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

a Find the resultant force exerted on the charge q2by the two charges q1and q3.. Find the resultant force exerted on the charge q2by the two charges q1and q3.. Find the resultant force e

19.3 Coulomb’s Law 643Both the inverse square laws describe a property of interacting objects wherecharges are involved in one case and masses in the other The laws differ in thatthe electrostatic forces between two charged particles may be either attractive orrepulsive, but gravitational forces are always attractive.

Fig 19.6 Newton’s law of universal gravitation states that the gravitational force between two objects

of masses m1and m2is attractive The magnitude of the force F12 exerted on object 1 by object 2 is equal

to the magnitude of the force F21exerted on object 2 by object 1 Note that →F

12 = −→F21.

The electrostatic constant k in Coulomb’s law has the value:

k = 8.9875 × 109N.m2/C2≈ 9 × 109N.m2/C2 (19.3)For historical reasons and for the aim of simplifying many other formulas, the con-

stant k is usually written as:

k= 1

where the quantity ◦ (called the permittivity constant of free space) has the

value:

Table 19.1 Charge and mass of the electron, proton and neutron.

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 q2by

the two charges q1and q3 (b) In a different layout (see Fig.19.8), q2experiences

a resultant force of zero Find the position of q2and find the magnitude of each

force exerted on q2

Solution: (a) Because q2is negative and both q1and q3are positive, the forces

→

F21and→

F23are both attractive as displayed in Fig.19.7 From Coulomb’s law we

can find F21as follows:

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.

Since F21is greater than F23, the resultant force F exerted on q2will 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 q2is zero, the magnitudes of F21and F23must

be equal Based on Fig.19.8, the equality of the two forces F21and F23leads tothe following steps:

From Coulomb’s law we can find either of the value of F21or the value of F23asfollows:

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

Proton Gravitational attraction

(0.053 × 10−9m)2 = 8.2 × 10−8N(b) From Newton’s law of gravitation, the magnitude of the gravitational force

19.3 Coulomb’s Law 647

Example 19.3

Two identical copper coins of mass m = 2.5 g contain about N = 2 × 1022atoms

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

Solution: (a) The upper coin is in equilibrium due to its weight and the electrostatic

repulsion between the two charged coins Therefore:

(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:

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 tensionalforce →

T from the string, the electric force→

F efrom the left sphere, and the

grav-itational force m→g After decomposing the tensional forceT→in the vertical andhorizontal directions, we apply the condition of equilibrium as follows:



F x = F e − T sin θ = 0 ⇒ F e = T sin θ



F y = T cos θ − mg = 0 ⇒ T cos θ = mg Eliminating T from the above two equations, we get the value of F e:

F e = mg tan θ = (2 × 10−3kg)(9.8 N/kg)(tan 5◦) = 1.7 × 10−3NFrom Fig.19.11a, we find the distance r between the two charges:

9× 109N.m2/C2 = 7.39 × 10−9CNote 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 q2by the

two charges q1and q3

is repulsive and the force →

F23 is attractive as displayed in Fig 19.12 From

Coulomb’s law we can find F21as follows:

We can also write the resultant force →

F in vector form as follows:

→

F = −F →i + F →j =−0.8→i + 0.6→j

N

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 q2by the two

F23is attractive as displayed in Fig.19.13 From Coulomb’s

law we can find F21and F23as follows:

We can also write the resultant force→

F in vector form as follows:

→

F = −F →i + F →j =−0.92→i + 1.44→j

N

19.4 Exercises 651

19.4 Exercises

Section 19.1 Electric Charge

(1) Explain what is meant by the following: (a) a neutral atom, (b) a negativelycharged 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 transferpositive charges from one of them to the other? Why so or why not?

Fig 19.14 See Exercise (2)

Before rubbing

Neutral rubber

Neutral fur

Section 19.2 Charging Conductors and Insulators

(3) If we repeat the experiment illustrated in Fig.19.3but instead of using a chargedplastic rod, we use a charged rubber rod, what will the final charge on the copperrod be?

(4) A charged plastic comb often attracts small bits of dry paper, as shown in theleft part of Fig.19.15 After a while, the bits of paper fall down, as shown inthe right part of Fig.19.15 Explain this observation

Fig 19.15 See Exercise (4)

(5) In an oxygen-enriched atmosphere (as in hospital operation rooms), workersmust 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 balloonfall afterwards?

