# Chapter 1 Coulomb’s Law:

Learning Objectives:

In this chapter you will basically learn:

$$\bullet$$ Fundamental concepts of electric charge.

$$\bullet$$ Distinguishing electrically neutral, negative, and positive charges and identifying excess charges.

$$\bullet$$ Distinguishing conductors, nonconductors (insulators), semiconductors, and superconductors.

$$\bullet$$ Describe the electrical properties of the particles inside an atomic nucleus.

$$\bullet$$ Identify conduction electrons and explain their role in making a negatively or positively charged conducting objects.

$$\bullet$$ Explain how a charged object can set up induced charge in a second object.

$$\bullet$$ Learn that charges with same electrical polarities repel each other and those with opposite electrical polarities attract each other.

$$\bullet$$ Learn that Coulomb’s law applies only to point-like particles.

$$\bullet$$ Learn if several forces act on a particle, find the net force by adding all the forces as vectors.

$$\bullet$$ Learn Shell Theorem-1: A charged particle outside a shell with charge uniformly distributed on its surface is attracted or repelled as if the shell’s charge were concentrated as a particle at its center.

$$\bullet$$ Learn Shell Theorem-2: A charged particle inside a shell with charge uniformly distributed on its surface has no net force acting on it due to the shell.

$$\bullet$$ Learn that if excess charge is put on a spherical conductor, it spreads out uniformly over the external surface area.

$$\bullet$$ Learn that if two identical spherical conductors touch or are connected by conducting wire, any excess charge will be shared equally.

$$\bullet$$ Identify that a nonconducting object can have any given distribution of charge, including charge at interior points.

$$\bullet$$ Identify current as the rate at which charge moves through a point.

$$\bullet$$ For current through a point, apply the relationship between the current, a time interval, and the amount of charge that moves through the point in that time interval.

## 1.1 Electrical Charge:

We observe that clothes come out of the laundry dryer, carry static electrical charges but we are not sure whether they are positively or negatively charged?

Charge is the fundamental quantity of electricity. Electricity is all about charge and we know how the charges interact. The classical study of electricity is generally divided into three general areas.

Electrostatics: the study of the forces acting between static charges.

Electric current: the study of the forms of energy associated with the flow of charge.

Electromagnetism: the study of the forces acting between charges in motion.

Electric charge comes in two and only two types.

positive (+) negative (−)

The term neutral does not refer to a third type of charge, but to the presence in a region of positive and negative charges in equal amount.

The presence of electrical charges both positive and negative can be easily verified with a very simple experiment as described below. A glass ball hang by means of a thread and rubbed with a silk cloth as shown in Fig. 1.1.a. Similarly, a second glass ball is hung exactly the same way as the first one and rubbed again with a silk cloth and bring it near the first hanging ball. The two hanging balls move away from each other. There must be a force of repulsion that acted in between the glass balls that pushes the balls away from each other.

In the second experiment a plastic ball is rubbed with fur and bring it to the first glass ball as shown in Fig. 1.1b. Now the glass ball and plastic ball moves towards each other with a force of attraction.

In the above experiments a glass ball is rubbed with a silk cloth, transfers a small amount of negative charge moves from the ball to the silk, leaving the ball with a small amount of excess positive charge. Similarly, rubbing the second ball with the silk cloth, it too becomes positively charged. So when we bring it near the first ball, the two balls repel each other.

Particles with the same sign of electrical charge repel each other. Particles with opposite signs attract each other.

We will find that, electrons are negative charges and protons are positive charges. An electron is considered the smallest quantity of negative charge and a proton the smallest quantity of positive charge.

It has been experimentally verified that two negative charges repel or two positive charges repel each other. A positive charge and a negative charge and vice versa attract each other.

When an atom loses an electron, the separated electron forms a negative charge, but the remaining that contains one less electron or consequently one more proton becomes a positively charged atom. A positive charge is not necessarily a single proton. In most cases, a positive charge is an atom that has lost one or more electron(s).

## 1.2 Conductors and Insulators:

Materials that are around us can be classified in terms of their ability to move the charge through them. Conductors are materials through which charge can move freely. All metals (such as gold, platinum, silicon, silver, titanium, copper, tungsten), the human body, and tap water are examples of Conductors.

Insulators are non-conductors through which charge cannot move freely. Common examples of insulating materials include rubber, plastic, glass, and pure water. However, all these insulating materials could be turned into conducting maerials depending on the addition of metallic power during processing.

Semiconductors in pure form are also insulators. Adding dopants in silicon, germanium or GaAs compound can be transformed into semiconductors that are intermediate between conductors and insulators. Semiconductors are used to manufacture ICs, transistors and diodes etc.

Superconductors are compound of many materials that allow charge carriers to pass without any electrical resistance.

The electronic structures of conductors and insulators differ due to arrangements of positively charged protons, negatively charged electrons, and electrically neutral neutrons. Atoms consist of the protons and neutrons that are packed tightly together in a central nucleus of an atom. However, negatively charged electrons encircling the nucleus determine the electical property of the materials. An electronically neutral material have equal number of electrons and protons. Electrons are held near the nucleus because they have the electrical sign opposite that of the protons in the nucleus and thus are attracted to the nucleus.

The electrons in the outer shells of a conductor are free to move within the solid, leaving behind positively charged atoms ( positive ions). We call the mobile electrons, conduction electrons. However, in the insulators there are few or no free electrons to conduct.

## 1.3 Static Electricity:

The of accumulation of negative or positive charges in a localized object without having ability to move or flow, causes localized charges and forms static electricity. This phenomena happens by transferring electrons to an insulator without having any movement. The lack of free electrons in the insulator, will keep the transferred electrons remain localized without having the ability to distribute itself in the insulator. The electrical phenomenon that forms due to non-movement of charges is called static electricity.

A conductor mounted on an insulator can also show the same phenomena with a given number of excess electrons that distribute themselves in that conductor without having any ability to flow of free electrons due to the insulating mount that acts as the barrier. Since electrons are having the same sign the redistribution of electrons on the conductor very much depend on the geometry of the conductor.

Figure-1-2 demonstrate how charges are distributed themselves on two different conductors mounted on insulating stands. In case of spherical sphere charges are evenly distributed over the surface of the sphere but for an oval-shaped conductor charges are accumulated on the edges.

