Introduction to Electrostatics with Solved Examples
Table of Contents
Electrostatics
Introduction to Electrostatics with Solved Examples- The study of electricity can be done under two topics; electrostatic and electrodynamics. In this article, we will study only electrostatics. Electrostatics is that branch of electricity, which deals with a static condition of charges. Electrostatics is also known as static electricity. Electrostatics is that branch of electricity which deals with charges at rest. In other words, it is that branch of science, in which static condition of electricity or charges is studied, is called static electricity or electrostatics.
The study of the behavior of the charges, when they are at rest, is called electrostatics. Electrostatics is that branch of science, which deals with the phenomena associated with electricity at rest.
We know that an atom tends to be electrically neutral, because the total number of protons (or the quantity of total positive charge) tends to be equal to the total number of electrons (or quantity of total negative charge) existing in any normal atom. For example, the simplest atom is that of hydrogen, which contains just one proton and one electron. Thus, it tends to be electrically neutral.
If electrons are omitted somehow from the outer shells or orbits of an atom of any substance, a positive charge generates on the body, and that body is called positively charged. On the contrary, if some electrons are entered into that body, a negative charge develops on the body and the body is said to be negatively charged. For example, when a glass rod is rubbed against a piece of silk cloth, then a positive charge on glass rod, whereas a negative charge produces on silk cloth. If two glass rods are rubbed with a silk cloth and then brought closer to each other, both these rods repel each other. Its reason is that a same charge generates on both the rods and as per rule, two similarly charged objects repel each other, whereas two negatively charged bodies attract each other. When a body is scratched, electrons present in it get heat energy. Hence, electrons become lighter, emit from this body and enter into another body (with which it has been rubbed). As a result, emission of electrons from a body and their inclusion into some other body, both bodies become electrically charged. A positive charge creates on the body from which electrons are expelled, whereas a negative charge emerges on the body, in which electrons penetrate or electrons are included.
Positive and Negative Charge
We know that atoms of all elements are electronically neutral, that’s an atom neither has a positive charge nor a negative charge. Because, the number of protons in any atom turn out to be exactly equal to the number of electrons existing in it. In other words, negative charge found on the electrons of any atom, tends to be equal to the positive charge existing on the protons of that atom. As such, resultant charge turns out to be zero and every atom is called electrically neutral. If some electrons are extricated somehow from the outer shell of an element’s atom, then balance of the atom disturbs. If electrons are emitted from an element ‘s atom, the sum of such emitted electrons of the atom is called negative charge, and a positive charge creates on the element or that body. On the contrary, if one or a few electrons are added to an atom (or if some electrons are included into the atom), then negative charge found on the atom will increase or there will be an excess in the negative charge.
The substance or element, electrons are included or added into the atom of which, becomes a negatively charged body.
In short, it may be said that an element or body becomes positive electrostatics as a result of deficiency of electrons in that element or body, and as a result of excess of electrons, the body becomes negative electrostatics. The total deficiency or excess of electrons in a body is known as its charge.
Absolute and Relative permittivity of a Medium
Before discussing about the status of electrostatic, it is very essential to know about a very specific characteristic of a medium, which is known as permittivity. This characteristic plays an extremely significant role in electrostatics. Every medium consists of the following two permittivityies.
(i). Absolute Permittivity
(ii). Relative Permittivity
The capacity of concentrating the electric flux in a dielectric or insulator is called permittivity.
(i). Absolute Permittivity
The value of resulting flux density (D) at any given value of electric field’s strength depends on the medium, in which flux creates. The ratio between D and E is known as absolute permittivity of the medium, i.e.,
ԑ = D/E = ԑ0 ԑr
The unit of absolute permittivity is Farad per meter (F/m). if the medium is free space, then the ratio of D/E in free space for any electric field, is known as permittivity or electric space constant, which is denoted by ԑ0. The unit of free space also tends to be farad per meter.
The ratio of D/E for an electric field in vacuo is called the permittivity of free space or the electric space constant, symbol ԑ0. The value of absolute permittivity in free space tends to be as follow;
ԑ0 = 1 / 36 π x 10 -9 F/m = 8.854 x 10 -12 F/m
(ii). Relative Permittivity
To measure absolute permittivity, vacuum or free space is used as a reference medium. Relative permittivity is a ratio of the absolute permittivity (ԑ) of a material to the absolute permittivity (ԑ0) of a vacuum, i.e.,
ԑr = ԑ / ԑ0
∴ ԑ = ԑ0 ԑr F/m
Its value in free space and air is unity, i.e., ԑr = 1. As a result of unity ratio between two similar quantities, there is no unit for relative permittivity. The permittivity value of other material ranges between 1 to 10. For example, paper’s permittivity tends to be between 2 to 2.5, mica 3 to 7, and glass 5 to 10. Only water and Titanate compound’s values cross this range.
