Content check — University Physics 2, Chapter 5: Electric Charges and Fields

Every problem in the chapter beside the book’s answer, a chapter at a time as you scroll.

Chapter 5: Electric Charges and Fields

69 problems · 22 figures

Check Your Understanding 5.1 §5.3 source ↗

What would be different if the electron also had a positive charge?

Answer

The force would point outward.

Check Your Understanding 5.2 §5.3 source ↗

What would be different if

q_{1}

were negative?

Answer

The net force would point

58 °

below the −x-axis.

Check Your Understanding 5.3 §5.4 source ↗

What is the electric field due to a single point particle?

Answer

\vec{E} =

\frac{1}{4 π ε_{0}} \frac{q}{r^{2}} \hat{r}

Check Your Understanding 5.4 §5.5 source ↗

How would the strategy used above change to calculate the electric field at a point a distance z above one end of the finite line segment?

Answer

We will no longer be able to take advantage of symmetry. Instead, we will need to calculate each of the two components of the electric field with their own integral.

Check Your Understanding 5.5 §5.5 source ↗

How would the above limit change with a uniformly charged rectangle instead of a disk?

Answer

The point charge would be

Q = σ a b

where a and b are the sides of the rectangle but otherwise identical.

Check Your Understanding 5.6 §5.5 source ↗

What would the electric field look like in a system with two parallel positively charged planes with equal charge densities?

Answer

The electric field would be zero in between, and have magnitude

\frac{σ}{ε_{0}}

everywhere else.

5-1 §5.1 source ↗

There are very large numbers of charged particles in most objects. Why, then, don’t most objects exhibit static electricity?

Answer

There are mostly equal numbers of positive and negative charges present, making the object electrically neutral.

5-3 §5.1 source ↗

A positively charged rod attracts a small piece of cork. (a) Can we conclude that the cork is negatively charged? (b) The rod repels another small piece of cork. Can we conclude that this piece is positively charged?

Answer

a. no; b. yes

5-5 §5.1 source ↗

How would you determine whether the charge on a particular rod is positive or negative?

Answer

Take an object with a known charge, either positive or negative, and bring it close to the rod. If the known charged object is positive and it is repelled from the rod, the rod is charged positive. If the positively charged object is attracted to the rod, the rod is negatively charged.

5-7 §5.2 source ↗

When a glass rod is rubbed with silk, it becomes positive and the silk becomes negative—yet both attract dust. Does the dust have a third type of charge that is attracted to both positive and negative? Explain.

Answer

No, the dust is attracted to both because the dust particle molecules become polarized in the direction of the silk.

5-9 §5.2 source ↗

Does the uncharged conductor shown below experience a net electric force?

A sphere is shown suspended by a thread from the ceiling. A negatively charged rod is brought near the sphere.

Answer

Yes, polarization charge is induced on the conductor so that the positive charge is nearest the charged rod, causing an attractive force.

5-11 §5.2 source ↗

Compare charging by conduction to charging by induction.

Answer

Charging by conduction is charging by contact where charge is transferred to the object. Charging by induction first involves producing a polarization charge in the object and then connecting a wire to ground to allow some of the charge to leave the object, leaving the object charged.

5-13 §5.2 source ↗

Trucks that carry gasoline often have chains dangling from their undercarriages and brushing the ground. Why?

Answer

This is so that any excess charge is transferred to the ground, keeping the gasoline receptacles neutral. If there is excess charge on the gasoline receptacle, a spark could ignite it.

5-15 §5.2 source ↗

Why do some clothes cling together after being removed from the clothes dryer? Does this happen if they’re still damp?

Answer

The dryer charges the clothes. If they are damp, the presence of water molecules suppresses the accumulation of excess charge by transferring electrons from negatively charged items to positively charged items, resulting in zero (or negligible) charges on every item.

5-17 §5.2 source ↗

Suppose someone tells you that rubbing quartz with cotton cloth produces a third kind of charge on the quartz. Describe what you might do to test this claim.

Answer

There are only two types of charge, positive and negative. All charged objects that are attracted to the cotton will be repelled by the quartz and all charged objects that are repelled by the cotton will be attracted to the quartz. Furthermore, bringing objects that are attracted to the quartz into contact with it will neutralize the charge on the object and quartz.

5-19 §5.2 source ↗

Suppose you place a charge q near a large metal plate. (a) If q is attracted to the plate, is the plate necessarily charged? (b) If q is repelled by the plate, is the plate necessarily charged?

Answer

a. No, since a polarization charge is induced. b. Yes, since the polarization charge would produce only an attractive force.

5-21 §5.3 source ↗

An atomic nucleus contains positively charged protons and uncharged neutrons. Since nuclei do stay together, what must we conclude about the forces between these nuclear particles?

Answer

The force holding the nucleus together must be greater than the electrostatic repulsive force on the protons.

