Introduction
Analogy is a powerful tool, and we use it often in physics. Remember, we started this journey with the static of mechanical objects. Now we move further from macroscopic to microscopic and analyze interaction between charges and the effect on the surrounding medium.
Electric Charge and Charge Conservation
A basic description of substances leads us to the concepts of electrical charges and the fundamental particle—the electron. Empirically, it has been found that a piece of amber gemstone, rubbed with a piece of animal fur, will create a new type of interaction; it will be able to attract to its surface small pieces of material and dust particles. The material is said to be electrically charged, and the smallest particle of charge is called the electron. The electron was assigned a negative charge of 1.6 · 10–19 Coulombs. These electrons are considered one of the elementary subatomic particles. Electrons (negatively charged) are constituents of atoms, which are electrically neutral (zero net charge). The atom also contains positive charges, or protons, which are positioned in the central part of the atom, the nucleus. The nucleus is also composed of neutral particles, called neutrons.
Contrary to the previous example where the amber was charged up with electrons, other materials tend to charge positively (with protons). This has been summarized in the electrostatic series. Materials such as amber, rubber, and polyethylene charge negatively, whereas paper, cat's fur, and nylon charge positively. The difference between the types of charge can be easily shown by letting two of these materials interact. Materials from the same group repel each other, and materials from the opposite group attract each other. This defines the electrical interaction.
Materials are considered to be conductors, semiconductors, or insulators. The nomenclature is revealing. Conductors will let charge spread in the volume. Semiconductors will vary their electrical properties and sometimes behave as conductors while at other times behaving as insulators. Insulators will localize the charge acquired; their electrons bond strongly to the atoms and make it hard for electrical charge to flow through the material.
A last observation. Charging an object can be achieved by different means: friction, conduction, or induction. Charging by friction involves rubbing two materials against each other and in the process electrons are transferred from one object to the other. Conduction involves a conductive material in which electrons are free to move in the volume. Induction is established between two materials that do not touch, but where the material 'that is charged interacts with particles in the uncharged material, creating a distribution of charge on the surface close to it.
In the neutral state, the atom has an equal number of electrons and protons, and the net charge is zero.
n = p
where n is the number of electrons and p is the number of protons. If an electron is given sufficient energy to be able to break the bond with the atom and leave it, then the atom becomes charged, as it now has one more proton than electrons. The atom becomes positively charged, and we call this new particle a positive ion (n < p). If by friction between two objects, one of the objects is charging positively (losing electrons), we typically retrieve the electrons on the second object. Then the second object, initially neutral, becomes negatively charged. This ion is called a negative ion (n > p). If the two objects are isolated from the exterior (that is, isolated from other objects), then the charges are moving from one to the other, and the total number of electrons and protons will stay the same; the ions simply rearrange between the two objects. We can define this as a new law of conservation, similar to the conservation of energy and conservation of momentum in mechanics and to the conservation of thermal energy in thermodynamics. This new law is the conservation of charge.
After the charge is rearranged between the objects in contact, the state remains unchanged as long as there is no other interaction. This defines a situation of equilibrium similar to previous cases of mechanical and thermal equilibrium where the charge, whether positive or negative, is measured in Coulombs (1 C), and its usual symbol is q. As mentioned previously, an electron has negative charge:
qe = –1.6 · 10–19 C
The proton is equal in charge but has more mass:
qp = +1.6 · 10–19 C
Example
Consider two objects, one made of rubber and the other of nylon that is electrically neutral. You rub the two objects together and the rubber becomes charged with –2.88 · 10–16 Coulombs. What is the nylon's charge, and how many electrons have been shifted to the rubber from the nylon in the process?
Solution
Initially, both objects have a zero net charge, and, because the system will be considered to be isolated, the same amount of charge will be retrieved in the final state.
qrubber = –2.88 · 10–16 C
qnylon = ?
N =?
qrubber + qnylon = 0 C
–2.88 · 10–16 C + qnylon = 0 C
qnylon = – (–2.88 · 10–16 C)
qnylon = + 2.88 · 10–16 C
The number of electrons that make up the charge of the rubber is:
N = qrubber/qe
N = (–2.88 · 10–16 C)/(–1.6 · 10–19 C) = 1,800 electrons pass from nylon to rubber.
Electric Forces and Coulomb's Law
We have talked about interaction between materials in the electrostatic series. We can summarize that there are two types of electrical interactions: attraction and repulsion. Attraction forces are established between particles of different charge sign (positive and negative charges). Repulsion forces act between charges of the same kind.
The quantitative expression of the electrical force is known as Coulomb's law (from Charles de Coulomb, 1736–1806). If one considers the charges to be pointlike (that is, all charges gather in one point), then the law says: The magnitude of the electrical force exerted by a point-like charge q1 on a charge q2 is proportional to the product of the charges and inversely proportional to the square of the distance between the charges.
