Hafez a radi, john o rasmussen auth principles of physics for scientists and engineers 2 27

19.3 Coulomb’s Law

643

Both the inverse square laws describe a property of interacting objects where
charges are involved in one case and masses in the other. The laws differ in that
the electrostatic forces between two charged particles may be either attractive or
repulsive, but gravitational forces are always attractive.

−F

r

r

r
+
q1

+
q2

F

−F

–
q2

–
q1

(a)

F

+
q1

−F

F

(b)

–
q2

(c)

Fig. 19.5 In (a) and (b), two charged particles of the same sign repel. In (c), two charged particles of
different signs attract each other. Notice that in all cases, the exerted forces are equal in magnitude but
opposite in direction

r
m2

m1

F 21
F12

Fig. 19.6 Newton’s law of universal gravitation states that the gravitational force between two objects
of masses m1 and m2 is attractive. The magnitude of the force F12 exerted on object 1 by object 2 is equal
→

→

to the magnitude of the force F21 exerted on object 2 by object 1. Note that F 12 = −F 21 .

The electrostatic constant k in Coulomb’s law has the value:
k = 8.9875 × 109 N.m2 /C2 ≈ 9 × 109 N.m2 /C2

(19.3)

For historical reasons and for the aim of simplifying many other formulas, the constant k is usually written as:
k=
where the quantity
value:

◦

1
4π

◦

(19.4)

(called the permittivity constant of free space) has the

644

19 Electric Force
◦

= 8.8542 × 10−12 C2 /N.m2

(19.5)

Any positive or negative charge q that can be detected is written as:
q = ne, n = ±1, ±2, ±3, . . .

(19.6)

where e is the smallest unit charge in nature1 and has the value:
e = 1.60219 × 10−19 C

(19.7)

As introduced earlier, the charge of an electron is −e and of a proton is +e. Therefore,
the number N of electrons or protons in 1 C is:
N=

1C
= 6.24 × 1018 electrons or protons
1.60219 × 10−19 C

(19.8)

Table 19.1 lists the charges and masses of the three elementary particles: the
electron, the proton, and the neutron.
Table 19.1 Charge and mass of the electron, proton and neutron.
Particle

Charge (C)
(= −1.60219 × 10−19

Mass (kg)
9.1095 × 10−31

Electron (e)

−e

Proton (p)

+e (= +1.60219 × 10−19 C)

1.67261 × 10−27

Neutron (n)

0

1.67492 × 10−27

C)

Example 19.1

Consider the three point charges q1 = +2 μC, q2 = −5 μC, and q3 = +8 μC that
are shown in Fig. 19.7. (a) Find the resultant force exerted on the charge q2 by
the two charges q1 and q3 . (b) In a different layout (see Fig. 19.8), q2 experiences
a resultant force of zero. Find the position of q2 and find the magnitude of each
force exerted on q2 .
Solution: (a) Because q2 is negative and both q1 and q3 are positive, the forces
→
and F23 are both attractive as displayed in Fig. 19.7. From Coulomb’s law we
can find F21 as follows:

→
F21

1

No charge smaller than e has yet been detected on a free particle. Recent theories propose the
existence of particles called quarks having charges −e/3 and +2e/3 inside nuclear matter. Although
a significant number of recent experiments indicate the existence of quarks inside nuclear matter,
free quarks have not been detected yet.


19.3 Coulomb’s Law

645

1m

+

5m

–

F 21

q1 = +2 μC

+

F 23

q2 = −5 μ C

q3 = + 8 μ C

Fig. 19.7

6m - x

x

+

F21

q 1 = +2 μ C

–
q2 =− 5 μ C

+

F23

q 3 = +8 μ C

Fig. 19.8

F21 = k

−6 C)(2 × 10−6 C)
|q2 | |q1 |
9
2 2 (5 × 10
=
9
×
10
N.m
/C
= 0.09 N
r2
(1 m)2

Also, we can find F23 as follows:
F23 = k

−6 C)(8 × 10−6 C)
|q2 | |q3 |

9
2 2 (5 × 10
=
9
×
10
N.m
/C
= 0.0144 N
r2
(5 m)2

Since F21 is greater than F23 , the resultant force F exerted on q2 will be toward
the charge q1 , i.e. to the left. Therefore:
F = F21 − F23 = 0.09 N − 0.0144 N = 0.0756 N
(b) When the resultant force on q2 is zero, the magnitudes of F21 and F23 must
be equal. Based on Fig. 19.8, the equality of the two forces F21 and F23 leads to
the following steps:
F21 = F23 =⇒ k

|q2 | |q1 |
|q2 | |q3 |
|q1 |
|q3 |
=k
=⇒ 2 =
x2
(6 m − x)2
x
(6 m − x)2

We can now substitute the given values of q1 and q3 and this yields the
following:
2 × 10−6 C 8 × 10−6 C
=
=⇒ (6 m − x)2 = 4x 2 =⇒ 6 m − x = 2x =⇒ x = 2 m
x2
(6 m − x)2


