Question

Explain why the electric flux through a closed surface with a given enclosed charge is independent of the size or shape of the surface.

Answer 1

Hi there!

Recall that:

`\Phi_E = \oint E \cdot dA = (Q_(enclosed))/(\epsilon_(0))`

The electric flux is PROPORTIONAL to the total ENCLOSED charge and inversely proportional to the permittivity of free space.

Using the equivalent equation:

`\Phi_E = (Q_(enclosed))/(\epsilon_0)`

ε₀ is a constant, so the electric flux is only dependent on the total enclosed charge. Increasing the enclosed charge increases the electric flux and vice-versa.