulsion of the protons and
permit a stable molecule as we’ll see in Chap.9)
This problem is similar to problem 2 in that we are asked to find the force between the two
protons. We are given the distance between the protons and we know the charge is +1.6 X
10-19 C from our reading.
Again, the relevant formula is : F = K*(Q1Q2)/R2 where K is the electric force constant and
K = 9 X109 N*m2/C2
plugging in to this formula we get:
F = K*(Q1Q2)/R2 = [(9 X 109 N*m2/C2) X (1.6 X 10-19 C * 1.6 X 10-19 C)]/(7.42 X 10-11 m)2
F = [(9 X 109 N*m2/C2) X (2.56 X 10-38 C2)]/ 5.51 X 10-21 m2
F = [2.3 X 10-28 N*m2]/5.51 X 10-21 m2 = 5.44 X 10-8 N
10. Sensitive instruments can detect the passage of as few as 60 electrons/sec.
To what
current does this correspond?
They are asking for current (charge per unit time) which is expressed as amperes (1 ampere
= 1 coulomb/sec). We are given a flow rate of 60 electrons per second. We know from
reading that 1 electron has a charge of -1.6 X 10-19 C.
The relevant formula is I = Q/t (current = charge transferred/time)
plugging in to the equation:
I = Q/t = 60 electrons*(-1.6 X 10-19 C/electron)]/1 sec = 9.6 X 10-18 C/sec
I = 9.6 X 10-18 amperes
12. A 120-V electric coffeepot draws a current of 0.6 A. What is the resistance of its heating
element?
We are asked to solve for resistance (expressed as ohms []) and are given the voltage (120
V) and the current (0.6 A).
The relevant formula is: V = I*R (voltage = current X resistance) which we can rearrange
to solve for the resistance (R = V/I)
plugging in to the formula:
R = V/I = 120 V/0.6 A = 200  (ohms)
20. A 120-V electric motor draws 2.5 A. (a) How many coulombs of charge pass through it in
15 min? (b) How many joules of energy does it use in 15 min?
(a) We are asked to find the charge that passes through the motor in 15 min. We are given
the time (15 min) and the current (2.5 A). Recall, current is a measure of charge transferred
per unit time and 1 A = 1 C/sec. Converting 15 minutes into seconds:
15 min X 60 sec/min = 900 sec. so the time equals 900 sec.
The relevant formula is I = Q/t (current = charge transferred/time)
Rearranging to solve for charge:
Q = I*t = 2.5 A X 900 sec = 2.5 C/sec X 900 sec = 2250 C (coulombs)
(b) We are asked to solve for work (joules) done over a 15 min period. We are given the
voltage (120 V) and the current (2.5 A). Since voltage X current = power and power =
work/time, we can combine these two equations and solve for the amount of work done over
the time given (15 min which we can convert to seconds as we did in part a)(15 min = 900
sec)
Relevant formulas:
P = V*I (power = voltage X current)
P = W/t (power = work/time)
Setting power = power we get
V*I = W/t
Now, solving for work we get
W = V*I*t (work = voltage X current X time) = (120 V) X (2.5 A) X (900 sec)
W = (300 watts) X (900 sec) = (300 j/sec) X (900 sec) = 270,000 j (joules)
25. A transformer has a 600-turn primary coil and a 200-turn secondary coil.
If the
secondary voltage is 80 V, what is the primary voltage? If the transformer delivers 300 W,
what is the primary current?
(a) we are asked to solve for the primary voltage (V1) of the transformer and are given the
number of turns for the primary coil (N1 = 600 turns) and the number of turns for the
secondary coil (N2 = 200 turns), along with the secondary voltage (V2 = 80 V).
The relevant formula is
V2 = (N2/N1) X (V1) (secondary voltage = (secondary turns/primary turns) X primary voltage)
Rearranging to solve for V1
V1 = (N1/N2) X V2 = (600/200) X 80 V = 3 X 80 V = 240 V
(b) We are asked to solve for the current and are given the voltage (240 V) and the power
(300 W)
The relevant formula is
P = V*I (power = voltage X current)
Rearranging to solve for current
I = P/V = 300 W/240 V = 1.25 A (amperes)