1. State the law in electrostatic that relates the electric flux passing through the surface with the charge enclosed.

Question

Two closed surface S₁ and S₂ enclose two charges q₁ and q₂ as shown in the figure.

1. State the law in electrostatic that relates the electric flux passing through the surface with the charge enclosed.

2. If q₁ = +6µC and q₂ = -4µC find the ratio of the flux passing through surfaces S₁ and S₂ .

3. Let the surface S₂ expands to double its area while S₁ remains as such. What will happen to the above ratio?

Answer 1

1. The net flux of electric field passing through a closed surface is equal to \(\frac{1}{\varepsilon_0}\) times the charge enclosed by the surface.

2. By Gauss’ theorem flux passing through a

surface ϕ = \(\frac{q}{\varepsilon_0}\)

Flux passing through S₁ ϕ₁ = \(\frac{6\times10^{-6}}{\varepsilon_0}\)

Flux passing through S₁ ϕ₂ = \(\frac{(6-4)\times10^{-6}}{\varepsilon_0}\)

\(\frac{\phi_1}{\phi_2}=\frac{6}{2}\)

\(\frac{\phi_1}{\phi_2}=3\)

3. Remains the same (because electric flux is independent of the size and shape of the enclosed surface).