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, which states that the electric flux (Φ) through a closed surface is proportional to the charge (Q) enclosed by that surface. The formula for electric flux is given by:

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

where:
- \( \Phi \) is the electric flux,
- \( Q \) is the total charge enclosed,
- \( \varepsilon_0 \) is the permittivity of free space (vacuum).

### Step-by-Step Solution:

1. **Identify the Charge**: We have a unit positive charge, which means \( Q = 1 \, \text{C} \) (coulomb).

2. **Use Gauss's Law**: According to Gauss's Law, the electric flux through a closed surface surrounding this charge is given by:
\[ \Phi = \frac{Q}{\varepsilon_0} \]

3. **Substitute the Values**: Substitute \( Q = 1 \, \text{C} \) into the equation:
\[ \Phi = \frac{1}{\varepsilon_0} \]

4. **Conclusion**: The total electric flux associated with a unit positive charge in vacuum is:
\[ \Phi = \frac{1}{\varepsilon_0} \]

### Final Answer:
The total electric flux associated with a unit positive charge in vacuum is \( \frac{1}{\varepsilon_0} \).

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