Part D: Problem-Solving
37. Two objects are located along the y-axis. Object A has a charge of +25 µC and is located at the origin. Object B has a
charge of -16 µC and is located a distance of 62 cm above object A. Determine the y-coordinate location where the
electric field is zero.
Answer: 310 cm
This problem involves determining a location along the axis where two vectors are directed in opposite direction and have
the same magnitude. At such a location the electric field created by object A equals the electric field created by object B.
EA = E B
kQA/RA2 = kQB/RB2
A diagram is always a useful way of approaching the problem. Since A is positively
charged, the electric field created by A is directed away from it. Since B is negativelycharged, the electric field created by B is directed towards. Thus, some directional
reasoning would lead to the conclusion that the location along the axis must
be outside of the two charges. And since object B has the smaller charge and creates the
smaller E field, the location will be closer to object B than to object A. That is, what
object B lacks in terms of charge it must make up for in terms of distance in order to
have an equal field to object A's. The results of this reasoning are shown in the diagram
at the right.
The location of 0 electric field is a distance of x cm from the -16 µC charge and a distance of 62+x cm from the +25µC
charge. These distances and charge values can be inserted into the equation above and algebra can be used to solve for x.
(Note that the + and - sign has been dropped from the equations; directional meaning has been accounted for in the
reasoning above.)
k•(25 µC)/(62 + x)2 = k•(16 µC)/(x)2
Divide each side through by k.
(25 µC)/(62 + x)2 = (16 µC)/(x)2
Take the square root of each side of the equation and then cross-multiply.
5/(62 + x) = 4/x
5x = 4•(62 + x) = 248 + 4x
x = 248
Since x represents the distance from object B and since object B is at a y-coordinate of 62 cm, the answer to the problem
is 310 cm.
38. Three objects are located along the x-axis. Object A with a charge of +5.6 µC is located at the origin. Object B has a
charge of -4.2 µC and is located at the -1.2 m location. Object C has a charge of +7.7 µC and is located at the +2.4 m
location. Determine the magnitude and direction of the net electric force acting upon object A.
Answer: 0.21 N, left
Object A experiences two forces - the force of object B attracting it and pulling it leftward (FBA) and the force of object C
repelling it and pushing it leftward (FCA). The diagram below depicts the situation and the forces.
The net electric force upon object A is simply the vector sum of these two forces. The strategy involves computing each
force (FBA and FCA ) individually and then adding them together as vectors. The calculations of the two forces are shown
below. (Note that the + and - signs have been dropped from the values on the charge since the directional information
has already been accounted for in the analysis above.)
FBA = k•QA•QB/d2 = (9.0x109 N/m2/C2)•(5.6x10-6 C)•(4.2 x10-6 C)/(1.2 m)2 = 0.147 N, left
FCA = k•QA•QC/d2 = (9.0x109 N/m2/C2)•(5.6x10-6 C)•(7.7 x10-6 C)/(2.4 m)2 = 0.067 N, left
Now the net electric force on object A can be determined by adding the two forces resulting from the interactions with
A and with B. The net force is 0.21 N, left. (0.214 N rounded to two significant digits)
39. Object A has a x-y coordinate position of (+5.0, 0.0). Object B has a x-y coordinate location of (0.0, +4.0). If object A
has a charge of -5.8 µC and object B has a charge of +8.9 µC, then what is the resultant electric field strength at the
origin.
Answer: 5.4 x103 N/C, 293 degrees counter-clockwise from east
Both charged object A and charged object B create an electric field, the strength of
which is dependent upon the distance from the charged object. This problem
involves computing the electric field strength as created by both charges individually
and as experienced at the origin (EA and EB). Since electric field strength is a vector, it
can be added using rules of vector addition. An electric field vector always has a
direction associated with the direction which a + test charge would be forced. Since
object A is a negatively charged vector, the vector EA is directed towards it (i.e.,
rightwards). Since object B is a positively charged vector, the vector EB is directed
away from it (i.e., downwards). Once computed individually, EA and EB will be added.
The diagram at the right depicts the situation. The calculations below represent the mathematics.
EA = kQA/RA2 = (9.0x109 N/m2/C2)•(5.8x10-6 C)/(5.0 m)2 = 2.088 x 103 N/C
EB = kQB/RB2 = (9.0x109 N/m2/C2)•(8.9x10-6 C)/(4.0 m)2 = 5.006 x 103 N/C
Now the two E values will be added as vectors. The sketch at the right depicts the addition. Since they
are at right angles to each other, the Pythagorean theorem will be used to determine the magnitude
and trigonometry will be used to determine the direction.
Enet = SQRT[(2.088 x 103 N/C)2 + (5.006 x 103 N/C)2]
Enet = SQRT[2.942 x 106 N2/C2] = 5.4 x103 N/C
Theta = invtan(EA/EB) = invtan[(2.088 x 103 N/C)/(5.006 x 103 N/C)] = 22.6 deg.
So the direction of the vector is ~23 degrees to the right of down (east of south) or simply 293 degrees counterclockwise from east.
40. For the situation described in Question #39, what would be the magnitude of the net electric force upon a +1.3 µC
charge if placed at the origin.
Answer: 0.0070 N
This is simply a matter of using the definition of electric field - the electric field is the force per coulomb of charge.