Charging by Contact:

When a charged object is brought into contact with an uncharged (electrically neutral) object, part of its charges flow onto the uncharged object and make it partially charged. The transfer proportion depends on the shapes of the two objects. For example, if the two objects are two identical metal spheres with insulator mountings, they share the charge equally. For asymmetric and unequal objects, the reasoning is more complicated and involved. The following figure shows the simple case of two identical metal spheres on insulator mountings a) before contact, b) during contact, and c) after separation.

Charging by Induction:

Charging by induction means charging without contact. The Earth may be considered as being electrically neutral. Adding a certain number of positive or negative charges to the Earth does not affect its neutrality. Earth is so huge that the charges on the objects do not count at all compared to the charges that the Earth contains. That is why Earth is electrically neutral for our experiments. We can easily transfer some charges to it or take from it and it will not be affected. If an electrically charged sphere (on an insulator mounting) is connected with a conductor (a metal wire) to the ground, it gets discharged either by transferring some electrons to the Earth or pulling some from it. The following figure shows how a positively charged sphere and a negatively charged one become discharged by being connected to the Earth.

The two shell theories for electrostatics that follow:

Shell theorem 1: A charged particle outside a shell with charge uniformly distributed on its surface is attracted or repelled as if the shell’s charge were concentrated as a particle at its center.

Shell theorem 2: A charged particle inside a shell with charge uniformly distributed on its surface has no net force acting on it due to the shell.

Charge on a conducting spherical shell spreads uniformly over the (external) surface.

SI Unit of Electric Charge: Coulomb

An accumulation of electric charges at a point (a tiny volume in space) is called a point charge.

If two like point charges are separated by 1 m and the repulsion force between them is $$9.0 \times 10^9$$N, each charge is called 1 Coulomb (1 C). It has been shown that it takes $$6.25 \times 10^{18}$$ electrons to form 1 C of negative electricity.

The SI unit of charge is the coulomb, the coulomb unit is derived from the SI unit ampere for electric current i. So, the electric current i is the rate dq/dt at which charge moves past a point.

$$$i=\frac{dq}{dt} \tag{1.1}$$$

Rearranging Eq. 21.6 and replacing the symbols with their units (coulombs C, amperes A,and seconds s) we see that 1 C = (1 A)(1 s). The elementary charge e is one of the important constants of nature.

## 1.4 Charge Is Quantized:

Electric charge is quantized means that the charge of a particle can be written as the multiple of the elementary charge, as ne, where n is a positive or negative integer and e is the elementary charge. Almost anything we see around us are made of discrete atoms and molecules. Charges are also made of discrete electrons and protons.

$$$q = ne, n = \pm{1},\pm{2},\pm{3}........... \tag{1.2}$$$

The magnitude of the charge of the electron and proton is

$$$e = 1.6 \times 10^{-19}C \tag{1.3}$$$

Example Problem 1-1:

It takes $$6.25 \times 10^{18}$$ protons to form 1 C of positive charge. Find the charge of each proton.

Solution: $q = ne$ $e=\frac{q}{n}=\frac{1 C}{6.25 \times 10^{18}}=1.6\times 10^{-19}\space C\space (Answer)$ Example Problem 1-2:

How many electrons are there in (a) -1 mC, (b) -1 nC, and (c) -1 pC? Solution:

$q = ne$ (a) $n=\frac{q}{e}=\frac{|-1\times 10^{-3}\space C|}{1.6\times 10^{-19}\space C}=6.25\times 10^{15}\space (Answer)$ (b) $n=\frac{q}{e}=\frac{|-1\times 10^{-9}\space C|}{1.6\times 10^{-19}\space C}=6.25\times 10^{9}\space (Answer)$ (c) $n=\frac{q}{e}=\frac{|-1\times 10^{-12}\space C|}{1.6\times 10^{-19}\space C}=6.25\times 10^{6}\space (Answer)$

Detecting Electric Charges with Gold Leaf Electroscope:

An electroscope (See Figure 1-5) is a device that detects the existence of electric charges on charged objects, positive or negative. An electroscope is made of a small cylindrical compartment made of glass or plastic with a metal rod inserted into it through an insulating cap. The end of the rod that is inside the compartment has two small metal foils (aluminum , gold, or any other metal) hinged to it that are free to open up like the wings of a butterfly. The outer end is connected to a metal sphere or a disk. When a charged object (no matter positive or negative) is brought into contact with the outer sphere or pan, some of the charges get transferred to the foils via the metal rod. The foils become charged up with like charges that repel each other causing the foils to separate and open up. That is how the foils indicate that some electric charges are transferred to them. Even if a charged object is held near the sphere or the pan with no physical contact, the foils still open up, but if the object is taken away from the pan, the foils drop down again. Why?

## 1.5 Coulomb’s Law:

The formulation of an equation that applies to Coulomb’s law, is based on the assumption that the force of attraction and repulsion in between charged particles.

If two charged particles are brought near each other, they each exert an electrostatic force on the other. The direction of the force vectors depends on the signs of the charges. If the particles have the same sign of charge, they repel each other. That means that the force vector on each is directly away from the other particle (Figs. 1-6a and b). If we release the particles, they accelerate away from each other. If, instead, the particles have opposite signs of charge, they attract each other.That means that the force vector on each is directly toward the other particle (Fig. 1-6c). If we release the particles, they accelerate toward each other.

The forces acting on the charged particles is given by Charles-Augustin de Coulomb, based on his experiments in 1785 and the eqution is known as Coulomb’s law. The formulation of this equation is based on the vector diagram of two charged particles as shown in Fig. 1.7, where particle 1 has charge $$q_{1}$$ and particle 2 has charge $$q_{2}$$. We assume that the force acted on particle 1 and is directed in terms of a unit vector $$\vec{r}$$ that points along a radial axis extending through the two particles, radially away from particle 2. Similar to other unit vectors, $$\hat{r}$$ has a magnitude of exactly 1 and without any unit.

