Gauss’s Law Part 1 : Electric Flux
Understanding electric flux is mandatory for understanding Gauss’s Law. Before we dive into electric flux or magnetic flux, let us begin with an analogy.
Let, a rectangular wire is placed in a stream of water. Water is moving at a constant velocity v from left to right. The area between the rectangular wire is A.
Let, after Δt time period , the volume of water that will pass through the wire is ΔV. At Δt time period, s = v*Δt is the distance that water will cross.
So, the volume of the crossed water will be,
volume = length * area = s * A = v Δt * A.
So , ΔV = v*A Δt
Or , ΔV/Δt = v*A
So, the rate of water flow is velocity times Area. In this case, the wire is placed perpendicular to the velocity of the water. Now, if we tilt the wire by an angle of θ then the rate of fluid flow through the wire will be changed.
In the above figure, both wires have same area but the right one is tilted by an angle θ. Though both wires have same area, less water will go through the second wire than the first one. As you can see this from the following figure.
If the wire is tilted by an angle θ , then the actual area that the water will cross is A cos θ.
This is less than the actual area of the wire A. Because,
cos θ < 1 for θ > 0
So, A cos θ < A
As the effective area decreases, less water will go through the tilted wire. If the wire is tilted by 90 degree, then no water will pass through the wire.
So, the fluid flow rate ,
v is a vector quantity. The above expression can be written as dot product of v and A if we consider the area “A” as a vector quantity.
Now, what will be the direction of A? From the rule of dot product , θ is the angle between v and A. As, we already know the direction of v , so we can figure out the direction of A.
From the figure, we can see that, the direction of A is perpendicular to the plane of rectangular wire. Now, we have completed the analogy of fluid flow. In this case, the fluid flux is the rate of fluid flow .
So, fluid flux = v A cos θ
Or, fluid flux = v . A
Now , for electric flux, think the electric field vector E in place of v. Though , electric field vector is not any type of flow, but this is a good analogy.
Electric field vector E represents the strength and direction of electric field at any place. The magnitude of E at a point is how much force will be applied on 1 coulomb positive charge in that point and the direction of E is the direction of that force. Electric field lines also represents the strength of electric field. If the field lines are dense, the electric field is strong. So, the magnitude of E is proportional to the density of electric field line. Let E represents how many electric fields are in 1 square meter area (though it is not the correct definition of E, but our analogy is correct as the magnitude of E is proportional to the density of electric field line). Let we place a rectangular wire in the field.
So, the number of electric field lines that will cross the area is E*A.
Let, φ = E*A
Here , φ is the electric flux (just like previous analogy where v*A was the fluid flux. In this case imagine E in v’s place ). If the wire is tilted at angle θ , the effective area will be reduced.
Now , the actual area that the electric field crosses is A cos θ. So, the electric flux ,
φ = E*A cos θ
E is a vector quantity . If we consider A as a vector quantity like the previous analogy, this equation can be written as a dot product.
So, φ = E . A.
If the surface A is curved or an irregular shape, then we have to divide the surface into smaller pieces dA .
In this case the tiny flux dφ that passes through the tiny area dA is , dφ=E*dA*cos θ
Or, dφ = E . dA
To get the flux for the whole surface , we have to integrate the equation on both sides.
So, φ = ∫ E . dA
In the next part of this series we will learn about the Gauss’s Law.
See the next part Gauss’s Law : The First Of Four Maxwell’s Equations