646

19 Electric Force

From Coulomb’s law we can find either of the value of F21 or the value of F23 as
follows:
F23 = F21 = k

−6 C)(2 × 10−6 C)
|q2 | |q1 |
9
2 2 (5 × 10
=
9
×
10
N.m
/C
= 0.0225 N
x2

(2 m)2

Example 19.2

In the classical model of the hydrogen atom proposed by Niels Bohr, the electron
rotates around a stationary proton in a circular orbit with an approximate radius
r = 0.053 nm, see Fig. 19.9. (a) Find the magnitude of the electrostatic force of
attraction, Fe , between the electron and the proton. (b) Find the magnitude of the
gravitational force of attraction, Fg , between the electron and the proton, and then
find the ratio Fe /Fg .
Fig. 19.9

Electron

Electron
me

r

Fe
Fe

Proton
Electrostatic
attraction

r
mp

Fg

Fg

Proton
Gravitational
attraction

A classical model of the hydrogen atom

Solution: (a) From Coulomb’s law, the magnitude of the electrostatic force of
→
attraction Fe between the electron and the proton is:
Fe = k

−19 C)2
| − e| |e|
9
2 2 (1.6 × 10
=
9
×
10
N.m
/C
= 8.2 × 10−8 N
r2
(0.053 × 10−9 m)2

(b) From Newton’s law of gravitation, the magnitude of the gravitational force
→
of attraction Fg between the two particles is:

Fg = G

me mp
r2

= 6.67 × 10−11 (N.m2 /kg2 )

(9.11 × 10−31 kg)(1.67261 × 10−27 kg)
(0.053 × 10−9 m)2

= 3.6 × 10−47 N
The ratio Fe /Fg ≈ 2.3 × 1039 . Thus, for elementary particles the gravitational
force is negligible compared to the electrical forces.


19.3 Coulomb’s Law

647

Example 19.3

Two identical copper coins of mass m = 2.5 g contain about N = 2 × 1022 atoms
each. A number of electrons n are removed from each coin to acquire a net positive
charge q. Assume that when we place one of the coins on a table and the second
above the first, the second coin stays at rest in air at a distance of 1 m, see Fig.
19.10. (a) Find the value of q that keeps the two coins in that configuration. (b)
Find the number of removed electrons n from each coin. (c) Find the fraction of
the copper atoms that lost those n electrons in each coin. Assume that each copper
atom loses only one electron.
Fig. 19.10

Coin

F
q

1m

mg
Coin
q

Solution: (a) The upper coin is in equilibrium due to its weight and the electrostatic
repulsion between the two charged coins. Therefore:
mg=k
q=

q×q
r2
m g r2
=
k

(2.5 × 10−3 kg)(9.8 N/kg)(1 m)2
= 1.65 × 10−6 C
9 × 109 N·m2 /C2

This small charge leads to a measurable force between large bodies.
(b) From the electronic charge (−e) and the total charge q on each coin, we
can find the number of removed electrons n as follows:
n=

q
1.65 × 10−6 C
=
≈ 1013 electrons (Very big number)
e
1.6 × 10−19 C

(c) The fraction of the copper atoms that loses the n electrons is:
f =

1013
n
=
= 5 × 10−10 (Very small fraction)
N
2 × 1022


648

19 Electric Force

Example 19.4

Two identical tiny spheres of mass m = 2 g and charge q hang from non-conducting
strings, each of length L = 10 cm. At equilibrium, each string makes an angle
θ = 5◦ with the vertical, see Fig. 19.11a. Find the magnitude of the charge on
each sphere.
Fig. 19.11

L

L

θ
q

θ

θ

T

x

q

T cos θ

T sin θ
q

r

Fe

mg
(b)

(a)

Solution: To analyze this problem, we draw the free-body diagram for the right
sphere as shown in Fig. 19.11b. This sphere is in equilibrium under the tensional
→

→

force T from the string, the electric force Fe from the left sphere, and the grav→
itational force m→
g . After decomposing the tensional force T in the vertical and
horizontal directions, we apply the condition of equilibrium as follows:
Fx = Fe − T sin θ = 0

⇒

Fy = T cos θ − mg = 0

⇒

Fe = T sin θ
T cos θ = mg

Eliminating T from the above two equations, we get the value of Fe :
Fe = mg tan θ = (2 × 10−3 kg)(9.8 N/kg)(tan 5◦ ) = 1.7 × 10−3 N
From Fig. 19.11a, we find the distance r between the two charges:
r = 2 x = 2 L sin θ = 2(0.1 m)(sin 5◦ ) = 0.017 m
Applying Coulomb’s law, we find the magnitude of the charge to be:
Fe = k

|q| |q|
r2

⇒

|q| =

Fe r 2
=
k

(1.7 × 10−3 N)(0.017 m)2
9 × 109 N.m2 /C2

= 7.39 × 10−9 C

Note that the charges of the two spheres could be positive or negative.


