Talha's Physics Academy
Electric Flux and Electric Flux Through a Sphere
Video Lecture
Watch the complete video lecture below to understand electric flux, maximum and minimum conditions, and the step-by-step derivation of electric flux through a sphere.
Electric Flux
Definition
Alternatively: "The scalar or dot product between electric intensity and area vector."
Mathematically, electric flux ($\Phi_e$) is written as:
$$\Phi_e = \vec{E} \cdot \vec{A}$$
$$\Phi_e = E A \cos\theta \quad \text{--- (i)}$$
Where $\theta$ is the angle between the electric field intensity vector $\vec{E}$ and the normal vector to the surface area $\vec{A}$.
Maximum Flux
If $\theta = 0^\circ$ (the electric field lines are parallel to the area vector / normal to the surface), equation (i) becomes:
$$\Phi_e = E A \cos(0^\circ)$$
$$\Phi_e = E A (1) = E A$$
Thus, the electric flux is maximum when the surface is held perpendicular to the electric field lines.
Minimum Flux
If $\theta = 90^\circ$ (the electric field lines are parallel to the surface, perpendicular to the area vector), equation (i) becomes:
$$\Phi_e = E A \cos(90^\circ)$$
$$\Phi_e = E A (0) = 0$$
Thus, the electric flux is minimum (zero) when the surface is held parallel to the electric field lines.
Electric Flux Through a Sphere
Consider an isolated point charge $+q$ placed at the center of an imaginary sphere of radius $r$. The electric lines of force from $q$ spread uniformly in space around it radially, cutting the surface of the sphere normally at all portions.
The total electric flux through the sphere is given by:
$$\Phi_e = \vec{E} \cdot \vec{A}$$
$$\Phi_e = E A \cos\theta$$
Since the electric lines pass normally through the surface area of the sphere, the angle $\theta = 0^\circ$, so $\cos(0^\circ) = 1$:
$$\Phi_e = E A \quad \text{--- (ii)}$$
The electric field intensity $E$ at distance $r$ from point charge $q$ is given by Coulomb's law:
$$E = \frac{1}{4\pi \varepsilon_0} \frac{q}{r^2}$$
And the total surface area of the sphere is:
$$A = 4\pi r^2$$
Substituting the values of $E$ and $A$ into equation (ii):
$$\Phi_e = \left( \frac{1}{4\pi \varepsilon_0} \frac{q}{r^2} \right) (4\pi r^2)$$
$$\Phi_e = \frac{q}{\varepsilon_0}$$
Thus, the total electric flux through a closed spherical surface depends solely upon the charge enclosed inside the sphere ($q$) and the permittivity of free space ($\varepsilon_0$), which forms the foundation of Gauss's Law.
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