Explain measurement of flux with examples

(a) If the Surface is Plane and the Electric Field is Uniform

(i) If the surface is perpendicular to the electric field the electric flux (φE) is given by the product of the electric field (E) and the surface are (A).

φE = EA

(ii) If the surface is parallel to the electric field [Fig (b)], no field lines cross that area and as such the flux in this case is zero, i.e.,

φE = 0 …. (2)

(iii) If the surface is neither perpendicular nor parallel to the electric field but makes an angle Ɵ with the field (E) as shown in the area that matters is the one which is perpendicular to the field lines and is equal to A cos Ɵ. Thus,

φE = E(A cos Ɵ) = EA cosƟ …… (3)

The results obtained in equations. (1), (2) and (3), may be combined into a single equation if we make use of the dot product of two vectors. Before we do so, the area A must be given a vector representation.

The area vector is denoted by a vector $overrightarrow{A}$ whose direction is perpendicular to the plane of the area and the magnitude of this vector is equal to A itself. The electric flux is then given by

${{phi }_{E}}=overrightarrow{E}cdot overrightarrow{A}=EAcos theta$ …… (4)

Here, Ɵ is the angle which the area vector $(overrightarrow{A})$ makes with electric field $overrightarrow{E}$.

Obviously, when $theta =0{}^circ ,{{phi }_{E}}=EAcos 0{}^circ =EA$ and when$theta =90{}^circ ,{{phi }_{E}}=EAcos 90{}^circ =0$ .

(iv) If the Surface is Curved and /or the Electric Field is Non-uniform

When the electric field varies either in direction or in magnitude over a surface (which may be plane or curved), we consider it to be divided into a large number of very small elements. Let us consider one such element of area $overrightarrow{dS,},where,overrightarrow{dS}$ is a vector whose direction is perpendicular to the element. The direction of $overrightarrow{dS,}$ by convention, is taken to be the direction of outward normal. The variation of electric field $(overrightarrow{E})$ over such an element can be neglected, i.e., $overrightarrow{E}$ can be taken to be uniform over the area $overrightarrow{dS}$.

Electric flux linked with the element of area $overrightarrow{dS}$ which makes an angle Ɵ with $overrightarrow{E}$ is given by

$d{{phi }_{E}}=overrightarrow{E}cdot overrightarrow{dS}$ …… (5)

The electric flux over the whole surface is given by

${{phi }_{E}}=int{_{S}}overrightarrow{E}cdot overrightarrow{dS}$ ….. (6)

Here, $int{_{S}}$ is known as the surface integral.

We are usually interested in calculating the flux through the closed surface (i.e., a surface with no holes in it). In such a case, the total flux through a closed surface is

${{phi }_{c}}=oint{_{S}}overrightarrow{E}cdot overrightarrow{dS}$ …. (7)

The symbol $oint{_{S}}$ represents the integral over a closed surface.

Thus,

${{phi }_{c}}=oint{_{S}}EdScos theta =oint{_{S}}(Ecos theta )ds=oint{_{S}}({{E}_{n}}dS)$ ….. (8)

Where En = Ecos Ɵ represents the component of the electric field normal to the surface. Here $oint{_{S}}{{E}_{n}}dS$ is called the surface integral of the normal component of the electric field over any closed surface. i.e. total normal flux.

Unit of φE

Since the unit of E is N/C and that of S is m2, the unit of φ is N m2/C.

Dimensional Formula for φE

As φE = ES cos Ɵ, dimensional formula for

${{phi }_{E}},is,[ML{{T}^{-3}}{{A}^{-1}}][{{L}^{2}}]=[M{{L}^{3}}{{T}^{-3}}{{A}^{-1}}]$