Where the proportionality constant is:
k = 9 · 109N · m2/C2
The electric force, as are all other forces, is a vector. The direction is given by the type of particles interacting, as shown in Figures 13.2 and 13.3 and defined previously.
Although the names given to the forces are different to show the action charge and the test-charge (for example, F21 indicates that charge 2 is acting on charge 1), the absolute value is the same.
F21 = F12
Also shown in the figures is the opposite direction of interaction.
In a vector form, the law is expressed as:
· r
One can see that the new formula shows that the direction of the electric force is the same as the direction between the two charges.
Coulomb's Law
The magnitude of the electrical force exerted by a point-like charge q1 on a charge q2 is proportional to the product of the charges and inversely proportional to the square of the distance between the charges.
Example 1
Two amber beads are brought in proximity after being charged with 10 and 50 electrons. They are placed at a distance of 20 cm from each other. Find the force acting on each bead.
Solution 1
First, convert to the units necessary to work with Coulomb's law and then find the charges. Then, calculate the forces on each object.
R = 20cm = 0.2m
q1 = 10 · (–1.6 · 10–19 C) = –1.6 · 10–18 C
q2 = 50 · (–1.6 · 10–19 C) = –8.0 · 10–18 C
F21 = F12 = ?
F21 = 
= 9 · 10–9 N · m2/C2 · 
F21 = 9 · 10–9 N · m2/C2 · 
F21 = 28.8 · 10–23N
F21 = F12 = 28.8 ·10–23N
The charges are the same sign, and therefore, the force is of repulsion.
If there are three or more objects charged and interacting, you will have to find the force that one charge is acted upon by all the other neighboring charges, and then find the net force. Remember that the force will be a vector, so the net force is the sum of vectors (value and direction).
Example 2
Consider three equal charges placed on the x-axis such that the distances between are 100 cm and 300 cm between charge 1 and charge 2 and between charge 1 and charge 3, respectively. Find whether the charge in the middle is at rest or not.
Solution 2
In order to find our answer, we start by converting all values to SI. Then, we will find the net force on the middle charge and see if it is zero or nonzero. In the case the force is zero, the charge is at rest (Newton's second law). If the net force is nonzero, the charge will be accelerated in the direction of the net force.
r12 = 100cm = 1m
r13 = 300cm = 3m
r23 =300 cm – 100 cm = 2 m = 2 · r12
q1 = q2 = q3 = q
Fnet = ?
Charge 2 is acted upon by two forces because of repulsion with the charges 1 and 3. The forces are F23 exerted by charge 3 on charge 2 and F21 exerted by charge 1 on charge 2. As you can see from the figure, they are in opposite directions.
If we consider the positive direction as in the figure, then F21 is positive and F23 is negative.
F21 = 
F23 =
As mentioned in the problem statement:
r23 = 2 · r12 = 2 · r21
The last term in the equality is possible to write because we are interested in the absolute value and not the direction (which was considered in drawing the figure).
F21 = 
Fnet = F23 + F21
Fnet = –F23 + F21
This result tells us that the charge is acted upon by a net force and subsequently will be accelerated by an acceleration that can be calculated based on Newton's second law if the mass of the charge is known:
Fnet = m · a
The force is in the direction of F21 , meaning the object is accelerated toward the positive x direction.
Electric Field and Potential
As we discovered previously, there is no need for contact in order to have electrical interaction, and this is not the only case we learned about so far. Gravitational force is carried out by the gravitational field, which changes the properties of the space around an object: The more massive the object, the larger the gravitational field and the greater the interaction with other objects found in proximity. Heating though radiation is another noncontact interaction we discussed. In the case of radiation, electromagnetic waves produced by a far-away source, such as the sun, can heat up an object, effectively increasing its temperature.
The space around an electrically charged object carries special properties, and we call this an electrical field. Any other charge in this region will interact with the electrical field, and we call this charge a testcharge. Hence, we have both a field-producing charge and a test-charge(s) that will be affected by the electrical field.
The electric field that propagates electrical interaction has vector properties because a positive charge will affect the charges around it in a different way than a negative charge will. The direction of the electric field and of the electric force is the same at any point around the charge.
The vectors starting or finishing on the charges are called electrical field lines, and they represent the direction of the electric field and electric force, as seen in Figure 13.7. The density of electrical field lines in a region represents the strength of the field in that area.
Depending on the position of the test-charge, there are many arrangements of the field lines encountered in applications. One is shown in the following example and figure.
The Electric Field
The electric field at a location around a charge q is defined to be proportional to the force at the specified location and inversely proportional to the charge by the field.