Under these assumptions, the electrostatic force between two charged particles either repulsion and attraction can be given as Coulomb’s law.

$$$\vec{F}=k\frac{|q_{1}||q_{2}|}{r^2} =\frac{1}{4\pi\varepsilon_0}\frac{|q_{1}||q_{2}|}{r^2} \space (Coulomb's\space Law) \tag{1.4}$$$

where r is the separation between the particles and k is a positive constant called the electrostatic constant or the Coulomb constant. The electrostatic constant can be written as

$$$k=\frac{1}{4\pi\varepsilon_0}={8.99 \times 10^9}{\frac{N.m^2}{C^2}} \tag{1.5}$$$

The quantity $$\varepsilon_{0}= 8.85 \times 10^{-12}{\frac{C^2}{N.m^2}}$$, is called the permittivity of free space, but inside the medium it is replaced with $$\varepsilon=\varepsilon_{0}\varepsilon_{r}$$. Here, $$\varepsilon_{r}$$ is called the relative permittivity or dielectric constant.

Analogous to the electrostatic force of Eq. 1.4, the gravitational force between two particles with masses $$m_{1}$$ and $$m_{2}$$ and separation r is known as the Newton’s equation of gravitation and is given by:

$$$\vec{F}=G\frac{m_{1}m_{2}}{r^2}\hat{r} \tag{1.6}$$$ where $$G = 6.67 \times 10^{-11}{\frac{N.m^2}{kg^2}}$$.

Multiple Forces. The electrostatic force obeys the principle of superposition. Suppose we have n charged particles near a chosen particle called particle 1; then the net force on particle 1 is given by the vector sum:

$$$\vec{F}_{1,net} = \vec{F}_{12} + \vec{F}_{13} + ........+ \vec{F}_{1n} \tag{1.7}$$$

in which, for example, $$\vec{F}_{13}$$ is the force on particle 1 due to the presence of particle 3.

Example Problem 1-3:

Find the magnitude and direction of the force between a 30.0 $$\mu$$C charge and a 40.0-μC charge when they are separated by a distance of 20.0 cm. Both are point charges.

Solution:

$F=\frac{1}{4\pi\varepsilon_0}\frac{|q_{1}||q_{2}|}{r^2}=\left({8.99 \times 10^9}{\frac{N.m^2}{C^2}}\right)\frac{|30\times 10^{-6}C||40\times 10^{-6}C|}{(0.20\space m)^2}=269.70\space N\space (Answer)$

F = 269.70 N, directed away from each other and can be seen in Figure-1-6(a).

Example Problem 1-4: As shown in Figure-1-8 (a), there are three charges $$q_1$$, $$q_2$$ and $$q_3$$ are placed with their respective charges and positions in the x-y plane. (b) The corresponding force diagram is shown. Find the force on charge $$q_3$$ with three significant figures.

Solution:

$r_{13} = r_{23} = \sqrt{5^2+5^2}=7.07\space m$ $F_{13}=\frac{1}{4\pi\varepsilon_0}\frac{|q_{1}||q_{2}|}{r_{13}^2}=\left({8.99 \times 10^9}{\frac{N.m^2}{C^2}}\right)\frac{|-20\times 10^{-6}C||50\times 10^{-6}C|}{(7.07\space m)^2}=0.180\space N\space (Answer)$ $F_{23}=\frac{1}{4\pi\varepsilon_0}\frac{|q_{1}||q_{2}|}{r_{23}^2}=\left({8.99 \times 10^9}{\frac{N.m^2}{C^2}}\right)\frac{|40\times 10^{-6}C||50\times 10^{-6}C|}{(7.07\space m)^2}=0.360\space N\space (Answer)$ $F_x = F_{13}\cos135^\circ+F_{23}\cos45^\circ=0.180\space N\cos135^\circ+0.360\space N\cos45^\circ=0.127\space N$ $F_y = F_{13}\sin135^\circ+F_{23}\sin45^\circ=0.180\space N\sin135^\circ+0.360\space N\sin45^\circ=0.382\space N$ $F=\sqrt{F_x^2+F_y^2}=\sqrt{(0.127\space N)^2+(0.382\space N)^2}=0.403\space N(Answer)$ Direction of the resultant force, F:$\theta=\tan^{-1}\frac{F_y}{F_x}=\tan^{-1}\frac{0.382}{0.127}=71.61^\circ\space(Answer)$

Example Problem 1-5: Finding the net force due to two other particles

Solution:

1. Figure 1-9a shows two positively charged particles fixed in place on an x axis. The charges are $$q_{1} = 1.60 \times 10^{-19}C$$ and $$q_{2} = 3.20 \times 10^{-19}C$$, and the particle separation is R = 0.0200 m. What are the magnitude and direction of the electrostatic force on particle 1 from particle 2?

$F_{12}=\frac{1}{4\pi\varepsilon_0}\frac{|q_{1}||q_{2}|}{R^2}=\left({8.99 \times 10^9}{\frac{N.m^2}{C^2}}\right)\frac{|1.6\times 10^{-19}C||3.20\times 10^{-19}C|}{(0.0200\space m)^2}=1.15\times 10^{-24}\space N$

Thus, force $$\vec F_{12}$$ has the following magnitude and direction (relative to the positive direction of the x axis): $1.15\times 10^{-24}\space N\space (Answer)$ and $180^\circ\space (Answer)$.

We can also write $$\vec F_{12}$$ in unit-vector notation as $\vec F_{12}=-(1.15\times 10^{-24}\space N)\hat i\space (Answer)$.

1. Figure 1-9b is identical to Fig. 1-7a except that particle 3 now lies on the x axis between particles 1 and 2. Particle 3 has charge $$q_{3} = -3.20 x 10^{-19}C$$ and is at a distance $$\frac{3}{4}R$$ from particle 1. What is the net electrostatic force $$\vec{F}_{1,net}$$ on particle 1 due to particles 2 and 3?

