The Theory Of Electric Field

Introduction:

Every electrical device of circuit has its own electric field. Electric field means a region around a charged particle. Every electrolytic capacitor has its own electric field. In this article we will discuss about the electric field of electrolytic capacitor.

Abstract: 

Every electrical device has its own electric field. It can also able to create a new electric field. Here we will discuss about the electric field of capacitor. 

Method: 

I have invented a formula to determine the value of electric field of electrolytic capacitor. Here’s the method-

We know that-    E= kQ÷r²

                             = kQ÷k²C²

                             = 1÷k×Q÷C²

                             = ΨᵦQ÷C²

So, E= ΨᵦQ÷C²_________(i)

If Q=2Q then-

      E'= Ψᵦ×2Q÷C²

         = 2ΨᵦQ÷C²

So, E'÷E= 2ΨᵦQ÷C²÷ ΨᵦQ÷C²

              = 2ΨᵦQ÷C²× C²÷ ΨᵦQ

              = 2

So, E'=2E

If Q=½Q then-

      E'= Ψᵦ×½Q÷C²

         = ½ΨᵦQ÷C²

So, E'÷E= ½ΨᵦQ÷C²÷ΨᵦQ÷C²

              = ½ΨᵦQ÷C²×C²÷ ΨᵦQ

              = ½

So, E'= ½E

If the value of the charge of a electrolytic capacitor increases, then the value of the force of the electrolytic capacitor increases proportionally. If the square value of the charge  of a electrolytic capacitor decreases, then the value of the force of the electrolytic capacitor decreases proportionally.

So, E∞Q__________(ii)

If C=2C then-

      E'= ΨᵦQ÷(2C)²

         = ΨᵦQ÷4C²

So, E'÷E= ΨᵦQ÷4C²÷ΨᵦQ÷C²

              = ΨᵦQ÷4C²×C²÷ΨᵦQ

              = ¼

So, E'=¼E

If C=½C then- 

      E'= ΨᵦQ÷(½C)²

         = ΨᵦQ÷¼C²

         = 4ΨᵦQ÷C²

So, E'÷E= 4ΨᵦQ÷C²÷ΨᵦQ÷C²

              = 4ΨᵦQ÷C²÷ C²÷ ΨᵦQ

              = 4

So, E'= 4E

If the square value of capacitance increases, the value of the power of that object decreases. Again, the value of capacitance decreases, the value of power of that object increases. 

So, E∞ 1÷C²____________(iii)

The graph of ii no. equation- 

The graph of iii no. equation- 

From ii and iii number equation- 

We get- E∞Q__________(ii)

              E∞ 1÷C²____________(iii)

So, E∞ Q×1÷C²

Or, E∞ Q÷C²

So, E= ΨᵦQ÷C²

Here, Ψᵦ- is a proportional constant. 

Constant's Calculation:

Here, Ψᵦ- is a proportional constant. 

Ψᵦ= 1÷k

     = 1÷1÷4πε₀

     = 4πε₀

.    = 4× 3.1416× 8.85×10⁻¹² Nm⁻² C²

     = 1.112×10⁻¹⁰ Nm⁻² C²

So, Ψᵦ= 1.112×10⁻¹⁰ Nm⁻² C²

Electric Field Of Capacitor: 

A electrolytic capacitor has its own electric field and it can able to create a new electric field. 

Here, MNOP- is a simple circuit. In this circuit, capacitor is a electric cell. From this capacitor, the electrons flow in the whole circuit. In this circuit, the electrons start flow from the positive(+)  point of the capacitor. The electrons flow positive point to M to N  point, N point to O point, O point to P  point, P point to the negative point of the capacitor. So, we can say that- “A electrolytic capacitor can able to create an electric field.” 

At last, The value of electric field of an electrolytic capacitor is proportional to the value of capacitor's charge and disproportional to the square value of the capacitance of the capacitor. 

Result:

The value of capacitor’s electric field is proportional to the value of capacitor’s charge and disproportional to the square value of the capacitor’s capacitance. 

Conditions Of This Law:  

There are some conditions of this law. Here’s that-

i) This law is only for capacitor. 

ii) The position of capacitor should be in a  circuit

Conclusion:

So, if we know the value of a capacitor’s charge and capacitance, we can determine the value of it's electric field.