If some other medium is taken instead of air or free space, and the relative permittivity of that medium with respect to free space is ԑr, then the value of absolute permittivity can be determined through the following formula;
ԑ = ԑ0 ԑr F/m
for example, if mica’s relative permittivity is 5, then its absolute permittivity will be as follows;
ԑ = ԑ0 ԑr = 8.854 x 10 -12 x 5 = 44.27 x 10 -12 F/m
Laws of Electrostatics
Charles Augustin Coulomb laid the foundation of electrostatics laws in the year 1785, which are also known as Coulomb’s laws. Remember nobody had any idea about the electric charge unit until 1784. Coulomb devised these laws as a result of continuous experiments on two masses on the basis of inverse square law. The detail of these laws is as below;
First Law;
Like charges of electricity repel each other, whereas unlike charges attract each other.
Second law
According to this law, force exerted between two small charged bodies (point charges) is directly proportional to the product of their charges and inversely proportional to the square of the distance between them OR the electric force of repulsion or attraction between two-point charges varies directly as the product of the charges and inversely as the square of the distances between them.
The above mentioned both laws are known as Coulomb’s laws, and algebraically, they can be expressed as below;
F α Q1 Q2 / d2
F = K Q1 Q2 / d2
Here, Q1 and Q2 means two-point charges, the mutual distance of which is “d”, as has been illustrated vide fig. 6.1. ᴋ is constant, the value of which depends on the units, in which F, Q and d are measured. Moreover, it also depends on the dielectric (insulating medium found between charged bodies) characteristics. Constant ᴋ is sometimes also denoted by Lambada λ.
Figure. 6.1 – Demonstration of electrostatics laws
If the medium of two-point charges is free space, and the unit system is MKS, i.e., F measured in Newtons, d in meters, and Q in coulombs, then the coulomb force of free space will be as follows;
F = 1 / 4 π Ԑ0. Q1 Q2 / d2
If medium consists of a dielectric or insulation, then the electrostatic force found between two charged bodies tends to be less than the free space due to relative permittivity of this dielectric. Thus, coulomb force of such medium tends to be as follows;
F = 1 / 4π Ԑ0 Ԑr Q1 Q2/ d2
Whereas, the value of Ԑ0 is 8.854 x 10 -12 and value of Ԑr (in case of air) 1.006. if the values of Q1 and Q2 are in coulombs, d in meters, and Ԑ in Farad per meter, then F tends to be in Newtons. Thus, coulomb’s law can also be expressed as follows;
F = Q1 Q2 / 4π Ԑ0 Ԑr d2
For example, in situation of air;
Ԑr = 1; 0Ԑ = 8.854 x 10 -12 F/m
Q1 = Q2 = 1 Coulomb
D = 1 meter
F = 1 x 1 / 4π x 8.854 x 10-12 x 12 = 8.997 x 109 N
= 9 x 109 N approx.
Hence, coulomb law can also be written as;
F = 9 x 109 Q1 Q2 / Ԑr d2 (in medium)
F = 9 x 109 Q1 Q2 / d2 (in air vacuum)
Thus, a coulomb charge can be defined as; a charge or magnitude of electricity which if placed in the air at a distance of one meter from a like and equal (or unequal) charge, it repels (or attracts) the same at a force of 9 x 109 newtons.
Calculations about Coulomb’s Law
Example 1: Calculate the electrostatic force of repulsion between two α – particles when at a distance of 10 -13 m from each other. Charge of α – particle is 3.2 x 10 -12 C. if mass of each particle is 6.68 x 10 -27 kg, compare this force with the gravitational force between them. Take the gravitational constant as equal to 6.67 x 10 -11 N-m2 / kg2.