Fig 19.16 See Exercise (6)

Charged balloon

(7) Using a charged rubber rod, show the steps of how two uncharged metallicspheres mounted on insulating stands can be electrostatically charged withequal 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 theseelectrons?

(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 beset 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? (Thenucleus can be treated as a point charge since the nuclear radius is of the order

of 10−14m).

(12) Two electrically neutral spheres are 0.1 m apart When electrons are movedfrom one of the spheres to another, an attractive force of magnitude 10−3N isestablished 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 Anelectrically neutral pin of silver has a mass of 10 g (a) Calculate the number ofelectrons in the silver pin (Avogadro’s number is 6.022 × 1026atoms/kmol).

(b) Electrons are added to the pin until the net charge is−1 mC How manyelectrons are added for every billion (109) electrons in the neutral atoms?

(14) Two protons in an atomic nucleus are separated by 2× 10−15m (a typical nuclear dimension) (a) Find the magnitude of the electrostatic repulsive forcebetween the protons (b) How does the magnitude of the electrostatic forcecompare to the magnitude of the gravitational force between the two protons?

inter-(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 inFig.19.17 Find the magnitude of the charge on each ball if they repel eachother with a force of magnitude 2× 10−5N Does the answer give you any hintabout the exact sign of each charge? Explain

Fig 19.17 See Exercise (16)

r

(17) Two point charges q1and q2are 3 m apart, and their combined charge is 40μC.

(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 4μC 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

the net force on any one of the three charges (b) If the charges are

posi-tive, i.e q1= q2= q3= +4 μC, would this change the magnitude calculated in

(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)

(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 q2has 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

(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 q1and q3 (b) Calculate the magnitude and direction of the

net force on the charge q2

Fig 19.23 See Exercise (24)

(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)

(26) In the Bohr model of the hydrogen atom, an electron of mass m = 9.11 ×

10−31kg revolves about a stationary proton in a circular orbit of radius

r = 5.29 × 10−11m, see Fig.19.25 (a) What is the magnitude of the electricalforce on the electron? (b) What is the magnitude of the centripetal acceleration

of the electron? (c) What is the orbital speed of the electron?

Fe

Fe

r

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) theeight Cs+ions?, (b) only seven Cs+ions?

(28) Two positive point charges q1and q2are set apart by a fixed distance d and have a sum Q = q1+ q2 For what values of the two charges is the Coulombforce maximum between them?

19.4 Exercises 657

Fig 19.26 See Exercise (27)

Cl Cs

a

(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 +qof mass m is in equilibrium at x= 0

and constrained to move only along the x-axis The charge +qis then displaced

from the origin to a small distance x  a and released, see Fig.19.27 (a) Showthat+qwill 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 × 10−25kg and a= 3 × 10−10m, find the oscillation period T.

Fig 19.27 See Exercise (29)

x a a

L = 10 cm, m = 2 g, and r = 1.7 cm, what is the value of q?

Fig 19.28 See Exercise (30)

θ θ

r x

(31) For the charge distribution shown in Fig.19.29, the long non-conducting

mass-less 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)

x L

while the second sphere has a positive charge 2 q, mass 3m, and makes a smaller

angleθ2with the vertical, see Fig.19.30 For small angles, take tanθ θ and

assume that the spheres only have horizontal displacements and hence theelectric force of repulsion is always horizontal (a) Find the ratioθ1/θ2 (b)

Find the distance r between the spheres.

Fig 19.30 See Exercise (32)

m q

3m 2q

L

r

L

1θ

1

T

Electric Fields 20

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 chargedparticle in a uniform electric field

20.1 The Electric Field

Based on the electric force between charged objects, the concept of an electric fieldwas developed by Michael Faraday in the 19th century, and has proven to havevaluable 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 asshown 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 aredetermined from both the magnitude and the direction of an electric force We definethe electric field vector →

E as follows:

Undergraduate Lecture Notes in Physics, DOI: 10.1007/978-3-642-23026-4_20,

© Springer-Verlag Berlin Heidelberg 2013

The electric field vector →

Eat a point in space is defined as the electricforce →

F acting on a positive test charge q◦ located at that point divided bythe magnitude of the test charge:

P due to the presence of q (c)

The force →F = q◦→E exerted

by →E on the test charge q

+ + + +

+

q

P

+ + + +

q

+

qoP

+ + + +

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 inthe 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,