Expressed mathematically,
E = F/q
and rearranged, the force is given by the equation
F = E•q
By substitution,
F = (5.4 x103 N/C)•(1.3x10-6 C) = 0.0070 N
41. Alpha particles (i.e., Helium nucleus) have a molar mass of 4.002 g/mol and consist of two protons and two neutrons.
a. Determine the charge of one alpha particle in units of Coulombs and the mass of one alpha particle in units of
kg.
b. Suppose that Ernie Rufferthord (not to be confused with the scientist of gold foil fame) wishes to suspend an
alpha particle in midair by attracting it to a bundle of electrons held a distance of 1.00-m above the alpha
particle. How many electrons would Ernie need in his bundle to accomplish such an amazing feat?
Answer: approx. 141 electrons
a. This involves a conversion using Avogadro's number and the molar mass:
mass = 1 alpha particle • (1 mol/6.02 x 1023 alpha particles) • (4.002 g/1 mol) • (1 kg/1000 g)
mass = 6.65 x 10-27 kg
b. Ernie wishes to use the force of electrical attraction between the alpha particle and the electron bundle to balance
the downward pull of gravity upon the alpha particle. So
Felect = Fgrav
k • Qalpha particle • Qe- bundle/R2 = malpha particle • g
The charge on the alpha particle can be determined from its composition - it contains 2 protons (and 2 neutral
neutrons), each having a charge of 1.6x10-19 C. So the charge of an alpha particle is 3.2x10-19 C. The only unknown
remaining is the charge on the electron bundle. Rearrangement of the equation yields the following expression for the
charge on the electron bundle:
Qe- bundle = malpha particle • g • R2 / (k • Qalpha particle)
Substitution of known values into this equation yields 2.26x10-17 C as the charge on the electron bundle. With each
electron contributed 1.6x10-19 C per electron, there must be approximately 141 electrons in the bundle.
42. A 1.19-gram charged balloon hangs from a 1.99-m string which is attached to the ceiling. A Van de Graaff generator
is located directly below the location where the string attaches to the ceiling and is at the same height as the balloon.
The string is deflected at an angle of 32.0 degrees from the vertical due to the presence of the electric field. Determine
the charge on the Van de Graaff generator if the charge on the balloon is 2.27x10-12 C.
Answer: 0.397 C
The balloon in this problem is experiencing an electric force resulting from the interaction with the Van de Graaff
generator. Diagram A illustrates the situation. The amount of charge on the Van de Graaff generator will be related to
the distance of separation (R), the electrical force of repulsion, and the charge on the balloon (which is given). So the
problem-solving strategy involves determining the electrical force of repulsion using a force analysis and then using
Coulomb's law to determine the charge on the Van de Graaff generator.
Diagram B above shows a free-body diagram for the balloon. The three forces must balance since the balloon is at rest.
Therefore, the x-component of the tension force (Fx in Diagram C) must equal the electrical force (Felect). And similarly,
the y-component of the tension force (Fy in Diagram C) must equal the gravity force (Fgrav). A trigonometric function
relating the angle of 32.0 degrees to these components of forces can be written.
tan(32.0 deg) = opposite/adjacent = Fx /Fy
And due to the equivalency between the components and the other two forces, it can also be stated that:
tan(32.0 deg) = Felect / Fgrav
The equation can be rearranged to solve for Felect; the expression for Fgrav and the related values can be substituted.
Felect = Fgrav • tan(32.0 deg) = m•g•tan(32.0 deg) = (0.00119 kg)•(9.8 m/s/s)•tan(32.0 deg)
Felect = 7.29 x 10-3 N
Now the separation distance (R) between the balloon and the generator can be determined. Observe in Diagram A that
the distance R is related to the length of the string and the angle with the vertical. Using the sine function, R can be
determined as follows:
sin(32.0 deg) = opposite/hypotenuse = R/1.99 m
R = 1.99 m • sin(32.0 deg) = 1.05 m
Now with Felect, R, and Q1 known, Q2 (the charge on the Van de Graaff generator) can be determined.
Felect = k•Q1•Q2/R2
By rearranging, the equation becomes
Q2 = Felect • R2 / (k•Q1) = (7.29 x 10-3 N) • (1.05)2 / [ (9.0x109 N•m2/C2) • (2.27x10-12 C)]
Q2 = 0.397 C
43. Ignoring trace elements, a typical elemental composition (by mass) of the human body is as follows:
10.5 % H
66.3 % O
19.9 % C
3.3 % N
a. Use these percentages, the molar mass values, Avogadro's number, and the atomic numbers to determine the
total number of electrons (and protons) in a 73-kg human body (160 pounds).
b. If these electrons and protons were placed 100 m apart (the distance of approximately one football field), then
what would be the force of electrical attraction between them.
Answer: See answers below.
a. How many electrons are in your body? It's kind of a neat question; and remarkably, a student of Chemistry and
Physics knows enough to come up with an estimate.
The strategy involves first using the mass of the body, the percent mass values, the atomic mass values, and Avogadro's
number to first determine the number of atoms of each element in the body; then use the atomic number of each
element to determine the number of electrons contributed by each element (assuming each atom is neutral); finally add
all the electrons to determine the total number of electrons. Here it goes:
When added, there are 24.3x1027 electrons (and a presumed equal number of protons).
b. The force of attraction of these protons and electrons at a separation distance of 100 m can easily be calculated if the
net + and net - charge is known. Simply multiply the number of electrons by the charge of one electron (1.6x10-19 C); do
similarly for the protons. The charge ends up being 3.9x109 C. Now Coulomb's law can be used to compute the electric
force of attraction:
Felect = k•Q1•Q2/R2 = (9.0x109 N•m2/C2)•(3.9x109 C)•(3.9x109 C)/(100 m)2 = 1.4x1025 N