5-23 §5.4 source ↗

When measuring an electric field, could we use a negative rather than a positive test charge?

Answer

Either sign of the test charge could be used, but the convention is to use a positive test charge.

5-25 §5.4 source ↗

If the electric field at a point on the line between two charges is zero, what do you know about the charges?

Answer

The charges are of the same sign.

5-27 §5.5 source ↗

Give a plausible argument as to why the electric field outside an infinite charged sheet is constant.

Answer

At infinity, we would expect the field to go to zero, but because the sheet is infinite in extent, this is not the case. Everywhere you are, you see an infinite plane in all directions.

5-29 §5.5 source ↗

Describe the electric fields of an infinite charged plate and of two infinite, charged parallel plates in terms of the electric field of an infinite sheet of charge.

Answer

The infinite charged plate would have

E = \frac{σ}{2 ε_{0}}

everywhere. The field would point toward the plate if it were negatively charged and point away from the plate if it were positively charged. The electric field of the parallel plates would be zero between them if they had the same charge, and E would be

E = \frac{σ}{ε_{0}}

everywhere else. If the charges were opposite, the situation is reversed, zero outside the plates and

E = \frac{σ}{ε_{0}}

between them.

5-31 §5.6 source ↗

If a point charge is released from rest in a uniform electric field, will it follow a field line? Will it do so if the electric field is not uniform?

Answer

yes; no

5-33 §5.6 source ↗

How would you experimentally distinguish an electric field from a gravitational field?

Answer

At the surface of Earth, the gravitational field is always directed in toward Earth’s center. An electric field could move a charged particle in a different direction than toward the center of Earth. This would indicate an electric field is present.

5-35 §5.6 source ↗

What is the ratio of the number of electric field lines leaving a charge 10q and a charge q?

Answer

5-37 §5.1 source ↗

Common static electricity involves charges ranging from nanocoulombs to microcoulombs. (a) How many electrons are needed to form a charge of −2.00 nC? (b) How many electrons must be removed from a neutral object to leave a net charge of

0.500 μ C

Answer

2.00 \times 10^{−9} C (\frac{1}{1.602 \times 10^{−19}} e/C) =

1.248 \times 10^{10} electrons

;
b.

0.500 \times 10^{−6} C (\frac{1}{1.602 \times 10^{−19}} e/C) =

3.121 \times 10^{12} electrons

5-39 §5.1 source ↗

To start a car engine, the car battery moves

3.75 \times 10^{21}

electrons through the starter motor. How many coulombs of charge were moved?

Answer

\frac{3.750 \times 10^{21} e}{6.242 \times 10^{18} e / C} =

–600.8 C

5-41 §5.1 source ↗

A 2.5-g copper penny is given a charge of

−2.0 \times 10^{−9} C

. (a) How many excess electrons are on the penny? (b) By what percent do the excess electrons change the mass of the penny?

Answer

2.0 \times 10^{−9} C (6.242 \times 10^{18} e / C) =

1.248 \times 10^{10} e

;

b.

9.109 \times 10^{−31} kg (1.248 \times 10^{10} e) =

1.137 \times 10^{−20} kg,

\frac{1.137 \times 10^{−20} kg}{2.5 \times 10^{−3} kg} =

4.548 \times 10^{−18} or 4.548 \times 10^{−16} %

5-43 §5.2 source ↗

Suppose a speck of dust in an electrostatic precipitator has

1.0000 \times 10^{12}

protons in it and has a net charge of −5.00 nC (a very large charge for a small speck). How many electrons does it have?

Answer

5.00 \times 10^{−9} C (6.242 \times 10^{18} e / C) =

3.121 \times 10^{10} e

;

3.121 \times 10^{10} e + 1.0000 \times 10^{12} e =

1.0312 \times 10^{12} e

5-45 §5.2 source ↗

A 50.0-g ball of copper has a net charge of

2.00 μ C

. What fraction of the copper’s electrons has been removed? (Each copper atom has 29 protons, and copper has an atomic mass of 63.5.)

Answer

atomic mass of copper atom times

1 u =

1.055 \times 10^{−25} kg

;
number of copper atoms

=

4.739 \times 10^{23} atoms

;
number of electrons equals 29 times number of atoms or

1.374 \times 10^{25} electrons

;

\frac{2.00 \times 10^{−6} C (6.242 \times 10^{18} e / C)}{1.374 \times 10^{25} e} =

9.083 \times 10^{−13} or 9.083 \times 10^{−11} %

5-47 §5.2 source ↗

How many coulombs of positive charge are there in 4.00 kg of plutonium, given its atomic mass is 244 and that each plutonium atom has 94 protons?

