The unit of electric permittivity is

Options

A. `"volt/m"^(2)`
B. `"joule/C"`
C. `"farad/m"`
D. `"henry/m"`

Answer

To determine the unit of electric permittivity, we can analyze the relationship between electric force, charge, and distance, and derive the unit step by step.

### Step-by-Step Solution:

1. **Understanding Electric Permittivity**:
Electric permittivity (\( \epsilon \)) is a measure of how much electric field (\( E \)) is reduced in a medium compared to a vacuum. The electric force between two point charges is given by Coulomb's law:
\[ F = \frac{1}{4 \pi \epsilon} \frac{q_1 q_2}{r^2} \]
where \( F \) is the force between the charges, \( q_1 \) and \( q_2 \) are the magnitudes of the charges, and \( r \) is the distance between them.

2. **Rearranging the Formula**:
Rearranging the formula to express \( \epsilon \):
\[ \epsilon = \frac{1}{4 \pi} \frac{q_1 q_2}{F r^2} \]

3. **Identifying Units**:
- The unit of charge (\( q \)) is the coulomb (C).
- The unit of force (\( F \)) is the newton (N).
- The unit of distance (\( r \)) is the meter (m).

4. **Substituting Units**:
Substituting the units into the equation for \( \epsilon \):
\[ \text{Unit of } \epsilon = \frac{C^2}{N \cdot m^2} \]

5. **Expressing Newton in Terms of Joules**:
We know that \( 1 \, \text{N} = 1 \, \text{kg} \cdot \text{m/s}^2 \) and \( 1 \, \text{J} = 1 \, \text{N} \cdot \text{m} \). Therefore:
\[ N = \frac{J}{m} \]
Thus, we can rewrite the unit of permittivity:
\[ \text{Unit of } \epsilon = \frac{C^2}{\frac{J}{m} \cdot m^2} = \frac{C^2 \cdot m}{J} \]

6. **Identifying the Correct Option**:
The derived unit of electric permittivity can also be expressed in terms of farads. The farad (F) is the unit of capacitance, and it is defined as:
\[ 1 \, \text{F} = 1 \, \frac{C^2}{J} \]
Therefore, we can express permittivity as:
\[ \epsilon = \frac{F \cdot m}{1} = \text{F/m} \]

7. **Conclusion**:
The unit of electric permittivity is:
\[ \text{F/m} \quad \text{(farads per meter)} \]