Electric Field related Formulas

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In the previous article Electric Field we had seen that if we gave a charge to any particular particle it experiences a force on it.
Now to understand the relationship between force and charge we did an experiment.
 
Suppose, the charge on a particular particle is q so then the charge particle will experience a force let’s call it as F.  Now if we doubled the charge on the same particle i.e. 2q it will experience double force on it than 1q. So the force on 2q will be 2F. Now we did it again we increased the charge three times i.e. 3q then the force experienced by that charged particle will naturally be 3F acting upon it. 

This experiment concludes that more the charge we add on the particle greater will be the force experienced by it. So the force increases as charge on that particle increases. 
Therefore, Force is directly proportional to the charge q
 
                          F∝ q                                  -------- (1) 

Now we had seen the behaviour of a charge in previous article Electric Field that if we keep a charged particle in a room it experiences a force suppose of 1 Newton. But if we take that same charged particle to some other place or room then the same charge will experience a force of 2 Newton. And at some other place the force acting on that charge is half Newton. 

The same particle the same charge but experiences the different forces at different places. So by our explanation the reason behind all this is Electric Field. 
Different densities of electric fields at different places applies different forces on same charged particle. 

The place where the charged particle experiences the force of 1 Newton the electric field is less but the place where the same charged particle is experiencing double amount of force i.e. 2 Newton the electric field is strong (dense). And the place where the charge is experiencing the force of half Newton the electric field is very weak. So we can conclude that the main reason behind the force experienced by the charged particle is none other than Electric Field (E) itself.

 Therefore, Force is directly proportional to Electric Field (E) 

                            F ∝ E                            ----------- (2) 

Therefore from (1) and (2) 
  
                        F ∝ q × E                        ------------ (3)

 So the force experienced by a particle depends upon two things:

1)  Amount of Electric Charge (Q) present on the particle.
2) Amount of Electric Field (E) present around. 

                       F = K.q.E                          ---------- (4) 

K = Proportionality constant and its value is 
          
           K= 1/(4π∈ ) = 1/(4π∈o. ∈r )

 ∈ =Absolute permittivity of the medium 
∈r =Relative permittivity of the medium.

Now according to equation number (4), 
If 1 Coulomb of charge (q) experiences a force (F) of 1 Newton then we can say that the electric field present is 1 Unit.
  Above statement means that, If Charge q = 1 and it experiences a force of 1 Newton i.e. F = 1 then the Electric field (E) = 1.

  Put q = 1, F = 1, and E = 1 in equation (4)
 Therefore, equation (4) will be 

                        1 = K × 1 × 1

 Therefore, K = 1 

Now put K = 1 in equation (4)

 Therefore the equation (4) will be 

                      F = 1 × q × E

                       F = q × E                       ------------ (5) 

This q is called as test charge

                             E= F/q                      ------------(6)

 If charge q = 1 coulomb then,
                  
                             E = F                       -------------(7) 

 Now according to coulomb’s Inverse square law,

           F = 1/(4π∈)×(Q × q)/r^2          ------------(8) 

          F = 1/(4π∈o)×(Q × q)/r^2 newtons 

 But we have derived that according to definition of electric field

                           F = q×E 

 So both the forces are same 

Therefore, by equating both the equations of coulomb’s law and formula of electric field
 By equating (5) and (8), we get

              1/(4π∈o ) × (Q × q)/r^2 = q × E 

Here Q is source charge Therefore, 

               E= 1/(4π∈o) × q1/r^2

 So this is the formula for electric field



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