Answer

244.00 u (1.66 \times 10^{−27} kg / u) =

4.050 \times 10^{−25} kg

;

\frac{4.00 kg}{4.050 \times 10^{−25} kg} =

9.877 \times 10^{24} atoms

;

9.877 \times 10^{24} (94) =

9.284 \times 10^{26} protons

;

9.284 \times 10^{26} (1.602 \times 10^{−19} C / proton) =

1.487 \times 10^{8} C

5-49 §5.3 source ↗

Two charges

+3 μ C

and

+12 μ C

are fixed 1 m apart, with the second one to the right. Find the magnitude and direction of the net force on a −2-nC charge when placed at the following locations: (a) halfway between the two (b) half a meter to the left of the

+3 μ C

charge (c) half a meter above the

+12 μ C

charge in a direction perpendicular to the line joining the two fixed charges

Answer

a. charge 1 is $3 μ C$ ; charge 2 is $12 μ C$ , $F_{31} =$ $2.16 \times 10^{−4} N$ to the left,
$F_{32} =$ $8.63 \times 10^{−4} N$ to the right,
$F_{net} =$ $6.47 \times 10^{−4} N$ to the right;
b. $F_{31} =$ $2.16 \times 10^{−4} N$ to the right,
$F_{32} =$ $9.59 \times 10^{−5} N$ to the right,
$F_{net} =$ $3.12 \times 10^{−4} N$ to the right,

Three charges are shown. Charge 1 is a 3 micro Coulomb charge at the bottom left. Charge 2 is a 12 micro Coulomb charge at the bottom right, 1 meter to the right of charge 1. Charge 3 is a minus 2 nano Coulomb charge 0.5 meters above charge 2. The charges define a right triangle, with charge 2 at the right angle. The angle at the vertex with charge one is theta. The forces on charge three are shown. F 3 1 points down and to the left, toward charge 1. Force F 3 2 points vertically down.

;
c. ${\vec{F}}_{31 x} =$ $−3.86 \times 10^{−5} N \hat{i}$ ,
${\vec{F}}_{31 y} =$ $−1.93 \times 10^{−5} N \hat{j}$ ,
${\vec{F}}_{32 y} =$ $−8.63 \times 10^{−4} N \hat{j}$
$\begin{matrix} {\vec{F}}_{net} & = −3.86 \times 10^{−5} N \hat{i} - 8.82 \times 10^{−4} N \hat{j} \end{matrix}$

5-51 §5.3 source ↗

Protons in an atomic nucleus are typically

10^{−15} m

apart. What is the electric force of repulsion between nuclear protons?

Answer

F = 230.7 N

5-53 §5.3 source ↗

Point charges

q_{1} = 50 μ C

and

q_{2} = −25 μ C

are placed 1.0 m apart. What is the force on a third charge

q_{3} = 20 μ C

placed midway between

q_{1}

and

q_{2}

Answer

F = 53.94 N

5-55 §5.3 source ↗

Two small balls, each of mass 5.0 g, are attached to silk threads 50 cm long, which are in turn tied to the same point on the ceiling, as shown below. When the balls are given the same charge Q, the threads hang at $5.0 °$ to the vertical, as shown below. What is the magnitude of Q? What are the signs of the two charges?

Two small balls are attached to threads which are in turn tied to the same point on the ceiling. The threads hang at an angle of 5.0 degrees to either side of the vertical. Each ball has a charge Q.

Answer

The tension is

T = 0.049 N

. The horizontal component of the tension is

0.0043 N

d =

0.088 m,

q =

6.1 \times 10^{−8} C

.
The charges can be positive or negative, but both have to be the same sign.

5-57 §5.3 source ↗

The net excess charge on two small spheres (small enough to be treated as point charges) is Q. Show that the force of repulsion between the spheres is greatest when each sphere has an excess charge Q/2. Assume that the distance between the spheres is so large compared with their radii that the spheres can be treated as point charges.

Answer

Let the charge on one of the spheres be nQ, where n is a fraction between 0 and 1. In the numerator of Coulomb’s law, the term involving the charges is

n Q (1 - n) Q .

This is equal to

(n - n^{2}) Q^{2}

. Finding the maximum of this term gives

1 - 2 n = 0 \Rightarrow n = \frac{1}{2}

5-59 §5.3 source ↗

A charge $q = 2.0 μ C$ is placed at the point P shown below. What is the force on q?

Answer

Define right to be the positive direction and hence left is the negative direction, then

F = −0.05 N

5-61 §5.3 source ↗

Two fixed particles, each of charge

5.0 \times 10^{−6} C,

are 24 cm apart. What force do they exert on a third particle of charge

−2.5 \times 10^{−6} C

that is 13 cm from each of them?

Answer

The particles form triangle of sides 13, 13, and 24 cm. The x-components cancel, whereas there is a contribution to the y-component from both charges 24 cm apart. The y-axis passing through the third charge bisects the 24-cm line, creating two right triangles of sides 5, 12, and 13 cm.

