Module – 01, Electrostatistic
Chapter – 01, Coulomb’s Law
In our daily life, we often notice that a television screen collects dust easily. In thunder storm, there are huge flashes of lightening. If we rub our nail with mosquito net, we may find a spark which is electric in nature. Sellotape and cling film sticks to everything. A metal rod is fixed at the top of a building etc. All these type of phenomena are related to this electrostatic.
The branch of physics which deals with the study of charges at rest and its properties and the phenomenon related to the charges is called as the Electrostatics.
Let us discuss about the charge.
The property of a body by virtue of which it attracts another body, is called as the charge. It is scalar quantity. It is of two types.
1 ) Positive Charge
2 ) Negative charge
Properties Of Charges :
* Like charges repel each other and opposite charges or unlike charges attract each other.
* The charge can be quantized in terms of its smallest unit i.e electronic charge.
* The charges are additive in nature.
* The charges are scalar quantities as it has only magnitude and no direction.
* The charge is also a fundamental properties of the system as like that of mass.
* The charge is conserved quantity.
Unit of charges :
1 ) SI unit → Coulomb
2 ) CGS unit → esu of charge or stat – coulomb
3 ) Other unit → emu of charges
Relationship between the different units of charges :
1 ) 1 C = 3 * 109 esu of cahrge
2 ) 1 emu of charge = 10 C
3 ) 1 emu of charge = 3 * 1010 esu of charge
Coulomb’s Law :
Statement : ” The electrostatic force of attraction or repulsion between two point charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between the charges. ”
Mathematical Expression :
Let q1 and q1 are the two point charges.
r = distance between the charges
F = Force of attraction or repulsion
According to Coulomb’s Law, we may have
F ∝ q1q2 —————————- ( 1 )
F ∝ 1 / r2 ————————— ( 2 )
On combing above two equations, we get
F ∝ q1q2 / r2
F = K q1q2 / r2
Where K is proportionality constant which is given by
K = 1 / ( 4πε0 ) = 9 * 109 Nm2C-2
where ε0 is the absolute permitivity of free space.
Value of K :
1 ) In SI system → 9 * 109 Nm2C-2
2 ) In CGS system → 1
Expression Of K :
K = 1 / ( 4πε0 )
Dimension Of K :
[ K ] = [ Fr ] / [ q * q ]
⇒[ K ] = [ MLT-2 * L2 ] / [ AT * AT ]
⇒ [ k ] = [ ML3T-4A-2 ]
About of ε0
1 ) Value of ε0 = 8.85 * 10-12 N-1m-2C2
2 ) Units of ε0 → Farad -meter-1 or Fm-1 or N-1m-2C2
3 ) Dimension = [ M-1L-3T4A2 ]
Vector form of the Coulomb’s Law :
Let r1 and r2 are the position vectors of the point charges q1 and q2 and i is the unit vector of the displacement vector of the point charges i.e r = ( r2 – r1 ). Here bold letters indicate vector quantity and unbold letters indicate the magnitude of the bold letters.
i = r / r
F = ( K q1q2 / r2 ) i
F = ( K q1q2 / r2 ) ( r / r )
F = ( K q1q2 / r3 ) r
Similarities between Coulomb’s Law and Newton’s Law of Gravitation :
* Both obeys inverse square law.
* Both involves the conservative forces.
* Both have central force.
* Both obeys newton’s third law of motion.
Dissimilarity between the Electrostatic force and Gravitational force.
|Electrostatic force||Gravitational force|
|1 ) This is between two point charges.||1 ) This is between two point masses.|
|2 ) This force is both attractive and repulsive in nature.||2 ) This force is always attractive in nature.|
|3 ) This force operates over distance which are not very large.||3 ) This force operates over a distance which are very very large.|
|4 ) This forces is explained by Coulomb’s law of electrostatic.||4 ) This force is explained by Newton’s law of gravitation.|
|5 ) This force is dependent of nature of surrounding.||5 ) This force is independent of nature of surrounding.|
Quantisation Of Charges :
It is the property of charges by virtue of which all the free charges are integral multiple of the basic unit of charge called electronic charge.
Q = ± ne
where n = number of free charges
e = electronic charges = 1.6 * 10-19 C
Conservation of charges :
The total amount of charges in an isolated system remains constant.
a ) When a glass rod is rubbed with silk, a positive charge is acquired by the glass where as the silk acquires the negative charge. Thus, the net charge on the rod – silk system is zero before and after rubbing.
b ) whenever an energetic gama photon interacts with the nuclear electric field, electron ( negative charge ) and positron ( positive charge ) are produced. Hence, in this system , the net charge on the system is zero before and after interaction with the field.
Electrostatic Induction :
When any charged body is brought to near another uncharged body, the opposite charge is appears at the uncharged body at the nearer end and at the farthest end the same charge is appeared. This phenomenon is called as the Electrostatic Induction.
In the above figure, we can see that the positively charged body is brought to near an uncharged body. At the nearest end of the uncharged body, opposite charge that is negative charge is appeared and at the farthest end, the same charge i.e positive charge is appeared. These appearance of charges on the uncharged body is nothing but the electrostatic induction. The charges appeared on the charged body are called as the induced charges and the charges on the charged body are termed as the inducing charges.
There two types of induced charges. One is free charge and other is bound charge.