19.3 Coulomb’s Law

649

Example 19.5

Consider three charges q1 = +12 μC, q2 = +6 μC, and q3 = −4 μC are setup
as shown in Fig. 19.12. Find the resultant force exerted on the charge q2 by the
two charges q1 and q3 .
q3 = − 4 μ C

y

F23

F
F21

–

θ

60 cm

x
+ q1 = + 12 μ C

+
q2 = + 6 μ C

90 cm

Fig. 19.12

→

Solution: Because q1 and q2 are positive, while q3 is negative, the force F21
→
is repulsive and the force F23 is attractive as displayed in Fig. 19.12. From
Coulomb’s law we can find F21 as follows:
F21 = k

−6 C)(12 × 10−6 C)
|q2 | |q1 |
9
2 2 (6 × 10
=
9
×
10
N.m
/C
r2
(0.9 m)2

= 0.8 N
Similarly, we can find F23 as follows:
F23 = k

(6 × 10−6 C)(4 × 10−6 C)
|q2 | |q3 |
= 9 × 109 N.m2 /C2
2
r
(0.6 m)2

= 0.6 N
→

→

Since F21 is perpendicular to F23 , we can use the Pythagorean theorem to find
→

the magnitude of the resultant force F , and we can use Fig. 19.12 to find its
direction. Thus:
F=

2 + F2 =
F21
23

θ = tan−1

F23
F21

(0.8 N)2 + (0.6 N)2 =

0.64 N2 + 0.36 N2 = 1 N

= tan−1 (0.75) = 36.9◦
→

We can also write the resultant force F in vector form as follows:
→

→

→

→

→

F = −F21 i + F23 j = −0.8 i + 0.6 j

N


650

19 Electric Force

Example 19.6

Consider three charges q1 = +5 μC, q2 = +10 μC, and q3 = −2 μC are setup as
shown in Fig. 19.13. Find the resultant force exerted on the charge q2 by the two
charges q1 and q3 .
q1 = + 5 μ C

+

y
x

50 cm
40 cm

q3 = − 2 μ C

q2 = + 10 μ C

θ

–

30 cm

+

F23

θ

F21 sinθ

F2x

F21 cos θ

φ

q2

+
F2y

F21

F

Fig. 19.13

→

Solution: Because q1 and q2 are positive, while q3 is negative, the force F 21 is
→
repulsive and the force F 23 is attractive as displayed in Fig. 19.13. From Coulomb’s
law we can find F21 and F23 as follows:
−6 C)(5 × 10−6 C)
|q2 | |q1 |
9
2 2 (10 × 10
=
9
×
10
N.m
/C
= 1.8 N
r2
(0.5 m)2
−6 C)(2 × 10−6 C)
|q2 | |q3 |
9
2 2 (10 × 10
=k
=
9
×

10
N.m
/C
=2N
r2
(0.3 m)2

F21 = k
F23

Using the coordinate system shown in Fig. 19.13, we have:
30 cm
− 2 N = −0.92 N
50 cm
40 cm
= −1.44 N
= −F21 sin θ = −(1.8 N)
50 cm

F2x = F21 cos θ − F23 = (1.8 N)
F2y

Resultant: F =

2 + F 2 = (−0.92 N)2 + (−1.44 N)2 = 1.7 N
F2x
2y
|F
2y |
= tan−1 (1.565) = 57.4◦

Direction: φ = tan−1
|F2x |
→
We can also write the resultant force F in vector form as follows:
→

→

→

→

→

F = −F2x i + F2y j = −0.92 i + 1.44 j

N


19.4 Exercises

19.4

651

Exercises

Section 19.1 Electric Charge
(1) Explain what is meant by the following: (a) a neutral atom, (b) a negatively
charged atom, and (c) a positively charged atom.