F_{y} = 2.56 N

in the negative y-direction since the force is attractive. The net force from both charges is

{\vec{F}}_{net} =

−5.12 N \hat{j}

5-63 §5.3 source ↗

What is the force on the charge q at the lower-right-hand corner of the square shown here?

Charges are shown at the corners of a square with sides length a. All of the charges are positive and all are magnitude q.

Answer

The diagonal is

\sqrt{2} a

and the components of the force due to the diagonal charge has a factor

cos θ = \frac{1}{\sqrt{2}}

;

{\vec{F}}_{net} =

[k \frac{q^{2}}{a^{2}} + k \frac{q^{2}}{2 a^{2}} \frac{1}{\sqrt{2}}] \hat{i} -

[k \frac{q^{2}}{a^{2}} + k \frac{q^{2}}{2 a^{2}} \frac{1}{\sqrt{2}}] \hat{j}

5-65 §5.4 source ↗

A particle of charge

2.0 \times 10^{−8} C

experiences an upward force of magnitude

4.0 \times 10^{−6} N

when it is placed in a particular point in an electric field. (a) What is the electric field at that point? (b) If a charge

q =

−1.0 \times 10^{−8} C

is placed there, what is the force on it?

Answer

E = 2.0 \times 10^{2} \frac{N}{C}

up;
b.

F = 2.0 \times 10^{−6} N

down

5-67 §5.4 source ↗

Consider an electron that is

10^{−10} m

from an alpha particle

(q = 3.2 \times 10^{−19} C) .

(a) What is the electric field due to the alpha particle at the location of the electron? (b) What is the electric field due to the electron at the location of the alpha particle? (c) What is the electric force on the alpha particle? On the electron?

Answer

E =

2.88 \times 10^{11} N / C

;
b.

E =

1.44 \times 10^{11} N / C

;
c.

F =

4.61 \times 10^{−8} N

on alpha particle;

F =

4.61 \times 10^{−8} N

on electron

5-69 §5.4 source ↗

What is the electric field at a point where the force on a

−2.0 \times 10^{−6} −C

charge is

(4.0 \hat{i} - 6.0 \hat{j}) \times 10^{−6} N?

Answer

E =

(−2.0 \hat{i} + 3.0 \hat{j}) N/C

5-71 §5.4 source ↗

The electric field in a particular thundercloud is

2.0 \times 10^{5} N/C .

What is the acceleration of an electron in this field?

Answer

F =

3.204 \times 10^{−14} N

a =

3.517 \times 10^{16} m / s^{2}

5-73 §5.4 source ↗

If the electric field is

100 N/C

at a distance of 50 cm from a point charge q, what is the value of q?

Answer

q =

2.78 \times 10^{−9} C

5-75 §5.4 source ↗

(a) What is the electric field of an oxygen nucleus at a point that is

10^{−10} m

from the nucleus? (b) What is the force this electric field exerts on a second oxygen nucleus placed at that point?

Answer

E =

1.15 \times 10^{12} N / C

;
b.

F =

1.47 \times 10^{−6} N

5-77 §5.4 source ↗

Point charges

q_{1} = 50 μ C

and

q_{2} = −25 μ C

are placed 1.0 m apart. (a) What is the electric field at a point midway between them? (b) What is the force on a charge

q_{3} = 20 μ C

situated there?

Answer

If the

q_{2}

is to the right of

q_{1},

the electric field vector from both charges point to the right. a.

E =

2.70 \times 10^{6} N / C

;
b.

F = 54.0 N

5-79 §5.4 source ↗

Point charges

q_{1} =

q_{2} =

4.0 \times 10^{−6} C

are fixed on the x-axis at

x = −3.0 m

and

x = 3.0 m .

What charge q must be placed at the origin so that the electric field vanishes at

x = 0, y = 3.0 m ?

Answer

There is

45 °

right triangle geometry. The x-components of the electric field at

y = 3 m

cancel. The y-components give

E (y = 3 m) =

2.83 \times 10^{3} N / C

.
At the origin we have a a negative charge of magnitide

q =

−2.83 \times 10^{−6} C

5-81 §5.5 source ↗

Calculate the magnitude and direction of the electric field 2.0 m from a long wire that is charged uniformly at

λ =

4.0 \times 10^{−6} C/m .

Answer

\vec{E} (z) =

3.6 \times 10^{4} N / C \hat{k}

5-83 §5.5 source ↗

The charge per unit length on the thin rod shown below is $λ$ . What is the electric field at the point P? (Hint: Solve this problem by first considering the electric field $d \vec{E}$ at P due to a small segment dx of the rod, which contains charge $d q = λ d x$ . Then find the net field by integrating $d \vec{E}$ over the length of the rod.)

