The dimensions of the electric susceptibility are :

Options

A. `[M^(0)L^(0)T^(0)A^(0)]`
B. `[M^(0)L^(-2)T^(0)A^(2)]`
C. `[MLT^(-2)A]`
D. `[M^(0)L^(-1)T^(0)A^(2)]`

Explanation

To determine the dimensions of electric susceptibility, we start by understanding its relationship with polarization and the electric field.

### Step-by-Step Solution:

1. **Understanding the Relationship**:
Electric susceptibility (χ) is defined in the context of polarization (P) and the electric field (E). The relationship is given by:
\[ P = \chi \cdot E \]
where:
- \( P \) is the polarization vector (dipole moment per unit volume),
- \( E \) is the electric field,
- \( \chi \) is the electric susceptibility.

2. **Rearranging the Equation**:
From the equation \( P = \chi \cdot E \), we can express electric susceptibility as:
\[ \chi = \frac{P}{E} \]

3. **Identifying Dimensions**:
- The dimension of polarization \( P \) can be expressed as:
\[ [P] = \frac{\text{Charge}}{\text{Volume}} = \frac{[Q]}{[L^3]} = [A][T]/[L^3] \]
- The dimension of electric field \( E \) is given by:
\[ [E] = \frac{\text{Force}}{\text{Charge}} = \frac{[M][L][T^{-2}]}{[Q]} = \frac{[M][L][T^{-2}]}{[A][T]} = \frac{[M][L]}{[A][T^3]} \]

4. **Calculating Dimensions of Electric Susceptibility**:
Now substituting the dimensions of \( P \) and \( E \) into the equation for \( \chi \):
\[ [\chi] = \frac{[P]}{[E]} = \frac{\frac{[A][T]}{[L^3]}}{\frac{[M][L]}{[A][T^3]}} \]
Simplifying this gives:
\[ [\chi] = \frac{[A][T]}{[L^3]} \cdot \frac{[A][T^3]}{[M][L]} = \frac{[A]^2[T^4]}{[M][L^4]} \]

5. **Conclusion**:
Since electric susceptibility is a ratio of two quantities with the same dimensions, it is dimensionless. Therefore, we can conclude that:
\[ [\chi] = M^0 L^0 T^0 A^0 \]

### Final Answer:
The dimensions of electric susceptibility are dimensionless, represented as \( M^0 L^0 T^0 A^0 \).