(2) A neutral rubber rod is rubbed with fur as shown in Fig. 19.14. After rubbing,
what would be the charge on each of these items? Is it possible to transfer
positive charges from one of them to the other? Why so or why not?
Fig. 19.14 See Exercise (2)

Neutral
rubber

Neutral
fur

Before rubbing

Section 19.2 Charging Conductors and Insulators
(3) If we repeat the experiment illustrated in Fig. 19.3 but instead of using a charged
plastic rod, we use a charged rubber rod, what will the final charge on the copper
rod be?
(4) A charged plastic comb often attracts small bits of dry paper, as shown in the
left part of Fig. 19.15. After a while, the bits of paper fall down, as shown in
the right part of Fig. 19.15. Explain this observation.
Fig. 19.15 See Exercise (4)


652

19 Electric Force

(5) In an oxygen-enriched atmosphere (as in hospital operation rooms), workers
must wear special conducting shoes and avoid wearing rubber-soled shoes.
Explain the reason behind this.

(6) A negatively charged balloon clings to a wall as shown in the right part of Fig.
19.16. Does this mean that the wall is positively charged? Why does the balloon
fall afterwards?
Fig. 19.16 See Exercise (6)

Charged
balloon

Before

After

(7) Using a charged rubber rod, show the steps of how two uncharged metallic
spheres mounted on insulating stands can be electrostatically charged with
equal amount of charges, but opposite in sign.

Section 19.3 Coulomb’s Law
(8) How many electrons exist in a −1 C charge? What is the total mass of these
electrons?
(9) Find the magnitude of the electrostatic force between two 1 C charges separated
by a distance (a) 1 cm, (b) 1 m, and (c) 1 km, if such a configuration could be
set up. Are these forces substantial forces? Do they indicate that the coulomb
is a very large unit of charge?
(10) Find the magnitude of the force between two electrons when they are separated
by 0.1 nm (a typical atomic dimension).
(11) The uranium nucleus contains 92 protons. How large a repulsive force would
a uranium proton experience when it is 0.01 nm from the nucleus center? (The
nucleus can be treated as a point charge since the nuclear radius is of the order
of 10−14 m).
(12) Two electrically neutral spheres are 0.1 m apart. When electrons are moved

from one of the spheres to another, an attractive force of magnitude 10−3 N is
established between them. How many electrons were transferred?


19.4 Exercises

653

(13) Silver has 47 electrons per atom and a molar mass of 107.87 kg/kmol. An
electrically neutral pin of silver has a mass of 10 g. (a) Calculate the number of
electrons in the silver pin (Avogadro’s number is 6.022 × 1026 atoms/kmol).
(b) Electrons are added to the pin until the net charge is −1 mC. How many
electrons are added for every billion (109 ) electrons in the neutral atoms?
(14) Two protons in an atomic nucleus are separated by 2 × 10−15 m (a typical internuclear dimension). (a) Find the magnitude of the electrostatic repulsive force
between the protons. (b) How does the magnitude of the electrostatic force
compare to the magnitude of the gravitational force between the two protons?
(15) Two particles have an identical charge q and an identical mass m. What must
the charge-mass ratio, q/m, of the two particles be if the magnitude of their
electrostatic force equals the magnitude of the gravitational force.
(16) Two equally charged pith balls are at a distance r = 3 cm apart, as shown in
Fig. 19.17. Find the magnitude of the charge on each ball if they repel each
other with a force of magnitude 2 × 10−5 N. Does the answer give you any hint
about the exact sign of each charge? Explain.
Fig. 19.17 See Exercise (16)

q

q

r

(17) Two point charges q1 and q2 are 3 m apart, and their combined charge is 40 μC.
(a) If one repels the other with a force of 0.175 N, what are the two charges?
(b) If one attracts the other with a force of 0.225 N, what are the two charges?
(18) Two point charges q1 = +4 μC and q2 = +6 μC are 10 cm apart. A point charge
q3 = +2 μC is placed midway between q1 and q2 . Find the magnitude and
direction of the resultant force on q3 .
(19) Three 4 μC point charges are placed along a straight line as shown in Fig. 19.18.
Calculate the net force on each charge.
(20) Three point charges q1 = q2 = q3 = −4 μC are located at the corners of an
equilateral triangle as shown in Fig. 19.19. (a) Calculate the magnitude of


654

19 Electric Force

the net force on any one of the three charges. (b) If the charges are positive, i.e. q1 = q2 = q3 = +4 μC, would this change the magnitude calculated in
part a?
Fig. 19.18 See Exercise (19)

q 1 =+ 4 C

q 2 =+ 4 C

q 3 =+ 4 C

+

+

+

3m

3m

Fig. 19.19 See Exercise (20)

q1

–
1m

1m

–
q

3

–

1m

q

2

(21) Three point charges q1 = +2 μC, q2 = −3 μC, and q3 = + 4 μC are located at
the corners of a right angle triangle as shown in Fig. 19.20. Find the magnitude

and direction of the resultant force on q3 .
Fig. 19.20 See Exercise (21)

q1 +

y
x

5cm
q2 –

30°

+ q3

(22) Three equal point charges of magnitude q lie on a semicircle of radius R as
√
shown in Fig. 19.21. Show that the net force on q2 has a magnitude kq2 / 2R2
and points downward away from the center C of the semicircle.
(23) Four equal point charges, q1 = q2 = q3 = q4 = +3 μC, are placed at the four
corners of a square that has a side a = 0.4 m, see Fig. 19.22. (a) Find the force