Answer

d E =

\frac{1}{4 π ε_{0}} \frac{λ d x}{{(x + a)}^{2}},

E =

\frac{λ}{4 π ε_{0}} [\frac{1}{l + a} - \frac{1}{a}]

5-85 §5.5 source ↗

Two thin parallel conducting plates are placed 2.0 cm apart. Each plate is 2.0 cm on a side; one plate carries a net charge of

8.0 μ C,

and the other plate carries a net charge of

−8.0 μ C .

What is the charge density on the inside surface of each plate? What is the electric field between the plates?

Answer

σ =

0.02 C / m^{2} E =

2.26 \times 10^{9} N / C

5-87 §5.5 source ↗

A total charge q is distributed uniformly along a thin, straight rod of length L (see below). What is the electric field at $P_{1} ? At P_{2} ?$

A horizontal rod of length L is shown. The rod has total charge q. Point P 1 is a distance a over 2 above the midpoint of the rod, so that the horizontal distance from P 1 to each end of the rod is L over 2. Point P 2 is a distance a to the right of the right end of the rod.

Answer

P_{1}

\vec{E} (y) =

\frac{1}{4 π ε_{0}} \frac{λ L}{y \sqrt{y^{2} + \frac{L^{2}}{4}}} \hat{j} \Rightarrow \frac{1}{4 π ε_{0}} \frac{q}{\frac{a}{2} \sqrt{{(\frac{a}{2})}^{2} + \frac{L^{2}}{4}}} \hat{j} =

\frac{1}{π ε_{0}} \frac{q}{a \sqrt{a^{2} + L^{2}}} \hat{j}

P_{2} :

Put the origin at the end of L.

d E =

\frac{1}{4 π ε_{0}} \frac{λ d x}{{(x + a)}^{2}},

\vec{E} =

-

\frac{q}{4 π ε_{0} l} [\frac{1}{l + a} - \frac{1}{a}] \hat{i}

5-89 §5.5 source ↗

Charge is distributed along the entire x-axis with uniform density

λ_{x}

and along the entire y-axis with uniform density

λ_{y} .

Calculate the resulting electric field at (a)

\vec{r} = a \hat{i} + b \hat{j}

and (b)

\vec{r} = c \hat{k} .

Answer

\vec{E} (\vec{r}) =

\frac{1}{4 π ε_{0}} \frac{2 λ_{y}}{a} \hat{i} +

\frac{1}{4 π ε_{0}} \frac{2 λ_{x}}{b} \hat{j}

; b.

\frac{1}{4 π ε_{0}} \frac{2 (λ_{x} + λ_{y})}{c} \hat{k}

5-91 §5.5 source ↗

A proton moves in the electric field

\vec{E} =

200 \hat{i} N/C .

(a) What are the force on and the acceleration of the proton? (b) Do the same calculation for an electron moving in this field.

Answer

\vec{F} =

3.2 \times 10^{−17} N \hat{i}

\vec{a} =

1.92 \times 10^{10} m / s^{2} \hat{i}

;
b.

\vec{F} =

−3.2 \times 10^{−17} N \hat{i}

\vec{a} =

−3.51 \times 10^{13} m / s^{2} \hat{i}

5-93 §5.5 source ↗

A spherical water droplet of radius

25 μ m

carries an excess 250 electrons. What vertical electric field is needed to balance the gravitational force on the droplet at the surface of the earth?

Answer

m =

6.5 \times 10^{−11} kg

E =

1.6 \times 10^{7} N / C

5-95 §5.5 source ↗

Shown below is a small sphere of mass 0.25 g that carries a charge of $9.0 \times 10^{−10} C .$ The sphere is attached to one end of a very thin silk string 5.0 cm long. The other end of the string is attached to a large vertical conducting plate that has a charge density of $30 \times 10^{−6} {C/m}^{2} .$ What is the angle that the string makes with the vertical?

A small sphere is attached to the lower end of a string. The other end of the string is attached to a large vertical conducting plate that has a uniform positive charge density. The string makes an angle of theta with the vertical.

Answer

E =

1.70 \times 10^{6} N / C

F =

1.53 \times 10^{−3} N T cos θ =

m g T sin θ =

q E

tan θ =

0.62 \Rightarrow θ =

32.0 °

,
This is independent of the length of the string.

5-97 §5.5 source ↗

Positive charge is distributed with a uniform density $λ$ along the positive x-axis from $r to \infty,$ along the positive y-axis from $r to \infty,$ and along a $90 °$ arc of a circle of radius r, as shown below. What is the electric field at O?

A uniform distribution of positive charges is shown on an x y coordinate system. The charges are distributed along a 90 degree arc of a circle of radius r in the first quadrant, centered on the origin. The distribution continues along the positive x and y axes from r to infinity.