The magnitude of force $$\vec F_{13}$$$can be calculated as we have done in (a), as follows: $F_{13}=\frac{1}{4\pi\varepsilon_0}\frac{|q_{1}||q_{3}|}{(\frac{3}{4}R)^2}=\left({8.99 \times 10^9}{\frac{N.m^2}{C^2}}\right)\frac{|1.6\times 10^{-19}C||-3.20\times 10^{-19}C|}{{(\frac{3}{4}}\times0.0200\space m)^2}=2.05\times 10^{-24}\space N$ We can also write $$\vec F_{12}$$ in unit-vector notation as $\vec F_{13}=(2.05\times 10^{-24}\space N)\hat i\space (Answer)$. The net force $$\vec F_{1,net}$$ on particle 1, is the vector sum of $$\vec F_{12}$$ and $$\vec F_{13}$$ we can write the net force $$\vec F_{1,net}$$ on particle 1 in unit-vector notation as $\vec F_{1,net}=\vec F_{12}+\vec F_{13}=-(1.15\times 10^{-24}\space N)\hat i+(2.05\times 10^{-24}\space N)\hat i=(9.0\times 10^{-25}\space N)\hat i\space (Answer)$ Thus, $$\vec F_{1,net}$$ has the following magnitude and direction (relative to the positive direction of the x axis): $$9.00 \times 10^{-25}$$N and $$0^\circ$$. (Answer) 1. Figure 1-9c is identical to Fig. 1-9a except that particle 4 is now included. It has charge $$q_{4} = -3.20 x 10^{-19}C$$ is at a distance $$\frac{3}{4}R$$ from particle 1, and lies on a line that makes an angle $$60^o$$ with the x axis. What is the net electrostatic force $$\vec{F}_{1,net}$$ on particle 1 due to particles 2 and 4? The magnitude of force $$\vec F_{14}$$$ can be calculated as we have done in (a), as follows:

$F_{14}=\frac{1}{4\pi\varepsilon_0}\frac{|q_{1}||q_{4}|}{(\frac{3}{4}R)^2}=\left({8.99 \times 10^9}{\frac{N.m^2}{C^2}}\right)\frac{|1.6\times 10^{-19}C||-3.20\times 10^{-19}C|}{{(\frac{3}{4}}\times 0.0200\space m)^2}=2.05\times 10^{-24}\space N$ In vector notation $\vec F_{14}=\vec F_{14}\cos\theta \hat i+\vec F_{14}\cos\theta \hat j=(2.05\times 10^{-24}\space N)(\cos60^\circ) \hat i+(2.05\times 10^{-24}\space N)(\sin60^\circ) \hat j$

$\vec F_{14}=(1.025\times 10^{-24}\space N) \hat i+(1.775\times 10^{-24}\space N) \hat j$ The net force $$\vec F_{1,net}$$ on particle 1, is the vector sum of $$\vec F_{12}$$ and $$\vec F_{14}$$ we can write the net force $$\vec F_{1,net}$$ on particle 1 in unit-vector notation as $\vec F_{1,net}=\vec F_{12}+\vec F_{14}=-(1.15\times 10^{-24}\space N)\hat i+(1.025\times 10^{-24}\space N) \hat i+(1.775\times 10^{-24}\space N) \hat j\space$ $\vec F_{1,net}= -(1.25\times 10^{-25}\space N)\hat i+(1.775\times 10^{-24}\space N) \hat j\space (Answer)$ Example Problem 1-6: Equilibrium of two forces on a particle Figure 1-10a shows two particles fixed in place: a particle of charge $$q_1 = 8q$$ at the origin and a particle of charge $$q_2 = 2q$$ at x = L. At what point (other than infinitely far away) can a proton be placed so that it is in equilibrium (the net force on it is zero)? Is that equilibrium stable or unstable? (That is, if the proton is displaced, do the forces drive it back to the point of equilibrium or drive it farther away?)

Solution:

$\frac{1}{4\pi\varepsilon_0}\frac{8qq_{p}}{x^2}=\frac{1}{4\pi\epsilon_{0}}\frac{2qq_{p}}{(x-L)^2}$ $\left(\frac{x-L}{x}\right)^2=\frac{1}{4}$ $\left(\frac{x-L}{x}\right)=\frac{1}{2}$ $x = 2L\space (Answer)$ Example Problem 1-7: Mutual electric repulsion in a iron nucleus in between protons. The nucleus in an iron atom has a radius of about $$4.0 \times 10^{-15}$$m and contains 26 protons. (a) What is the magnitude of the repulsive electrostatic force between two of the protons that are separated by $$4.0 \times 10^{-15}$$m?

__Solution: (a) The protons can be treated as charged particles, so the magnitude of the electrostatic force on one from the other is given by Coulomb’s law. Table 1-1 tells us that the charge of a proton is +e.Thus, Eq. 1-3 gives us

$F=\frac{1}{4\pi\varepsilon_0}\frac{e^2}{r^2}=\left({8.99 \times 10^9}{\frac{N.m^2}{C^2}}\right)\frac{(1.6\times 10^{-19}C)^2}{(4.0\times 10^{-15}m)^2}=14\space N\space (Answer)$

1. What is the magnitude of the gravitational force between those same two protons? With $$m_p$$ (= $$1.67 \times 10^{-27}$$ kg) representing the mass of a proton, Eq. 1-6 gives us

$F=G\frac{m_p^2}{r^2}=\left(6.67 \times 10^{-11}{\frac{N.m^2}{kg^2}}\right)\frac{(1.67\times 10^{-27}\space kg)^2}{(4.0\times 10^{-15}m)^2}=1.2\times 10^{-35}\space N(Answer)$ The electron and proton both have a charge of magnitude e (Table 2.1).

#### Table 1-1 Particle, Symbol and Charge.

Particle Symbol Charge
Electron e or $$e^-$$ -e
Proton p +e
Neutron n 0

Quarks:

Quarks are postulated to be the elementary building blocks of the heavy particles, the baryons (protons and neutrons), their existence so far cannot be detected individually.

According to quantum chromodynamics mesons are formed of pairs of quark and antiquark, and nucleons of the triplets of up and down quarks with fractional charges $$\pm{2e/3}$$ and $$\pm{e/3}$$ respectively. The integer electric charges, 0 and +1, of neutron N° and proton N+ are the sum total of the charges of the constituent quarks as shown in Figure 1-11:

## 1.6 Charge Is Conserved:

Benjamin Franklin put together the hypothesis of conservation of charge, that works both for large-scale charged bodies and for atoms, nuclei, and elementary particles.

