Question: What is electric flux ? Give an expression for electric flux. Give S.I unit and dimensional formula for electric flux.
Solution:
Electric Flux: Electric flux linked with any surface is defined as the total number of electric field lines passing through that surface.
Consider a surface of area S. Let electric field lines representing electric field $\vec{E}$ pass normally through the plane of area S.
Then the electric flux through the surface of area S is given by
$\phi = ES \qquad \dots(i)$
Thus, electric flux may be defined as the product of the magnitude of the electric field (E) and the surface area (S) perpendicular to the electric field.
Let $\theta$ be the angle made by the electric field $\vec{E}$ with the area vector $d\vec{S}$ of the surface element. (E cos $\theta$) is the component of the electric field perpendicular to the surface area. Then, the electric flux through the surface element is given by :
$d\phi = (E \cos \theta) dS$
or
$d\phi = E dS \cos \theta = \vec{E} \cdot d\vec{S} \qquad \dots(ii)$
Total electric flux through the surface is given by :
$\int d\phi = \int E dS \cos \theta = \int_S \vec{E} \cdot d\vec{S}$
or
$\Phi = \int_S \vec{E} \cdot d\vec{S} = \int_S E dS \cos \theta$
Thus, total electric flux through a surface in an electric field may be defined as the surface integral of the electric field over that surface.
Special cases :
(1). If $\theta = 0^\circ$ i.e., the electric field $\vec{E}$ is perpendicular to the plane of the surface, then electric flux through the surface is given by
$\Phi = \int E \, dS \cos 0^\circ$
$\phi= \int E \, dS = E \int dS$
$\phi= ES$
$\because \cos 0^\circ = 1 \text{ and } \int dS = S)$
Thus, electric flux (or the total number of electric field lines) through the surface is maximum, if electric field is perpendicular to the plane of the surface.
(2). If $\theta = 90^\circ$ i.e., the electric field $\vec{E}$ is parallel to the plane of the surface, then electric flux through the surface is given by
$\phi = \int E \, dS \cos 90^\circ = 0 (\because \cos 90^\circ = 0)$
Thus, electric flux through the surface is zero, if electric field is parallel to the plane of the surface.
(3). If \(\theta = 180^\circ\) i.e., electric field is anti-parallel to the area vector of the surface, then
$\phi = \int E \, dS \cos 180^\circ = -E \int dS = -ES$
Thus, electric flux is negative.
SI unit of electric flux
$\phi = ES$
$N C^{-1}m^{2} or Nm^{2}C^{-1} or J m C^{-1}$
Dimensional formula of electric flux =$ [Electric field \times Area]$
$= \dfrac{[\text{Force}]}{[\text{charge}]} \times [\text{Area}]$
$= \dfrac{[\text{Force}] [\text{Area}]}{[\text{current}] [\text{time}]}$
$= [ML^3 T^{-3} A^{-1}]$