Answer

circular arc

d E_{x} (− \hat{i}) =

\frac{1}{4 π ε_{0}} \frac{λ d s}{r^{2}} cos θ (− \hat{i})

{\vec{E}}_{x} =

\frac{λ}{4 π ε_{0} r} (− \hat{i})

d E_{y} (− \hat{i}) =

\frac{1}{4 π ε_{0}} \frac{λ d s}{r^{2}} sin θ (− \hat{j})

{\vec{E}}_{y} =

\frac{λ}{4 π ε_{0} r} (− \hat{j})

;
y-axis:

{\vec{E}}_{x} =

\frac{λ}{4 π ε_{0} r} (− \hat{i})

;
x-axis:

{\vec{E}}_{y} =

\frac{λ}{4 π ε_{0} r} (− \hat{j})

\vec{E} =

\frac{λ}{2 π ε_{0} r} (− \hat{i}) +

\frac{λ}{2 π ε_{0} r} (− \hat{j})

5-99 §5.5 source ↗

A particle of mass m and charge

− q

moves along a straight line away from a fixed particle of charge Q. When the distance between the two particles is

r_{0}, − q

is moving with a speed

v_{0} .

(a) Use the work-energy theorem to calculate the maximum separation of the charges. (b) What do you have to assume about

v_{0}

to make this calculation? (c) What is the minimum value of

v_{0}

such that

− q

escapes from Q?

Answer

W =

\frac{1}{2} m (v^{2} - v_{0}^{2})

\frac{Q q}{4 π ε_{0}} (\frac{1}{r} - \frac{1}{r_{0}}) =

\frac{1}{2} m (v^{2} - v_{0}^{2}) \Rightarrow r_{0} -

r =

\frac{4 π ε_{0}}{Q q} \frac{1}{2} r r_{0} m (v^{2} - v_{0}^{2})

; b.

r_{0} - r

is negative; therefore,

v_{0} > v

r \to

\infty,

and v \to

0 : \frac{Q q}{4 π ε_{0}} (- \frac{1}{r_{0}}) =

-

\frac{1}{2} m v_{0}^{2} \Rightarrow v_{0} =

\sqrt{\frac{Q q}{2 π ε_{0} m r_{0}}}

5-101 §5.6 source ↗

In this exercise, you will practice drawing electric field lines. Make sure you represent both the magnitude and direction of the electric field adequately. Note that the number of lines into or out of charges is proportional to the charges.

(a) Draw the electric field lines map for two charges $+ 20 μ C$ and $−20 μ C$ situated 5 cm from each other.

(b) Draw the electric field lines map for two charges $+ 20 μ C$ and $+ 20 μ C$ situated 5 cm from each other.

Answer

Figure a shows a positive 20 micro Coulomb charge on the left, a negative 20 micro Coulomb charge on the right, and the field lines due to the charges. The field lines come out of the positive charge and converge coming into the negative charge. The outer field lines extend beyond the drawing area and so we see them bending to the right, toward the negative charge, but only see part of the line. The density of lines coming out of the positive is the same as the density going into the negative. Figure b shows a positive 20 micro Coulomb charge on the left, a positive 20 micro Coulomb charge on the right, and the field lines due to the charges. The field lines come out of the positive charges and diverge, bending away from the far charge. The density of lines is the same near each of the charges. Figure c shows a positive 20 micro Coulomb charge on the left, a negative 30 micro Coulomb charge on the right, and the field lines due to the charges. The field lines come out of the positive charge. More lines go into the negative 20 micro Coulomb charge than come out of the positive 20 micro Coulomb charge. All of the lines coming out of the positive charge terminate at the negative, while the outer lines going into the negative start at infinity.

5-103 §5.6 source ↗

Two charges of equal magnitude but opposite sign make up an electric dipole. A quadrupole consists of two electric dipoles that are placed anti-parallel at two edges of a square as shown.

Four charges are shown at the corners of a square. At the top left is positive 10 nano Coulombs. At the top right is negative 10 nano Coulombs. At the bottom left is negative 10 nano Coulombs. At the bottom right is positive 10 nano Coulombs.

Draw the electric field of the charge distribution.

Answer

5-105 §5.7 source ↗

Consider the equal and opposite charges shown below. (a) Show that at all points on the x-axis for which $| x | ≫ a,$ $E \approx$ ${Q a / 2 π ε}_{0} x^{3} .$ (b) Show that at all points on the y-axis for which $| y | ≫ a,$ $E \approx$ ${Q a / π ε}_{0} y^{3} .$

Two charges are shown on the y axis of an x y coordinate system. Charge +Q is a distance a above the origin, and charge −Q is a distance a below the origin.

