No, electric flux does not depend on the radius of the Gaussian surface.
Electric flux is a measure of the electric field passing through a given surface. According to Gauss's Law, the electric flux through a closed surface is proportional to the enclosed electric charge, regardless of the shape or size of the surface. This means that as long as the same amount of charge is enclosed within a Gaussian surface, the electric flux through that surface will remain constant, irrespective of its radius.
Explanation:
Gauss's Law states:
∮ E ⋅ dA = Qenclosed / ε0
Where:
- E is the electric field.
- dA is the differential area vector of the surface.
- Qenclosed is the total charge enclosed by the surface.
- ε0 is the permittivity of free space.
As the equation shows, the electric flux (∮ E ⋅ dA) is directly proportional to the enclosed charge (Qenclosed) and inversely proportional to the permittivity of free space (ε0), which is a constant. The radius of the Gaussian surface does not appear in this equation. Therefore, changing the radius while keeping the enclosed charge constant will not affect the electric flux.
Example:
Imagine a point charge +q at the center of two concentric spherical Gaussian surfaces, one with radius r1 and another with radius r2 (where r2 > r1). Both surfaces enclose the same charge +q. According to Gauss's Law, the electric flux through both surfaces will be the same:
Φ1 = q / ε0
Φ2 = q / ε0
Therefore, Φ1 = Φ2, even though the radii are different.
In summary: The electric flux depends solely on the enclosed charge and not on the dimensions (like radius) of the Gaussian surface used to calculate it. The choice of the Gaussian surface is a mathematical tool used for easier calculation, but it does not influence the physical quantity of electric flux itself.