In the above figure, if the uncharged body having induced charges is earthing, then the charges at the farthest end ( positive charges ) are neutralized by the free electrons and the charges at the nearest end i.e the negative charges are spread over the uncharged body. Here, the charges at the nearest end are referred to as the bound charges and the charges at the farthest end is termed as the free charge. After removing, the earthing, the uncharged body is no more uncharged. It is charged permanently.
Limitation of Coulomb’s Law :
1 ) This law does not hold in all situation. It holds when the charges at rest and they are of point size.
2 ) It does not explain the charged bodies of finite size because in this case, the charge is distributed uniformly when two bodies are brought together.
Important Points for remembering :
1 ) when the two identical shaped bodies carrying charges q and q are touched then the net charge on the them is equal to the average charge between them.
2 ) Proton and neutron are themselves built up of more elementary units called quarks.
3 ) A proton and electron combine to produce a gama ray photon. This is called as the Annihilation of matter.
4 ) The charge on the body is not affected by its motion.
Questions – Answer Zone
1 ) State Coulomb’s law of electrostatic.
Ans : The electrostatic force of attraction or repulsion between two point charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between the charges.
2 ) Define Electrostatic induction.
Ans : When any charged body is brought to near another uncharged body, the opposite charge is appears at the uncharged body at the nearer end and at the farthest end the same charge is appeared. This phenomenon is called as the Electrostatic Induction.
3 ) Give an example of inverse square law.
Ans : Coulomb’s law of electrostatic is an example of inverse square law.
4 ) Define free charge.
Ans : If a charged body is brought to near an uncharged body, the opposite charge is appeared at the nearest end while same charge is appeared at the farthest end. The charge at the farthest end is called as free charge as this can be neutralized by earthing.
5 ) what is meant by bound charge ?
Ans : If a charged body is brought to near an uncharged body, the opposite charge is appeared at the nearest end while same charge is appeared at the farthest end. The charge at the nearest end is termed as the bound charge.
6 ) Coulomb is the unit of which quantity ?
Ans : Coulomb is the unit of charge.
7 ) The coulomb’s law of electrostatic stands for which physical quantity ?
Ans : It stands for charge.
8 ) Why electrostatic force is central force ?
Ans : As electrostatic force acts along the line joining the two point charges.
9 ) If the force of attraction between two oppositely charged bodies is 2 N. These bodies are separated by a distance of 0.5 m. What is the amount of charge on the bodies if the charges are equal in magnitude?
Ans : Given that electrostatic force ( F ) = 2 N
distance between the bodies ( r ) = 0.5 m
Let q is equal charge on the body then form Coulomb’s law, we get
F = K q2 / r2
where K = 9 * 10 Nm2C-2
q2 = F r2 / K
q2 = ( 2 * 0.5 * 0.5 ) / ( 9 * 109 )
q = 0.000236 C
q = 2.36 * 10-4 C
10 ) If two charged bodies having charges 8 μC and -5μC are separated by a distance of 10 cm. What is the electrostatic force between these two bodies ?
Ans : Given that
First charge ( q1 ) = 8μC
Second charge ( q2 ) = -5 μC
distance between the charges ( r ) = 10 cm = 0.1 m
Required force = K q1q2 / r2
F = ( 9 * 109 * 8 * 10-6 * 5 * 10-6 ) / ( 0.1 )2
F = 0.36 / 0.01
F = 36 N
11 ) If the electrostatic force between 6μC and 7μC is 0.042 N. Find the distance by which the charges are separated by ?
First charge ( q1 ) = 6μC
Second charge ( q2 ) = 7 μC
distance between the charges ( r ) = ??
force ( F ) = 0.042 N
From Coulomb’s law of electrostatic
F = K q1q2 / r2
r2 = K q1q2 / F
r2 = ( ( 9 * 109 * 6 * 10-6 * 7 * 10-6 ) / 0.042
r2 = 9
r = 3 m
Hence, the required distance between the charges is 3 m.
12 ) If the difference of two charges is 1 μC and the sum of the square of the charges is 13 ( μC )2 . If the electrostatic force between the charges is 0.6 N, find the distance between the charges.
Ans : Let Q and q are the two charges
Force ( F ) = 0.6 N
Q – q = 1 μ —————————- ( 1 )
Q2 + q2 = 13 μ2 ————————– ( 2 )
( Q – q )2 + 2 Qq = 13 μ2
By using equation ( 1 ) we get
( 1 μ )2 + 2Qq = 13 μ2
2 Qq = 13μ2 – 1 μ2
2 Qq = 12μ2
Qq = 6 μ2
From Coulomb’s law of electrostatic
F = K Qq / r2
r2 = KQq / F
r2 = ( 9 * 109 ) * 6 μ2 / 0.6
r2 = ( 9 * 109 * 6 * 10-6 * 10-6 ) / 0.6
r2 = 0.054 / 0.06
r2 = 0.09
r = 0.3 m
13 ) If the 2μC is placed at each corner of a square of side 0.2 cm, find the force experienced by the -1μC placed at the center of the square.
14 ) At each corer of rectangle ABCD having length l and breadth, x μC is placed. Find the force experienced by the charge placed at A.
15 ) At three corners of a square of side a, 0.2μC is placed. What charge should be placed at the fourth corner so that the system is in equilibrium ?