Answer

E_{x} = 0,

E_{y} =

\frac{1}{4 π ε_{0}} [\frac{2 q}{(x^{2} + a^{2})} \frac{a}{\sqrt{(x^{2} + a^{2})}}]

\Rightarrow x ≫ a \Rightarrow \frac{1}{2 π ε_{0}} \frac{q a}{x^{3}}

E_{y} =

\frac{q}{4 π ε_{0}} [\frac{2 y a + 2 y a}{{(y - a)}^{2} {(y + a)}^{2}}]

\Rightarrow y ≫ a \Rightarrow \frac{1}{π ε_{0}} \frac{q a}{y^{3}}

5-107 §5.7 source ↗

A water molecule consists of two hydrogen atoms bonded with one oxygen atom. The bond angle between the two hydrogen atoms is $104 °$ (see below). Calculate the net dipole moment of a hypothetical water molecule where the charge at the oxygen molecule is −2e and at each hydrogen atom is +e. The net dipole moment of the molecule is the vector sum of the individual dipole moment between the two O-Hs. The separation O-H is 0.9578 angstroms.

A schematic representation of the outer electron cloud of a neutral water molecule is shown. Three atoms are at the vertices of a triangle. The hydrogen atom has positive q charge and the oxygen atom has minus two q charge, and the angle between the line joining each hydrogen atom with the oxygen atom is one hundred and four degrees. The cloud density is shown as being greater at the oxygen atom.

Answer

The net dipole moment of the molecule is the vector sum of the individual dipole moments between the two O-H. The separation O-H is 0.9578 angstroms:

\vec{p} =

1.889 \times 10^{−29} Cm \hat{i}

5-109 §5.AP source ↗

What is the force on the $5.0 - μ C$ charge shown below?

The following charges are shown on an x y coordinate system: Minus 3.0 micro Coulomb on the x axis, 3.0 meters to the left of the origin. Positive 5.0 micro Coulomb at the origin. Positive 9.0 micro Coulomb on the x axis, 3.0 meters to the right of the origin. Positive 6.0 micro Coulomb on the y axis, 3.0 meters above the origin.

Answer

{\vec{F}}_{net} =

[−8.99 \times 10^{9} \frac{3.0 \times 10^{−6} (5.0 \times 10^{−6})}{{(3.0 m)}^{2}} - 8.99 \times 10^{9} \frac{9.0 \times 10^{−6} (5.0 \times 10^{−6})}{{(3.0 m)}^{2}}] \hat{i}

−8.99 \times 10^{9} \frac{6.0 \times 10^{−6} (5.0 \times 10^{−6})}{{(3.0 m)}^{2}} \hat{j} =

−0.06 N \hat{i} -

0.03 N \hat{j}

5-111 §5.AP source ↗

Four charged particles are positioned at the corners of a parallelogram as shown below. If $q = 5.0 μ C$ and $Q = 8.0 μ C,$ what is the net force on q?

Four charges are positioned at the corners of a parallelogram. The top and bottom of the parallelogram are horizontal and are 3.0 meters long. The sides are at a thirty degree angle to the x axis. The vertical height of the parallelogram is 1.0 meter. The charges are a positive Q in the lower left corner, positive 2 Q in the lower right corner, negative 3 Q in the upper left corner, and positive q in the upper right corner.

Answer

Charges Q and q form a right triangle of sides 1 m and

3 + \sqrt{3} m .

Charges 2Q and q form a right triangle of sides 1 m and

\sqrt{3} m .

\begin{matrix} F_{x} & = 0.049 N, \end{matrix}

F_{y} = 0.093 N

{\vec{F}}_{net} =

0.051 N \hat{i} +

0.093 N \hat{j}

5-113 §5.AP source ↗

A charge

q = −2.0 μ C

is released from rest when it is 2.0 m from a fixed charge

Q = 6.0 μ C .

What is the kinetic energy of q when it is 1.0 m from Q?

Answer

W = 0.054 J

5-115 §5.AP source ↗

Find the electric field at P for the charge configurations shown below.

In figure a, positive charge q is on the left, negative charge q is a distance a to the right of it. Point P is a distance a to the right of the negative charge q. In figure b, positive charge q is on the left, and a positive charge q is a distance a to the right of it. Point P is below the midpoint, a distance a from each of the charges so that the two charges and point P are at the vertices of an equilateral triangle whose sides are length a. In figure c, four charges are at the corners of a square whose sides are length a. The two top corners each have positive charge q. The two bottom corners each have negative charge q. Point P is at the center of the square.

Answer

\vec{E} =

\frac{1}{4 π ε_{0}} (\frac{q}{{(2 a)}^{2}} - \frac{q}{a^{2}}) \hat{i}

; b.

\vec{E} =

\frac{\sqrt{3}}{4 π ε_{0}} \frac{q}{a^{2}} (− \hat{j})

; c.

\vec{E} =

\frac{2}{π ε_{0}} \frac{q}{a^{2}} \frac{1}{\sqrt{2}} (− \hat{j})

5-117 §5.AP source ↗

Point charges are placed at the four corners of a rectangle as shown below: $q_{1} =$ $2.0 \times 10^{−6} C,$ $q_{2} =$ $−2.0 \times 10^{−6} C,$ $q_{3} =$ $4.0 \times 10^{−6} C,$ and $q_{4} =$ $1.0 \times 10^{−6} C .$ What is the electric field at P?