Solution;
Here, Q1 = Q2 = 3.2 x 10 -19 C, d = 10 -13 m
F = 9 x 109 Q1 Q2 / d2 = 9 x 109 x 3.2 x 10-19 x 3.2 x 10 -19 / (10-13)2 = 9.2 x 10-2 N
The force of gravitational attraction between the two particles is given by;
F = G m1 m2 / d2 = 6.67 x 10 -11 x (6.68 x 10 -27)2 / (10 -13)2 = 2.97 x 10 -37 N
Obviously, this force is negligible as compared to the electrostatic force between the two particles.
Example 2; Calculate the distance of separation between two electrons (in vacuum) for which the electric force between them is equal to the gravitational force on one of them at the earth surface. Mass of electron = 9.1 x 10 -31 kg, Charge of electron = 1.6 x 10 -19 C.
Solution;
Gravitational force on one electron = mg newton = 9.1 x 10 -31 x 9.81 N
Electrostatic force between the electrons;
= 9 x 109 Q2 / d2 = 9 x 109 x (1.6 x 10-19)2 / d2 N
Equating the two forces, we get;
9 x 109 x 2.56 x 10-38 / d2 = 9.1 x 10-31 x 9.81 ∴ d = 5.08 m
Example 3: Two small identical conducting spheres have charges of 0.2 x 10-9 C and -0.5 x 10 -9 respectively. When they are placed 4 cm apart, what is the force between them? If they are brought into contact and then separated by 4 cm, what is the force between them?
Solution;
F = 9 x 10-9 Q1 Q2 / d2 = 9 x 10-9 x 2 x 10-9 x (-0.5 x 10-9) / 0.042 = 56.25 x 10 -7 N
When two identical spheres are brought into contact with each other and then separated, each gets half of the total charge. Hence,
Q1 = Q2 = [2 x 10-9 + (-0.5 x 10-9)] / 2 = 0.75 x 10-9 C
When they are separated by 4 cm.
F = 9 x 10-9 x (0.75 x 10 -9)2 / 0.042 = 0.316 x 10-5 N
Example 4: Two-point charges are separated by a distance of 20 cm in air. One of the charges is positive and has a magnitude of 8 micro – coulomb and the other is a negative charge of magnitude 12 micro – coulombs. Calculate the force between the charges. Is it a force of attraction or repulsion?
Solution;
The force between two charges is;
F = q1 q2 / 4 π ɛ r ɛ0 d2 newton OR F – 9 x 10 -9 Q1 Q2 / d2 newton
[1 / 4 π ɛ0 = 9 x 1099approx.)]
Since the charges are in air,
ɛr = 1 and ɛ0 = 8.854 x 10 -12
Here, q1 = 8 x 10 -6 coulombs, q2 = 12 x 10-6 coulombs and d = 20 cm = 0.2 meters
Substituting the values of the various quantities, we have;
F = 8 x 10-6 x 12 x 10-6 / 4 π x 1 x 8.854 x 10-12 x (0.02)2
= 20.7 newtons
Since the two charges are of opposite nature, therefore the force between them is of attraction.
Example 5; Two similar charges placed 10 cm apart in air, repel each other with a force of 10-3 newtons. Determine the quantity of each charge in coulomb.
Solution;
Let the magnitude of the charge be q coulombs. Then, the force of repulsion F is given by;
F = q x q / 4π Ԑ0 Ԑr. d2
q2 = 4π Ԑ0 Ԑr d2. F
since, Ԑ0 = 8.854 x 10-12
Ԑr = 1 (the medium being air)
d = 10 CMS = 0.1 meter
F = 10-3 newtons
q2 = 4π 8.854 x 10-12 x 1 x (0.1)2 x 10 -3
or
q = √4π x 8.85 x 10-12 x 0.01 x 10 -3
= 0.033 micro. Coulomb
Example 6: What is the charge, expressed in micro- coulombs, on two equally and similarly charged spheres placed in air with their center 20 cm apart repelling each other with a force of 1.07 x 10-3 N?
Solution;
F = 1.07 x 10-3 N
And Ԑr = 1 (medium being air)
Ԑ0 = 8.854 x 10-12
And d = 20 cm = 0.2 meter
Now F = q2 / 4π Ԑ0 Ԑr d2 N or q2 = F x 4π Ԑ0 Ԑr d2 coulombs, which on substituting the above values, gives, i.e.,
q2 = 1.07 x 10-3 x 4π x 8.854 x 1-12 x 1 x (0.2)2 = 4.762 x 10-15
q = √4.762 x 10 -15 = 6.9 x 10-8 coulombs, or = 6.9 x 10-8 x 106 micro-coulomb = 0.069 μ C
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