19.4 Exercises

655

on q1 . (b) Find the force exerted on a test charge of 1 C placed at the center P
of the square.
Fig. 19.21 See Exercise (22)

q1

q3

R

C

R

q2

Fig. 19.22 See Exercise (23)

q1

+

+ q4

y

x

P

q2

+ a = 0.4m + q 3

(24) Four equal point charges q1 = q2 = q3 = q4 = −1 μC, are located as shown in
Fig. 19.23. (a) Calculate the net force exerted on the charge q4 , which is located
midway between q1 and q3 . (b) Calculate the magnitude and direction of the
net force on the charge q2 .
Fig. 19.23 See Exercise (24)

q1

–

y
x

–

1m
q2

–

1m

q4

–

q3

(25) A negative point charge of magnitude q is located on the x-axis at point x = −a,

and a positive point charge of the same magnitude is located at x = +a, see Fig.
19.24. A third positive point charge q◦ is located on the y-axis with a coordinate


656

19 Electric Force

(0, y). (a) What is the magnitude and direction of the force exerted on q◦ when
it is at the origin (0,0)? (b) What is the force on q◦ when its coordinate is (0, y)?
(c) Sketch a graph of the force on q◦ as a function of y, for values of y between
−4a and +4a.
Fig. 19.24 See Exercise (25)

y

+ + qo
y

r−
-q

–

θ

r+
θ

+q

+

0

x

(a,0 )

(-a,0 )

(26) In the Bohr model of the hydrogen atom, an electron of mass m = 9.11 ×
10−31 kg revolves about a stationary proton in a circular orbit of radius
r = 5.29 × 10−11 m, see Fig. 19.25. (a) What is the magnitude of the electrical
force on the electron? (b) What is the magnitude of the centripetal acceleration
of the electron? (c) What is the orbital speed of the electron?
Fig. 19.25 See Exercise (26)

Electron

r

Fe
Fe

Proton

(27) In the cesium chloride crystal (CsCl), eight Cs+ ions are located at the corners
of a cube of side a = 0.4 nm and a Cl− ion is at the center, see Fig. 19.26. What
is the magnitude of the electrostatic force exerted on the Cl− ion by: (a) the

eight Cs+ ions?, (b) only seven Cs+ ions?
(28) Two positive point charges q1 and q2 are set apart by a fixed distance d and
have a sum Q = q1 + q2 . For what values of the two charges is the Coulomb
force maximum between them?


19.4 Exercises

657

Fig. 19.26 See Exercise (27)

a
Cs
Cl

(29) Two equal positive charges q are held stationary on the x-axis, one at x = − a
and the other at x = +a. A third charge +q of mass m is in equilibrium at x = 0
and constrained to move only along the x-axis. The charge +q is then displaced
from the origin to a small distance x a and released, see Fig. 19.27. (a) Show
that +q will execute a simple harmonic motion and find an expression for its
period T. (b) If all three charges are singly-ionized atoms (q = q = +e) each of
mass m = 3.8 × 10−25 kg and a = 3 × 10−10 m, find the oscillation period T.

Fig. 19.27 See Exercise (29)

+ q′

+q
0

+q

x

a

a

(30) Two identical small spheres of mass m and charge q hang from non-conducting
strings, each of length L. At equilibrium, each string makes an angle θ with
the vertical, see Fig. 19.28. (a) When θ is so small that tan θ sin θ , show that
the separation distance r between the spheres is r = (Lq2 /2π ◦ mg)1/3 . (b) If
L = 10 cm, m = 2 g, and r = 1.7 cm, what is the value of q?

Fig. 19.28 See Exercise (30)

θ

L

q

L

θ

x

x

r

q


658

19 Electric Force

(31) For the charge distribution shown in Fig. 19.29, the long non-conducting massless rod of length L (which is pivoted at its center) is balanced horizontally
when a weight W is placed at a distance x from the center. (a) Find the distance
x and the force exerted by the rod on the pivot. (b) What is the value of h when
the rod exerts no force on the pivot?
Fig. 19.29 See Exercise (31)

L
x
+2q

+q
h

h

W

Pivot

+Q

+Q

(32) Two small charged spheres hang from threads of equal length L. The first sphere
has a positive charge q, mass m, and makes a small angle θ1 with the vertical,
while the second sphere has a positive charge 2 q, mass 3m, and makes a smaller
angle θ2 with the vertical, see Fig. 19.30. For small angles, take tan θ θ and
assume that the spheres only have horizontal displacements and hence the
electric force of repulsion is always horizontal. (a) Find the ratio θ1 /θ2 . (b)
Find the distance r between the spheres.
Fig. 19.30 See Exercise (32)

L

T1

L

θ1

θ2

m

T2
3m

q

2q
r

20

Electric Fields

In this chapter, we introduce the concept of an electric field associated with a variety
of charge distributions. We follow that by introducing the concept of an electric field
in terms of Faraday’s electric field lines. In addition, we study the motion of a charged
particle in a uniform electric field.