A rectangle is shown with a charge at each corner. The rectangle is 4.0 centimeters high and 6.0 centimeters wide. At the top left is a positive charge q 1. At the top right is a negative charge q 2. At the lower left is a positive charge q 3. At the lower right is a positive charge q 4. Point P is in the middle of the upper edge, 3.0 centimeters to the right of q 1 and 3.0 centimeters to the left of q 2.

Answer

{\vec{E}}_{net} =

{\vec{E}}_{1} +

{\vec{E}}_{2} +

{\vec{E}}_{3} +

{\vec{E}}_{4} =

(4.65 \hat{i} + 1.44 \hat{j}) \times 10^{7} N/C

5-119 §5.AP source ↗

A positive charge q is released from rest at the origin of a rectangular coordinate system and moves under the influence of the electric field

\vec{E} =

E_{0} (1 + x / a) \hat{i} .

What is the kinetic energy of q when it passes through

x = 3 a ?

Answer

F =

q E_{0} (1 + x / a) W =

\frac{1}{2} m (v^{2} - v_{0}^{2})

\frac{1}{2} m v^{2} =

q E_{0} (\frac{15 a}{2}) J

5-121 §5.AP source ↗

Charge is distributed uniformly along the entire y-axis with a density

λ_{y}

and along the positive x-axis from

x = a to x = b

with a density

λ_{x} .

What is the force between the two distributions?

Answer

Electric field of wire at x:

\vec{E} (x) =

\frac{1}{4 π ε_{0}} \frac{2 λ_{y}}{x} \hat{i}

d F =

\frac{λ_{y} λ_{x}}{2 π ε_{0}} (ln b - ln a)

5-123 §5.AP source ↗

Calculate the electric field due to a uniformly charged rod of length L, aligned with the x-axis with one end at the origin; at a point P on the z-axis.

Answer

A rod of length L is shown, aligned with the x-axis with the left end at the origin. A point P is shown on the z axis, a distance a above the left end of the rod. A small segment of the rod is labeled as d x and is a distance x to the right of the left end of the rod. The line from dx to point P makes an angle of theta with the x axis. The vector d E, drawn with its tail at point P, points away from the segment d x.

$d E_{x} =$ $\frac{1}{4 π ε_{0}} \frac{λ d x}{(x^{2} + a^{2})} \frac{x}{\sqrt{x^{2} + a^{2}}}$ ,
${\vec{E}}_{x} =$ $\frac{λ}{4 π ε_{0}} [\frac{1}{\sqrt{L^{2} + a^{2}}} - \frac{1}{a}] \hat{i}$ ,
$d E_{z} =$ $\frac{1}{4 π ε_{0}} \frac{λ d x}{(x^{2} + a^{2})} \frac{a}{\sqrt{x^{2} + a^{2}}}$ ,
${\vec{E}}_{z} =$ $\frac{λ}{4 π ε_{0} a} \frac{L}{\sqrt{L^{2} + a^{2}}} \hat{k}$ ,
Substituting z for a, we have:
$\vec{E} (z) =$ $\frac{λ}{4 π ε_{0}} [\frac{1}{\sqrt{L^{2} + z^{2}}} - \frac{1}{z}] \hat{i} +$ $\frac{λ}{4 π ε_{0} z} \frac{L}{\sqrt{L^{2} + z^{2}}} \hat{k}$

5-125 §5.AP source ↗

The charge per unit length on the thin rod shown here is $λ .$ What is the electric force on the point charge q? (See the preceding problem.)

A rod of length l is shown. The rod lies on the horizontal axis, with its center at the origin, so the ends are a distance of l over 2 to the left and right of the origin. A positive charge q is on the y axis, a distance a to above the origin.

Answer

There is a net force only in the y-direction. Let

θ

be the angle the vector from dx to q makes with the x-axis. The components along the x-axis cancel due to symmetry, leaving the y-component of the force.

d F_{y} =

\frac{1}{4 π ε_{0}} \frac{a q λ d x}{{(x^{2} + a^{2})}^{3 / 2}}

F_{y} =

\frac{1}{2 π ε_{0}} \frac{q λ}{a} [\frac{l / 2}{{({(l / 2)}^{2} + a^{2})}^{1 / 2}}]

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Problems and answers are adapted from University Physics Volume 2 by Samuel J. Ling, Jeff Sanny and William Moebs, OpenStax, and are used under the CC BY-NC-SA 4.0 licence. Access for free at https://openstax.org/books/university-physics-volume-2/pages/1-introduction. Illustrations: Copyright Rice University, OpenStax, under CC BY-NC-SA 4.0 license. This adaptation is released under the same licence, is non-commercial, and is neither affiliated with nor endorsed by OpenStax or Rice University.