Electric Field Between Two Charged Plates
In this article we will use Gauss’s law to measure the electric field between two charged plates and the electric field of a capacitor.
Gauss’s Law states that :
The net electric flux through any hypothetical closed surface is equal to (1/ε0) times the net electric charge within that closed surface
Let we have a charged plane of infinite length and width. And let positive charges are equally distributed throughout the surface.
Though the plane in the picture doesn’t have infinite length and width , let us assume this as an infinite plane.
Let the charge density on the surface is λ coulomb/meter² . So, in 1m² area on the plane, there are λ coulomb charges.
The plane is symmetric. From the symmetricity of the system , we can say that the direction of electric field is perpendicular to the plane . If electric field is not perpendicular , then rotating the plane will break the symmetricity.
If the electric fields are perpendicular , the symmetricity will be preserved.
Also from the symmetricity , we can say that the magnitude of the electric field will be the same on equidistant distances from the plane.
Let us assume a hypothetical cylinder with height h and base area A. The outer surface of the cylinder is our Gaussian surface.
Now, we have to calculate flux through the Gaussian surface. The cylinder has 3 surfaces . Upper and lower bases and one curved surface.
Let , φ1 = flux through upper base
φ2 = flux through lower base
φ3 = flux through curved surface
So, φ1 = ∮E*dA*cos 0⁰ ………..[Direction between E and dA is 0⁰]
or, φ1 = ∮E*dA
or, φ1 = E ∮dA ……………[Because E is constant]
∮dA is the surface area of bases = A . So, the equation becomes :
φ1 = E * A
For lower base , the equations are the same . So, φ2 = E * A
For the curved surface the angle between E and dA is 90⁰. So, cos 90⁰ = 0.
So, φ3 = ∮E*dA*cos 90⁰
or, φ3 = ∮E*dA* 0
or φ3 = 0
So, the net flux,
φ = φ1 + φ2 + φ3
or, φ = E*A + E*A + 0
or, φ = 2*E*A ……………(1)
According to Gauss’s Law , φ = Σ q /ε₀
Here, Σ q = total charge on the plane inside the cylinder. So, Σ q = λ*A
So, φ = (λ*A) /ε₀ ……………..(2)
From equation (1) and (2) ,
(λ*A) /ε₀ = 2*E*A
or, λ /ε₀ = 2*E
or, E = λ /2ε₀
So, for a infinite plane with charge density λ , the electric field ,
Notice that, r is not present in the equation . So, E does not change over distance from the plate.
Two Parallel Plane with Same Charge
Let , we have two parallel infinite plate each positively charged with charge density λ.
Now we want to calculate the electric field of these two parallel plate combined. In the previous section we learnt about their individual electric field is E = λ /2ε₀
In the middle of the two plate , both electric fields are opposite to each other . So, the cancel each other and the net electric field inside is zero.
On left and right side, both electric fields are in the same direction. So, their vector sum = E = λ /ε₀
This is the case in parallel plate capacitor. Now what will happen if the two plates have opposite charge ? The question is left for the reader.
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