In our previous example of rubbing a glass rod with silk, a positive charge appears on the rod and a negative charge of equal magnitude appears on the silk. Therfore, rubbing a glass does not create charge but only transfers it from one object to another, disturbing the electrical neutrality of each object during the process.

As an example, a uranium-238 nucleus $${238}_U$$ transforms into a thorium-234 nucleus $${234}_{Th}$$ by emitting an alpha particle. Because that particle has the same makeup as a helium-4 nucleus, it has the symbol $${4}_{He}$$. Here the superscript in the symbol for the nucleus is called the mass number and is the total number of the protons and neutrons in the nucleus.

$$${238}_U -> {234}_{Th} + {4}_{He} \tag{1.8}$$$

## Solved Problems Coulombs-Law

1. [1] Of the charge Q initially on a tiny sphere, a portion q is to be transferred to a second, nearby sphere. Both spheres can be treated as particles and are fixed with a certain separation. For what value of q/Q will the electrostatic force between the two spheres be maximized?

Solution: After transferring a q amount of charge from an initial charge Q, now the charge distribution are Q-q and q. Both spheres can be treated as particles and are fixed with a separation of r.

The magnitude of the electrostatic force between the charges Q-q and q with a separation of r can be expressed using Coulomb’s law as given below:

$F=\frac{1}{4\pi\varepsilon_0}\frac{|(Q-q)||q|}{r^2}$ The electrostatic force between the two spheres be maximized by taking derivative of F with respect to q as follows:

$\frac{dF}{dq}=\frac{1}{4\pi\varepsilon_0r^2}\frac{d(Qq-q^2)}{ dq}=0$ $(Q-2q)=0$ $\frac{q}{Q}=\frac{1}{2}\space (Answer)$ 2. [3] What must be the distance between point charge $$q_1 = 26.0 ~\mu$$C and point charge $$q_2 = -47.0 ~\mu C$$ for the electrostatic force between them to have a magnitude of 5.70 N?

Solution:

To find the separation distance between the charges, we can use Coulomb’s Law:

$r=\sqrt{\frac{1}{4\pi\varepsilon_0}\frac{|q_1||q_2|}{F}}=\sqrt{\left({8.99 \times 10^9}{\frac{N.m^2}{C^2}}\right)\frac{|26\times 10^{-6}C||-47\times 10^{-6}C|}{5.70\space N}}=1.388\space m\space (Answer)$

1. [4] In the return stroke of a typical lightning bolt, a current of $$2.5 \times 10^4$$ A exists for $$20 \mu s$$. How much charge is transferred in this event?

Solution:

Using Eq. 1-1, $dq = idt = (2.5 \times 10^4\space A)(20\times 10^{-6}\space s)=500\times 10^{-3}\space C\space = 500\space mC(Answer)$

1. [7] In Figure 1-12, three charged particles lie on an x axis. Particles 1 and 2 are fixed in place. Particle 3 is free to move, but the net electrostatic force on it from particles 1 and 2 happens to be zero. If $$L_{23}$$ = $$L_{12}$$, what is the ratio $$q_1/q_2$$?

Solution:

According to the question $$F_{13}=-F_{23}$$ and assuming $$L_{23}$$ = $$L_{12}$$, we get

$\frac{1}{4\pi\varepsilon_0}\frac{|q_1||q_3|}{(2d)^2}=-\frac{1}{4\pi\epsilon_{0}}\frac{|q_2||q_3|}{d^2}$ Equating we get, $\frac{q_1}{q_2}=-4\space (Answer)$ 5. [10] In Fig. 1-13, four particles form a square.The charges are $$q_1 = q_4 = Q$$ and $$q_2 = q_3 = q$$. (a) What is Q/q if the net electrostatic force on particles 1 and 4 is zero? (b) Is there any value of q that makes the net electrostatic force on each of the four particles zero? Explain.

Solution: (a) We need to assume that Q > 0 and q < 0, so that we can fulfill electrostatic force on particles 1 and 4 is zero.

Considering the above figure, we get

$\vec F_{1,net} = \vec F_{12} + \vec F_{13} + \vec F_{14}=0$ $F_{1,net,x} = F_{12,x} + F_{13,x} + F_{14,x}$ $F_{1,net,x} = F_{12} + 0 - F_{14}\cos45^\circ$ $0 = \frac{1}{4\pi\varepsilon_0}\frac{Q|q|}{a^2} - \frac{1}{4\pi\varepsilon_0}\frac{Q^2}{2a^2}\cos45^\circ$ $\frac{Q}{|q|}=2\sqrt 2$ $\frac{Q}{q}=-2\sqrt 2=-2.828\space (Answer)$ Similarly, we can get the same result using the y-components of the forces $F_{1,net,y} = F_{12,y} + F_{13,y} + F_{14,y}$ $F_{1,net,y} = 0 - F_{13} + F_{14}\sin45^\circ$ $0 = -\frac{1}{4\pi\varepsilon_0}\frac{Q|q|}{a^2} + \frac{1}{4\pi\varepsilon_0}\frac{Q^2}{2a^2}\sin45^\circ$ $\frac{Q}{|q|}=2\sqrt 2$ $\frac{Q}{q}=-2\sqrt 2=-2.828\space (Answer)$ (b) The x-compoments of the net force on 2 due to other three particles are $F_{2,net,x} = F_{21,x} + F_{23,x} + F_{24,x}$ $F_{2,net,x} = -F_{21} -F_{23} + F_{24}$ $0 = -\frac{1}{4\pi\varepsilon_{0}}\frac{Q|q|}{a^2} + \frac{1}{4\pi\varepsilon_{0}}\frac{q^2}{2a^2}\cos45^\circ-0$

$\frac{1}{4\pi\varepsilon_{0}}\frac{Q|q|}{a^2} = \frac{1}{4\pi\varepsilon_{0}}\frac{q^2}{2a^2}\cos45^\circ$ $\frac{Q}{|q|}=\frac{1}{2\sqrt 2}$ $\frac{Q}{q}=-0.354$ The result is not same as (a), therefore it is not possible to get the net force zero on particle 2.