20.1

The Electric Field

Based on the electric force between charged objects, the concept of an electric field
was developed by Michael Faraday in the 19th century, and has proven to have
valuable uses as we shall see.
In this approach, an electric field is said to exist in the region of space around
→
any charged object. To visualize this assume an electrical force of repulsion F
between two positive charges q (called source charge) and q◦ (called test charge),
see Fig. 20.1a.
Now, let the charge q◦ be removed from point P where it was formally located as
→

shown in Fig. 20.1b. The charge q is said to set up an electric field E at P, and if q◦
→
is now placed at P, then a force F is exerted on q◦ by the field rather than by q, see
Fig. 20.1c.

Since force is a vector quantity, the electric field is a vector whose properties are
determined from both the magnitude and the direction of an electric force. We define
→
the electric field vector E as follows:

H. A. Radi and J. O. Rasmussen, Principles of Physics,
Undergraduate Lecture Notes in Physics, DOI: 10.1007/978-3-642-23026-4_20,
© Springer-Verlag Berlin Heidelberg 2013

659


660

20 Electric Fields

Spotlight

→

The electric field vector E at a point in space is defined as the electric
→
force F acting on a positive test charge q◦ located at that point divided by
the magnitude of the test charge:

→

→

F

E =
q◦

Fig. 20.1 (a) A charge q

(20.1)

q

→

+
+++
+

exerts a force F on a test

(a)

charge q◦ at point P. (b) The
→

electric field E established at

qo

+

F

P

P due to the presence of q. (c)
→

→

The force F = q◦ E exerted

q

→

by E on the test charge q◦

+
+++
+

(b)

q

+
+++
+

(c)

P

E

qo

+
P

F = qo E

This equation can be rearranged as follows (see Fig. 20.1c):
→

→

F = q◦ E

(20.2)

→

The SI unit of the electric field E is newton per coulomb (N/C).
→
The direction of E is the direction of the force on a positive test charge placed in
the field, see Fig. 20.2.

20.2

The Electric Field of a Point Charge

To find the magnitude and direction of an electric field, we consider a positive point
charge q as a source charge. A positive test charge q◦ is then placed at point P,


20.2 The Electric Field of a Point Charge

661

a distance r away from q, see Fig. 20.3. From Coulomb’s law, the force exerted on
q◦ is:
→

F =k

q q◦ →
rˆ
r2

(20.3)

q

(a)

+
+++
+

E
P

- -- -- -

q
(b)

E
P
→

Fig. 20.2 (a) If the charge q is positive, then the force F on the test charge q◦ (not shown in the figure)
→

at point P is directed away from q. Therefore, the electric field E at P is directed away from q. (b) If the
→

charge q is negative, then the force F on q◦ at point P is directed toward q. Therefore, the electric field

→

E at P is directed toward q

where →
rˆ is a unit vector directed from the source charge q to the test charge q◦ . This
force has the same direction as the unit vector →
rˆ .
→

→

Since the electric field at point P is defined from Eq. 20.1 as E = F /q◦ , then
according to Fig. 20.3, the electric field created at P by q is an outward vector given by:
→

E =k

qo

r

+

(20.4)

E

F

r

P

r

q

q→
rˆ
r2

+

P

r

q

+
→

Fig. 20.3 If the point charge q is positive, then both the force F on the positive test charge q◦ and the
→

electric field E at point P are directed away from q


662

20 Electric Fields
→

→

When the source charge q is negative, the force F on q◦ and the electric field E
at point P will be toward q, see Fig. 20.4.
Note that for both positive and negative charges, →
rˆ is a unit vector that is always

directed from the source charge q to the point P, see Figs. 20.3 and 20.4.