Example 1
Consider positive and negative spherical charges in proximity. Draw the field lines.
Solution 1
The field lines are similar to the ones we have shown previously, but because the two particles interact, the field created by each will affect the other particle. The final arrangement looks like that in Figure 13.8.
According to the definition for electric field, the unit for the electric field will be the Newton/Coulomb (N/C). Because force was previously defined by Coulomb's law, if we consider the charge ql creating a field E and a test-charge q2 to be present in the field, then the field at point r with respect to charge ql is:
where F21 is the force that charge q2 feels due to the field created by charge q1. In the last expression, the field created by the charge q1 is seen to be in the same direction as the vector position r if the charge is positive and opposite to r if the charge is negative, relative to the field lines drawn at the beginning of this section.
If we are interested in the value of the field created by charge q1 then:
If the field is created by two or more electrical charges, the vector expression is helpful in finding the net electric field at one location as we will see in the following example.
Example 2
Consider two charges, as shown in Figure 13.9, creating a net electric field. Find the direction and the value of the net field at point A. The charges are q1 = + 1.6 · 10–19 C and q2 = –1.6 · 10–19C, and the distance to the point is r1 = 22 cm and r2 = 12 cm.
Electric Potential Energy
The electric potential energy (EPE) per unit of charge defines the electric potential
Solution 2
First, convert the data into SI units. Then, find the field determined by each charge respectively at point A, and next find the net field.
r1 = 22cm = 0.22m
r2 = 12 cm = 0.12 m
q1 = +1.6 · 10–19 C
q2 = –1.6 · 10–19 C
Enet = ?
E1 = 
The vector position has both a horizontal and a vertical component, and they can be represented as:
r1 = (r1x, r1y) = (r1 · cos α ;r1 · sin α)
where α is the angle the vector makes with the positive x-axis. In the case of the vector r1 the angle is 360° – 60° = 300°
r1 = (r1x,r1y) = (r1 · cos 300°; r1 · sin 300°)
r1 = (r1x,r1y) = (0.22 · cos 300°; 0.22 · sin 300°)
r1 = (r1x,r1y) = (0.11;–0.19) m
E1 =
E1 = 1.4 · 10–7 – (0.11;–0.19) N/C
E1 = (0.15;–0.27) · 10–7 N/C
A similar calculation can be made for the components of the second charge distance and electric field, and it can be said that:
r2 = (r2x,r2y) = (r2x,r2y) = r2 · cos 0°;r2 · sin 0°)
r2 = (r2x,r2y) = (r2x,r2y) = 0.12 · cos 0°;0.12 · sin 0°)
r2 = (r2x,r2y) = (0.12;0) m
E2 = (1;0) · 10–7N/C
The net electric field is the vector sum of the two fields:
Enet = E1 + E2 = [(0.15;–0.27) + (1;0)] · 10–7N/C
Enet = E1 + E2 = (1.15;–0.27) · 10–7N/C
And the direction of the field can be found from the Ex and Ey components of the field:
θ = –13°
Similar to the situation of a moving object in a gravitational field (where a potential energy is associated with the position of the object in the field), an electric potential energy can be defined for a charge placed in an electric field. As a greater mass needs more energy to be moved in a gravitational field, so does a larger charge need more energy to be moved through an electric field.
Electric potential, known as V, is "measured" in joule/coulombs and is called a volt (from Alessandro Volta, 1745–1827). Due to the unit for electric potential, this quantity is also called voltage. The voltage is seen previously to depend on energy, which is a scalar quantity; therefore, voltage is also a scalar. Voltage will still be positive or negative depending on the type of charge. The quotation marks in the first sentence of this paragraph are intended to emphasize the fact that voltage based on electric potential energy is a relative quantity. Potential energy, if you remember from mechanics, is proportional to the position of an object but with respect to an origin. A more measurable quantity is the difference in potential energy, hence the difference of voltage between two positions.
If a charge q is moving in an electric field E between positions A and B, then the field does work on the charge that is equal to the variation of the electric potential energy (analogous to the work done by gravity: W = – ΔPE).
In this case, the work to move the charge from point A to B is:
W = – ΔEPE = –(EPEB – EPEA)
Dividing both sides by the charge q:
And considering the previous definition of voltage:
The work can be related to the voltage difference by:
Example 3
Consider a test-charge q = –1.6 pC move in an electric field E from point A to point B. The work done by the field to move the charge is 2.8 fJ. Find the potential difference between the two points.
Solution 3
First, convert the units and then proceed in formulating the solution.
W = 2.8 fJ = 2.8 · 10–15J
q = –1.6 · 10–12 C
ΔV = ?
ΔV = 1.7 · 10–3
Practice problems of this concept can be found at: Electrostatics Practice Questions
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