Similarly, The y-compoments of the net force on 2 due to other three particles $F_{2,net,y} = F_{21,y} + F_{23,y} + F_{24,y}$ $F_{2,net,y} = -F_{21} -F_{23} + F_{24}$ $0 = -0 + \frac{1}{4\pi\varepsilon_{0}}\frac{q^2}{2a^2}\sin45^\circ-\frac{1}{4\pi\varepsilon_{0}}\frac{qQ}{a^2}$

$\frac{1}{4\pi\varepsilon_{0}}\frac{Q|q|}{a^2} = \frac{1}{4\pi\varepsilon_{0}}\frac{q^2}{2a^2}\cos45^\circ$ $\frac{Q}{|q|}=\frac{1}{2\sqrt 2}$ $\frac{Q}{q}=-0.354$ Again, we found that the y-components of the forces on particle 2, is not same as we get in (a), therefore it is not possible to get the net force zero on particle 2.

1. [13] In Fig. 1-14, particle 1 of charge +1.0 $$\mu$$C and particle 2 of charge $$-3.0 \mu C$$ are held at separation L = 10.0 cm on an x axis. If particle 3 of unknown charge $$q_3$$ is to be located such that the net electrostatic force on it from particles 1 and 2 is zero, what must be the (a) x and (b) y coordinates of particle 3?

Solution: Let’s assume that the particle 3 be placed at a distance x from particle 1 and (L-x) distance from particle 2. At equilibrium, the net force along the x-axis is $F_{x,net}= \frac{1}{4\pi\varepsilon_{0}}\frac{q_1q_3}{|x|^2}+\frac{1}{4\pi\epsilon_{0}}\frac{q_2q_3}{|L-x|^2}=0$ $\frac{|L-x|^2}{|x|^2}=-\frac{q_2}{q_1}$ $|x|=\frac{L}{\sqrt{-\frac{q_2}{q_1}}-1}=\frac{10\space cm}{\sqrt{-\frac{-3\mu C}{1\mu C}}-1}=13.66\space cm$ $x=-13.66\space cm\space (Answer)$ (b) The y coordinate of particle 3 is $$0\space (Answer)$$.

1. [15] The charges and coordinates of two charged particles held fixed in an xy plane are $$q_1 = +3.0 \mu C$$, $$x_1 = 3.5$$ cm, $$y_1 = 0.50$$ cm, and $$q_2 = -4.0 \mu C$$, $$x_2 = -2.0 cm$$, $$y_2 = 1.5$$ cm. Find the (a) magnitude and (b) direction of the electrostatic force on particle 2 due to particle 1. At what (c) x and (d) y coordinates should a third particle of charge $$q_3 = +4.0 \mu C$$ be placed such that the net electrostatic force on particle 2 due to particles 1 and 3 is zero?

Solution:

1. The Euclidean distance between the two charges can be found using,

$r_{21}=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}=\sqrt{(-2.0\space cm-3.5\space cm)^2+(1.5\space cm-0.5\space cm)^2}=5.59\space cm$ magnitude of the electrostatic force on particle 2 due to particle 1 is

$F_{21}=\frac{1}{4\pi\varepsilon_{0}}\frac{q_1q_2}{r_{21}^2}=\left({8.99 \times 10^9}{\frac{N.m^2}{C^2}}\right)\frac{|(3\times 10^{-6}C)(-4\times 10^{-6}C)|}{(0.0559\space m)^2}=34.524\space N\space (Answer)$

1. direction of the electrostatic force on particle 2 due to particle 1 can be found using the slope of the line connecting the two charges,

$\theta = tan^{-1}\left(\frac{y_2-y_1}{x_2-x_1}\right)=tan^{-1}\left(\frac{1.5\space cm-0.5\space cm}{-2.0\space cm-3.5\space cm}\right)=-10.30^\circ\space (Answer)$ (c) At what (c) x coordinate should a third particle of charge $$q_3 = +4.0 \mu C$$ be placed such that the net electrostatic force on particle 2 due to particles 1 and 3 is zero?

$F_{net,2x}=F_{23x}+F_{21x}=0$ $\frac{1}{4\pi\varepsilon_{0}}\frac{|q_2||q_3|}{r_{23}^2}=\frac{1}{4\pi\varepsilon_{0}}\frac{|q_1||q_2|}{r_{21}^2}$ Solving, we get $|r_{23}|=\sqrt{\frac{|q_3|}{|q_1|}r_{21}^2}=\sqrt{\frac{|(4\times 10^{-6}C)}{|(3\times 10^{-6}C)|}(5.59\space cm)^2}=6.45\space cm$ Therefore, $x_3 = x_2-|r_{23}|\cos\theta= -2\space cm-|6.45|\space cm\times\cos(-10.30^\circ)=-8.346\space cm\space (Answer)$ (d) $y_3 = y_2+|r_{23}|\sin\theta= 1.5\space cm-|6.45|\space cm\times\sin(-10.30^\circ)=2.653\space cm\space (Answer)$

1. [22] Figure 1-15 shows an arrangement of four charged particles, with angle $$\theta = 30.0^\circ$$ and distance d = 2.00 cm. Particle 2 has charge $$q_2 = +8.00 \times 10^{-19}$$ C; particles 3 and 4 have charges $$q_3 = q_4 = -1.60 \times 10^{-19}$$ C. (a) What is distance D between the origin and particle 2 if the net electrostatic force on particle 1 due to the other particles is zero? (b) If particles 3 and 4 were moved closer to the x axis but maintained their symmetry about that axis, would the required value of D be greater than, less than, or the same as in part (a)?