In all previous and coming discussions, the positive test charge q◦ must be very
small, so that it does not disturb the charge distribution of the source charge q.
→

Mathematically, this can be done by taking the limit of the ratio F /q◦ when q◦
approaches zero. Thus:
→

→

F
E = lim
q◦ →0 q◦

(20.5)

qo
F

+

E

r

P

r
r

q

P

r

q

-

-

→

Fig. 20.4 If the point charge q is negative, then both the force F on the test positive charge q◦ and the
→

electric field E at point P are directed toward q, but the unit vector →
rˆ remains pointed toward P

The electric field due to a group of point charges q1 , q2 , q3 . . . at point P can
be obtained by first using Eq. 20.4 to calculate the electric field of each individual
charge, such that:
→

En = k

qn →
rˆn
rn2

(n = 1, 2, 3, . . .)

(20.6)

→

Then we calculate the vector sum E of the electric fields of all the charges. This sum
is expressed as follows:
→

→

→

→

E = E1 + E2 + E3 + . . . = k

n

qn →
rˆn
rn2

(n = 1, 2, 3, . . .)

(20.7)

20.2 The Electric Field of a Point Charge

663

where rn is the distance from the nth source charge qn to the point P and →
rˆn is a unit
vector directed away from qn to P.
It is clear that Eq. 20.7 exhibits the application of the superposition principle to
electric fields.
Example 20.1

√
Four point charges q1 = q2 = Q and q3 = q4 = −Q, where Q = 2 μC, are placed
at the four corners of a square of side a = 0.4 m, see Fig. 20.5a. Find the electric
field at the center P of the square.
a

+

q1

-

q4

a

+

q1

-

q4

E2 4

a
q2

45 °

a

P

-

+

q3

q2

P

+

E

45°
E13

-

x

q3

(b)

(a)
Fig. 20.5

Solution: The distance between each charge and the center P of the square is
√
a/ 2. At point P, the point charges q1 and q3 produce two diagonal electric field
→
→
vectors E 1 and E 3 , both directed toward q3 , see Fig. 20.5b. Hence, their vector
→

→

→

sum E 13 = E 1 + E 3 points toward q3 and has the magnitude:
E13 = E1 + E3 = k

Q

Q
Q
+k
= 4k 2
√
√
2
2
a
(a/ 2)
(a/ 2)
→

→

At point P, the charges q2 and q4 produce two diagonal electric fields E 2 and E 4 ,
→
→
→
both directed toward q4 , see Fig. 20.5b. Hence, their vector sum E 24 = E 2 + E 4
points toward q4 and has the magnitude:
E24 = E2 + E4 = k

Q
Q
Q
+k
= 4k 2
√
√

2
2
a
(a/ 2)
(a/ 2)
→

→

We now must combine the two electric field vectors E 13 and E 24 to form
→
→
→
the resultant electric field vector E = E 13 + E 24 which is along the positive
x-direction and has the magnitude:


664

20 Electric Fields

8Q
Q
1
= k√
E = E13 cos 45◦ + E24 cos 45◦ = 2 × 4k 2 × √
a
2
2a2
√

−6
8( 2 × 10 m)
= (9 × 109 N.m2 /C2 ) √
= 4.5 × 105 N/C
2 (0.4 m)2

Example 20.2

(Electric Dipole)

Consider two point charges q1 = −24 nC and q2 = +24 nC that are 10 cm apart,
forming an electric dipole, see Fig. 20.6. Calculate the electric field due to the two
charges at points a, b, and c.
Fig. 20.6

E 2c

Ec

c
E 1c

10 cm

q1

10 cm

60°

Ea a
4 cm

q2

6 cm

b

Eb

2 cm

Solution: At point a, the electric field vector due to the negative charge q1 , is
directed toward the left, and its magnitude is:
E1a = k

−9 C)
|q1 |
9
2
2 (24 × 10
=
(9
×
10
N.m
/C
)
= 135 × 103 N/C

2
(0.04 m)2
r1a

The electric field vector due to the positive charge q2 is also directed toward the
left, and its magnitude is:
E2a = k

|q2 |
(24 × 10−9 C)
= (9 × 109 N.m2 /C2 )
= 60 × 103 N/C
2
(0.06 m)2
r2a

Then, the resultant electric field at point a is toward the left and its magnitude is:
Ea = E1a + E2a = 135 × 103 N/C + 60 × 103 N/C
= 195 × 103 N/C

(Toward the left)


20.2 The Electric Field of a Point Charge

665

At point b, the electric field vector due to the negative charge q1 , is directed toward
the left, and its magnitude is:
E1b = k

−9 C)
|q1 |
9
2 2 (24 × 10
=
(9
×
10
N.m
/C
)
= 15 × 103 N/C
2
(0.12 m)2
r1b

In addition, the electric field vector due to the positive charge q2 is directed toward
the right, and its magnitude is:
E2b = k

−9 C)
|q2 |
9
2 2 (24 × 10
=
(9
×
10
N.m

/C
)
= 540 × 103 N/C
2
(0.02 m)2
r2b

Since E2b > E1b , the resultant electric field at point b is toward the right and its
magnitude is:
Eb = E2b − E1b = 540 × 103 N/C − 15 × 103 N/C
= 525 × 103 N/C

