Total electric flux associated with unit positive charge in vacuum is

Options

A. `4pi epsilon_(0)`
B. `(1)/(4pi epsilon_(0))`
C. `(1)/(epsilon_(0))`
D. `epsilon_(0)`

Answer

To find the total electric flux associated with a unit positive charge in vacuum, we can use Gauss's Law. Here’s the step-by-step solution:

### Step 1: Understand Gauss's Law
Gauss's Law states that the total electric flux (Φ) through a closed surface is equal to the charge (Q) enclosed by that surface divided by the permittivity of free space (ε₀). Mathematically, it is expressed as:

\[ \Phi = \frac{Q}{\varepsilon_0} \]

### Step 2: Identify the Charge
In this case, we are considering a unit positive charge, which is equal to 1 coulomb (Q = +1 C).

### Step 3: Substitute the Charge into Gauss's Law
Now, substituting the value of the charge into Gauss's Law, we have:

\[ \Phi = \frac{1 \, \text{C}}{\varepsilon_0} \]

### Step 4: Recognize the Value of ε₀
The permittivity of free space (ε₀) is a constant with a value of approximately \(8.854 \times 10^{-12} \, \text{C}^2/\text{N m}^2\). However, for the purpose of this question, we only need to express the flux in terms of ε₀.

### Step 5: Final Expression for Electric Flux
Thus, the total electric flux associated with a unit positive charge in vacuum is:

\[ \Phi = \frac{1}{\varepsilon_0} \]

### Conclusion
Therefore, the total electric flux associated with a unit positive charge in vacuum is \( \frac{1}{\varepsilon_0} \).