Solution: (a) The forces on charge particle is shown in figure below:

The net force on charged partices due to other charges is zero, that is

$\vec F_{1,net}=\vec F_{12}+\vec F_{13}+\vec F_{14}=0$ Also, components of the forces can also be written as

$F_{1x,net}=F_{12x}+F_{13x}+F_{14x}=0$ $-F_{12}+F_{13}\cos30^\circ+F_{14}\cos30^\circ=0$ $-\frac{1}{4\pi\varepsilon_{0}}\frac{|q_1||q_2|}{r_{12}^2}+\frac{1}{4\pi\varepsilon_{0}}\frac{|q_1||q_3|}{r_{13}^2}\cos30^\circ+\frac{1}{4\pi\epsilon_{0}}\frac{|q_1||q_4|}{r_{14}^2}\cos30^\circ=0$ With the given condition, $$q_3=q_4$$ and also $$r_{13}=r_{14}$$, the above equation can be written as $-\frac{1}{4\pi\varepsilon_{0}}\frac{|q_1||q_2|}{r_{12}^2}+2\frac{1}{4\pi\epsilon_{0}}\frac{|q_1||q_3|}{r_{13}^2}\cos30^\circ=0$ $-\frac{1}{4\pi\varepsilon_{0}}\frac{|q_2|}{(d+D)^2}+2\frac{1}{4\pi\epsilon_{0}}\frac{|q_3|}{(\frac{d}{\cos30^\circ})^2}\cos30^\circ=0$ $\frac{|8.0\times 10^{-19}\space C|}{(2+D)^2}=2\frac{|-1.6\times 10^{-19}\space C|}{(\frac{2}{\sqrt{3}/2})^2}(\sqrt{3}/2)$ $(\sqrt3)(6)(1.6)(2+D)^2=(8)(8)(4)$ $(2+D)=3.924$ $D=1.924\space cm\space (Answer)$ (b) $\frac{1}{4\pi\varepsilon_{0}}\frac{|q_2|}{(d+D)^2}=2\frac{1}{4\pi\varepsilon_{0}}\frac{|q_3|}{(\frac{d}{\cos\theta})^2}\cos\theta$

If particles 3 and 4 were moved closer to the x axis means $$\theta \to 0$$ and the right side of the above equation gets bigger. To keep the equilibrium left side must also be increased by decreasing the value of D. Therefore, the value of D be less than, as in part (a).

1. [25] How many electrons would have to be removed from a coin to leave it with a charge of $$-1.0 \times 10^{-7}$$ C?

Solution:

Using Eq. 1-2, the number of electrons to be removed is equal to: $n=\frac{q}{e}=\frac{|-1.0\times 10^{-7}\space C|}{1.6\times 10^{-19}\space C}=6.25\times 10^{11}\space electrons\space (Answer)$

1. [37] Identify X in the following nuclear reactions: (a) $$^1H + ^9Be \to X + n$$; (b) $$^{12}C + ^1H \to X$$; (c) $$^{15}N + ^1H \to ^4He + X$$. See Appendix F.

Solution: (a) $$^1H + ^9Be \to X + n$$ In the given nuclar reaction, the charge and mass number must be conserved in both sides of the equation. In $$^1H$$, we have one electron and one proton, and in $$^9Be$$, we have 4 electrons, 4 protons and 5 nutrons. So, X on the right must have to 5 protons and 4 nutrons to conserve the equation. Therfore, the final equal becomes

$^1H + ^9Be \to ^9B + n\space (Answer)$

1. $$^{12}C + ^1H \to X$$ In the given equation $$^{12}C$$ has 6 electrons, 6 protons and 8 nutrons, $$^1H$$ has one electron and one proton. So, on the right hand side X must be $$^{13}N$$. Therefore, the final equation will be given by

$^{12}C + ^1H \to ^{13}N\space (Answer)$

1. $$^{15}N + ^1H \to ^4He + X$$

In the given equation $$^{15}N$$ has 7 electrons, 7 protons 8 nutrons, and $$^1H$$ has one electron and one proton. On the right hand side $$^4He$$ has 2 protons, 2 nutrons and X must $$^{12}C$$ with 6 protons and 6 nutrons. Therefore, the final equation will be given by

$^{15}N + ^1H \to ^4He + ^{12}C\space (Answer)$

1. [42] In Fig. 1-16, two tiny conducting balls of identical mass m and identical charge q hang from nonconducting threads of length L. Assume that $$\theta$$ is so small that $$\tan\theta$$ can be replaced by its approximate equal, $$\sin\theta$$. (a) Show that $x=\left(\frac{q^2L}{2\pi\varepsilon_0 mg}\right)^{1/3}$ gives the equilibrium separation x of the balls. (b) If L = 120 cm, m = 10 g, and x = 5.0 cm, what is |q|?

Solution: (a) At equlibrium the sum of all forces at the left or right charge is equal to zero as can be seen in the figure below.

$\sum F_x = T\sin\theta-\frac{1}{4\pi\varepsilon_{0}}\frac{q^2}{x^2}=0$ (1)

$\sum F_y = T\cos\theta-mg=0$ (2)

Dividing (1) by (2), we get

$\tan\theta = \frac{\frac{1}{4\pi\varepsilon_{0}}\frac{q^2}{x^2}}{mg}$ With small approximation, $\tan\theta \approx \sin\theta=\frac{x/2}{L}$ $\tan\theta = \frac{\frac{1}{4\pi\varepsilon_{0}}\frac{q^2}{x^2}}{mg}$ $\frac{x/2}{L}=\frac{\frac{1}{4\pi\varepsilon_{0}}\frac{q^2}{x^2}}{mg}$ $x^3=\frac{q^22L}{4\pi\varepsilon_{0}mg}$ $x=\left(\frac{q^2L}{2\pi\varepsilon_{0}mg}\right)^{1/3}\space (Answer)$ (b) Given L = 120 cm, m = 10 g, and x = 5.0 cm, the value of |q| is $q=\sqrt{\frac{2x^3\pi\varepsilon_0 mg}{L}}=\sqrt{\frac{2(0.05\space m)^3\pi(8.85 \times 10^{-12}{\frac{C^2}{N.m^2}}) (0.01\space kg)(9.8\space m/s^2)}{1.2\space m}}=2.38\times 10^{-8}\space C\space (Answer)$

1. [50] Figure 1-17 shows a long, nonconducting, massless rod of length L, pivoted at its center and balanced with a block of weight W at a distance x from the left end. At the left and right ends of the rod are attached small conducting spheres with positive charges q and 2q, respectively.A distance h directly beneath each of these spheres is a fixed sphere with positive charge Q. (a) Find the distance x when the rod is horizontal and balanced. (b) What value should h have so that the rod exerts no vertical force on the bearing when the rod is horizontal and balanced?