(Toward the right)
→

→

At point c, the magnitudes of the electric field vectors E 1c and E 2c established
by q1 and q2 are the same because |q1 | = |q2 | = 24 nC and r1c = r2c = 10 cm.
Thus:
E2c = E1c = k

−9 C)
|q1 |
9
2 2 (24 × 10
=
(9
×
10

N.m
/C
)
= 21.6 × 103 N/C
2
(0.1 m)2
r1c

The triangle formed from q1 , q2 , and point c in Fig. 20.6 is an equilateral
triangle of angle 60◦ . Hence, from geometry, the vertical components of the
→

→

two vectors E 1c and E 2c cancel each other. The horizontal components are both
directed toward the left and add up to give the resultant electric field Ec at point c,
see the figure below.

E 2c

Ec

E 1c cos60°

E 2c cos60°

60°

c
60°

E 1c

Thus: Ec = E1c cos 60◦ + E2c cos 60◦ = 2E1c cos 60◦
= 2(21.6 × 103 N/C)(0.5) = 21.6 × 103 N/C

(Toward the left)


666

20.3

20 Electric Fields

The Electric Field of an Electric Dipole

Generally, the electric dipole introduced in Example 20.3 consists of a positive charge
q+ = +q and a negative charge q− = −q separated by a distance 2a, see Fig. 20.7.
In this figure, the dipole axis is taken to be along the x-axis and the origin of the
xy plane is taken to be at the center of the dipole. Therefore, the coordinates of q+
and q− are (+a, 0) and (−a, 0), respectively.

y

E

E+
P (x,y)

E−

r−

-q

r+

+q

r−

-

0

(-a,0)

(a,0)

y

r+

+

x
x-a

x+a

→

→

→

Fig. 20.7 The electric field E = E + + E − at point P(x, y) due to an electric dipole located along the
x-axis. The dipole has a length 2a

Let us assume that a point P(x, y) exists in the xy-plane as shown in Fig. 20.7. We
→
will call the electric field produced by the positive charge E+ and the electric field
→

produced by the negative charge E− .
Using the superposition principle, the total electric field at P is:
→

→

→

E = E+ + E− = k

q+ →
q− ˆ
rˆ+ + k 2 →
r−
2
r+

r−

(20.8)

2 = (x − a)2 + y2 and r 2 = (x + a)2 + y2 .
From the geometry of Fig. 20.7, we have r+
−
→
ˆ
In addition, r + is a unit vector directed outwards and away from the positive charge
q at (+a, 0). On the other hand, →
rˆ is a unit vector directed outwards and away
+

−

from the negative charge q− at (−a, 0). Accordingly, Eq. 20.8 becomes:
→

E =k

−q
q
→
→
rˆ+ +
rˆ−
2
2
(x − a) + y

(x + a)2 + y2

(20.9)


20.3 The Electric Field of an Electric Dipole

667

Therefore, the general electric field will take the following form:
→

E = kq

(x

→
rˆ+
− a)2

+ y2

−

(x

→
rˆ−
+ a)2

(20.10)

+ y2

The Electric Field Along the Dipole Axis
Let us first assume a point P exists on the dipole axis, i.e. y = 0, and satisfies the
→
→
condition x < −a, as shown in Fig. 20.8a. In this case, →
rˆ+ = →
rˆ− = − i , where i is
a unit vector along the x-axis.

E+ P

E−

-q

(a)
x < −a

+q

0

-a
-q

(b)

E+ E− P

-

− a< x< + a

a
(c)

-q

→

0

0

→

+

x

+a
+q

-a

x

+a
+q

-

+

+
+a

E−

P

E+

x

x > +a

→

Fig. 20.8 The electric field E = E+ + E− at different points along the axis of a dipole that has a length 2a

When P has an x-coordinate that satisfies −a < x < + a as in Fig. 20.8b, then
→
→
→
rˆ+ = − i and →

rˆ− = + i . When P satisfies x > + a as in Fig. 20.8c, then →
rˆ+ =
→
→
rˆ = + i . Substituting in Eq. 20.10, we get:
−

⎧
1
1
⎪
⎪
−kq
−
⎪
2
⎪
(x
−
a)
(x
+
a)2
⎪
⎪
⎨
→
1
1
+

E = −kq
⎪
(x − a)2
(x + a)2
⎪
⎪
⎪
⎪
1
1
⎪
⎩ kq
−
(x − a)2
(x + a)2

→

i

→

i

→

i

x < −a (Toward the right)
−a < x < +a (Toward the left)

x > +a (Toward the right)
(20.11)