Question

Flux coming out from a unit positive charge kept in air is

• A. $\varepsilon_0$
• B. $(\varepsilon_0)^{-1}$
• C. $(4\pi\varepsilon_0)^{-1}$
• D. $4\pi\varepsilon_0$

Hint:

Gauss’s Law: \[ \Phi_E = \fracQ_\textenc\varepsilon_0 \] Applies to any closed surface. For a unit charge, $\Phi_E = 1/\varepsilon_0$.

Answer 1

The Correct Option is B

According to Gauss's law, the total electric flux $\Phi_E$ through a closed surface is equal to the net charge enclosed divided by the permittivity of free space: \[ \Phi_E = \frac{Q}{\varepsilon_0} \] Here, $Q = +1~\text{C}$ (unit positive charge), and since the medium is air (which is very close to vacuum in behavior), we use $\varepsilon_0$. Thus: \[ \Phi_E = \frac{1}{\varepsilon_0} = (\varepsilon_0)^{-1} \] So, the electric flux through a closed surface enclosing a unit positive charge in air is $(\varepsilon_0)^{-1}$.

Answer 2

Step 1: Understanding Electric Flux
Electric flux is a measure of the number of electric field lines passing through a given surface. It quantifies how much the electric field "flows" through that surface. When dealing with point charges, the electric flux through a closed surface surrounding the charge depends only on the charge enclosed and is independent of the shape or size of the surface.

Step 2: Gauss's Law and Its Significance
Gauss's law is a fundamental principle in electrostatics which states that the total electric flux (Φ) through a closed surface is equal to the net charge (Q) enclosed divided by the permittivity of free space (ε₀). Mathematically, it is expressed as:
Φ = Q / ε₀
This law helps relate electric fields to their sources (charges) and simplifies the calculation of electric fields for symmetrical charge distributions.

Step 3: Applying Gauss's Law to a Unit Positive Charge in Air
Consider a point charge of +1 coulomb placed in air (which can be approximated as free space). The permittivity of free space, ε₀, is a constant that characterizes how electric fields propagate in vacuum or air.
Using Gauss's law, the electric flux through any closed surface surrounding this charge is:
Φ = Q / ε₀ = 1 / ε₀
This means the total flux depends only on the magnitude of the enclosed charge and the medium's permittivity.

Step 4: Physical Interpretation and Conclusion
The flux being equal to (ε₀)⁻¹ for a unit charge signifies that the number of electric field lines emanating from the charge is fixed and is inversely proportional to the permittivity of the medium. A lower ε₀ would mean stronger electric fields and more flux lines per unit area.
Thus, the electric flux coming out from a unit positive charge kept in air is exactly (ε₀)⁻¹, which is a fundamental constant in electrostatics.