Solution: (a) To find the distance x when the rod is horizontal and balanced we need to use the net torque about the pivot point for all three forces is zero. That is

$\sum\vec \tau = \vec r\times\vec F=0$ $[\hat jF\times(-\hat i\frac{L}{2})]+[-\hat jW\times \hat i(x-\frac{L}{2})]+[\hat jF\times\hat i(\frac{L}{2})]=0$ $-\hat kF(\frac{L}{2})\sin90^\circ-\hat kW(x-\frac{L}{2})\sin90^\circ+\hat k F(\frac{L}{2})\sin90^\circ=0$ $-\frac{1}{4\pi\epsilon_{0}}\frac{qQ}{h^2}\frac{L}{2}\sin90^\circ-W(\frac{L}{2}-L+x)\sin90^\circ+\frac{1}{4\pi\varepsilon_{0}}\frac{2qQ}{h^2}\frac{L}{2}\sin90^\circ =0$ $W(x-\frac{L}{2})= -\frac{1}{4\pi\varepsilon_{0}}\frac{qQ}{h^2}\frac{L}{2}+\frac{1}{4\pi\epsilon_{0}}\frac{2qQ}{h^2}\frac{L}{2}$ $Wx= -\frac{1}{4\pi\varepsilon_{0}}\frac{qQ}{h^2}\frac{L}{2}+\frac{1}{4\pi\varepsilon_{0}}\frac{2qQ}{h^2}\frac{L}{2}+W(\frac{L}{2})$ $x= -\frac{1}{4\pi\varepsilon_{0}}\frac{qQ}{h^2}\frac{L}{2W}+\frac{1}{4\pi\varepsilon_{0}}\frac{2qQ}{h^2}\frac{L}{2W}+\frac{L}{2}$ $x= \frac{1}{4\pi\varepsilon_{0}}\frac{qQ}{h^2}\frac{L}{2W}+\frac{L}{2}=\frac{L}{2} \left[\frac{1}{4\pi\varepsilon_{0}}\frac{qQ}{h^2W}+1\right]\space (Answer)$ (b) Here at equilibrium $$\sum\vec F=0$$ we find

$\frac{1}{4\pi\varepsilon_{0}}\frac{qQ}{h^2}-W+\frac{1}{4\pi\epsilon_{0}}\frac{2qQ}{h^2}=0$ $\frac{1}{4\pi\varepsilon_{0}}\frac{3qQ}{h^2}=W$ $h=\sqrt{\frac{1}{4\pi\varepsilon_{0}}\frac{3qQ}{h^2W}}\space (Answer)$

## Problems Coulombs-Law

Section 21-1 Coulomb’s Law

1. [5] A particle of charge $$+3.00 \times 10^{-6}$$ C is 12.0 cm distant from a second particle of charge $$-1.50 \times 10^{-6}$$ C. Calculate the magnitude of the electrostatic force between the particles.

2. [11] In Fig. 1-18, the particles have charges $$q_1= -q_2$$ = 100 nC and $$q_3 =-q_4$$ = 200 nC, and distance a = 5.0 cm. What are the (a) x and (b) y components of the net electrostatic force on particle 3?

1. [14] Three particles are fixed on an x axis. Particle 1 of charge $$q_1$$ is at x = -a, and particle 2 of charge $$q_2$$ is at x = +a. If their net electrostatic force on particle 3 of charge +Q is to be zero, what must be the ratio $$q_1/q_2$$ when particle 3 is at (a) x = +0.500a and (b) x=+1.50a?

2. [23] In Fig. 1-19, particles 1 and 2 of charge q1 = q2 = +3.20 ^{-19}\$ C are on a y axis at distance d = 17.0 cm from the origin. Particle 3 of charge $$q_3 = +6.40\times 10^{-19}$$ C is moved gradually along the x axis from x = 0 to x = +5.0 m. At what values of x will the magnitude of the electrostatic force on the third particle from the other two particles be (a) minimum and (b) maximum? What are the (c) minimum and (d) maximum magnitudes?

Section 21-2 Charge Is Quantized

1. [24] Two tiny, spherical water drops, with identical charges of $$-1.00 \times 10^{-16}$$ C, have a center-to-center separation of 1.00 cm. (a) What is the magnitude of the electrostatic force acting between them? (b) How many excess electrons are on each drop, giving it its charge imbalance?

2. [26] What is the magnitude of the electrostatic force between a singly charged sodium ion ($$Na^+$$, of charge +e) and an adjacent singly charged chlorine ion ($$Cl^-$$, of charge -e) in a salt crystal if their separation is $$2.82 \times 10^{-10}$$ m?

3. [32] Figure 1-20a shows charged particles 1 and 2 that are fixed in place on an x axis. Particle 1 has a charge with a magnitude of $$|q_1|$$ = 8.00e. Particle 3 of charge $$q_3$$ = +8.00e is initially on the x axis near particle 2. Then particle 3 is gradually moved in the positive direction of the x axis. As a result, the magnitude of the net electrostatic force $$\vec F_{2,net}$$ on particle 2 due to particles 1 and 3 changes. Figure 1-20b gives the x component of that net force as a function of the position x of particle 3. The scale of the x axis is set by $$x_s$$ = 0.80 m.The plot has an asymptote of $$F_{2,net} = 1.5 \times 10^{-25}$$ N as $$x \to \infty$$. As a multiple of e and including the sign, what is the charge $$q_2$$ of particle 2?

1. [33] Calculate the number of coulombs of positive charge in 250 $$cm^3$$ of (neutral) water. (Hint: A hydrogen atom contains one proton; an oxygen atom contains eight protons.)

2. [34] Figure 1-21 shows electrons 1 and 2 on an x axis and charged ions 3 and 4 of identical charge %q and at identical angles u. Electron 2 is free to move; the other three particles are fixed in place at horizontal distances R from electron 2 and are intended to hold electron 2 in place. For physically possible values of $$q \le 5e$$, what are the (a) smallest, (b) second smallest, and (c) third smallest values of $$\theta$$ for which electron 2 is held in place?

Section 21-3 Charge Is Conserved

1. [36] Electrons and positrons are produced by the nuclear transformations of protons and neutrons known as beta decay. (a) If a proton transforms into a neutron, is an electron or a positron produced? (b) If a neutron transforms into a proton,is an electron or a positron produced?