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Force between 2 charged particles is always q1q2/4pi*epsilonknot*r2 irrespective of the medium between charged particles. Plzz xplain how??

Force between 2 charged particles is always q1q2/4pi*epsilonknot*r2 irrespective of the medium between charged particles. Plzz xplain how??

Grade:11

2 Answers

Aishwarya Muralidharan
34 Points
9 years ago

According to newtons law of gravitation, gravitational force,

F=Gm1*m2/r2, where m1, m2 are the masses of the two bodies , r is the distance between them and G is the gravitational constant.

Similarly if we consider two point charges - q1 and q2, distance between them is r in vacuum, then the force acting between them i.e. the electrostatic force on any one of them due to the other is directly proportional to the product of the magnitude of the charges and inversely proportional to the square of the distance betweeen them. this was measured by Charles Augustine de Coloumb.

this force in magnitude, F IS DIRECTLY PROPORTIONAL TO q1*q2/r2

i.e.  F=(1/4PI EPSILONKNOT)  * ( q1*q2/r2)

where 1/4pi epsilon knot is the constant that we put on removing proportionality sign. it is called PERMITTIVITY OF FREE SPACE= 8.854*10-12C2N-1m-2.  this is the permittivity of vacuum but depending upon the material medium, its value changes. the electrostatic force is dependent of the medium between the charged particles.

The permittivity of a medium describes how much electric field (more correctly, flux) is generated per unit charge in that medium.

FITJEE
43 Points
8 years ago

 No,the statemennt that force between the two paricle is kq1q2/r^2 . because this coulomb is dependent on the type of medium.

k is a constant whose value is 1/4 pie epsilon naught epsilon

the epsilon determines the dielectric constant for medium. and for vacumm its value is 1 thats why its not written in the formula.


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