COULOMB'S LAW

STATEMENT:-

                           Electrostatic force of attraction or repulsion between two charges is directly proportional to product of their magnitude and inversely proportional to squares of distance between them.

EXPLANATION & DERIVATION:-

                                     

                         Consider two charges "q1" and "q2" placed at some distance "r" from each other, then by coulombs law coulombic force of attraction or repulsion "F" is given by

 F∝q1q2 ------(a)

 F∝1/r^2------(b)

joining (a) & (b)

we get

F∝q1q2/r^2

or

F=k	q1q2			or	F =	1		q1q2		----(c)
	r2					4πε0		r2

where k is proportionality or coulombs constant

k=      1       = 8.99 × 10⁹ Nm²/C²

      4πε₀  

ε_{0 = Permittivity of free Space=} 8.85 × 10⁻¹² C²/Nm²

however if any dielectric or any other medium  other than air or vacuum is present as shown in fig then force between the charges is given by


 --------(d)

whereas e_{r= Relative Permittivity of a medium}

_{and it depends upon nature of medium}

_{in vector form (c) & (d) can be written as} 

---------(e)

--------(f)

(d) or (f) shows that force decreases in the presence of dielectric medium of relative permittivity

e_r

IN CASE OF TWO FORCES:-

ATTRACTIVE FORCES:-

For the fig eq (e) can be written as

------(g)

------(h)

whereas

 Force exerted by charge q2

on charge q1

              Force exerted by charge q1

on charge q2

By Newtons third law

------(i)

that is force exerted by charge q2 on q1 is equal to the force exerted by charge q1 on q2

using (g) & (h) in (i) we get

after cancellation we get

---------(j)

Eq (j) shows for attractive forces unit vectors are opposite direction to each other
REPULSIVE FORCE:-

                                   If the medium between two charges is other than air or vaccume then (g) and (h) can be written as

-----(k)

                 

-----(l)

In general the magnitude of electrostatic force in the presence of any dielectric is written as

CONCLUSION:

 that is electrostatic force decreases in the presence of any dielectric medium.

RELATIVE PERMITTIVITY OF SOME MATERIALS AT ROOM TEMPERATURE:-

Material Dielectric Constant Vacuum 1 Glass 5-10 Mica 3-6 Mylar 3.1 Neoprene 6.70 Plexiglas 3.40 Polyethylene 2.25 Polyvinyl chloride 3.18 Teflon 2.1 Germanium 16 Strontiun titanate 310 Titanium dioxide (rutile) 173 perp 86 para Water 80.4 Glycerin 42.5 Liquid ammonia(-78°C 25 Benzene 2.284 Air(1 atm) 1.00059 Air(100 atm) 1.0548

NOTE:-

Attractive force exists between two charges and repulsive force exist outside but for both kinds of force either attractive or repulsive force of one charge on other is in opposite direction to the other and exists on same line which join their centres.

STATIC EQUILIBRIUM OF CHARGES:-

Following are the conditions for

• Three charges must be in straight line.
• Charge in the middle must be opposite to the charges at the corner to provide an attractive force between the charges to counteract with the attractive force between the charges at corner.
• The charge in the middle must be in between the charges at cornes otherwise two forces of attraction on it would not be in same direction.  

SOLVING METHOD OF STATIC EQUILIBRIUM OF CHARGES:-

Consider the following problem:

3 charges are in static equilibrium. Q1 has charge 4q. Q2 has charge q. Q3 has a small charge. where is Q3 and what is charge on Q3?

SOLUTION:-

Suppose charge Q1=+4q is at the origin and charge Q2=+q is at x=1. Charge Q3=-x is placed at x=a where 0<a<1. The force on each of the three charged bodies that are attractive and repulsive must be equal and opposite. We only need to consider the force on any two of the charged bodies to get two equations to solve for the two unknowns (x, a) in the third equation can be derived from the other two.

Consider the charge Q1 on it attractive forces given by Q3 are equal to repulsive force provided by Q2. So 

K4q.x/a^2=-K.4.q.q/1^2 since attractive force= -repulsive force or x is negative

 after solving we get 

x=-(a²).q----(a)

 for Q2 

Kqx/(1-a)²=K4qq/1^2

after solving we get

x=4q.(1-a)²-----(b)

equating (a) & (b) we get

4q.(1-a)²=q.a²

after solving we get

4.(1-2a+a²)=a²

4-8a+4a²=a²

3a²-8a+4=0

by applying quadratic formulae or any other method we get

a=2/3        and a=2

since a is not equal to 2 so a=2/3

that is Q3 should be placed at 2/3 of the distance from Q1 to Q2 that is closer to Q2

In order to calculate charge on Q3 we know attractive force on Q1 is equal but opposite to its repulsive forces

 so K.4q.x/(2/3)²=-K.4q.q/1²

after solving we get

x=4q/9 

hence we find that charge on the body at centre is 4q/9 & placed at 2/3 of the distance from Q1 to Q2

APPLICATIONS OF COULOMBS LAW:-

There are various applications of coulombs law. Coulombs law is also applied in batteries since water is used in lead storage batteries since water has very high relative permittivity and we know that

 because of this electrostatic force of attraction between the plates of batteries reduces so chances of shortage of plates of batteries decrease and consequently life of battery increases. Just like that coulombs law has many of applications.