The dimensions of electrical conductivity is …………

Explanation

To find the dimensions of electrical conductivity (σ), we can start from its definition and derive the dimensions step by step.

### Step-by-Step Solution:

1. **Understanding Electrical Conductivity**:
Electrical conductivity (σ) is defined as the ratio of current density (J) to the electric field (E). The formula is given by:
\[ \sigma = \frac{J}{E} \]

2. **Expressing Current Density (J)**:
Current density (J) is defined as the current (I) flowing per unit area (A). Therefore:
\[ J = \frac{I}{A} \]

3. **Expressing Electric Field (E)**:
The electric field (E) can be expressed in terms of force (F) and charge (Q):
\[ E = \frac{F}{Q} \]

4. **Substituting the Expressions**:
Now, substituting the expressions for J and E into the formula for σ:
\[ \sigma = \frac{I/A}{F/Q} = \frac{I \cdot Q}{A \cdot F} \]

5. **Relating Charge (Q) to Current (I)**:
Charge (Q) can be expressed as current (I) multiplied by time (T):
\[ Q = I \cdot T \]
Substituting this into the equation gives:
\[ \sigma = \frac{I \cdot (I \cdot T)}{A \cdot F} = \frac{I^2 \cdot T}{A \cdot F} \]

6. **Finding Dimensions of Each Quantity**:
- Current (I) has dimensions of [I].
- Area (A) has dimensions of [L^2].
- Force (F) can be expressed as mass (M) times acceleration (a), where acceleration has dimensions of [L T^{-2}]. Thus, the dimensions of force are:
\[ [F] = [M][L][T^{-2}] = [M L T^{-2}] \]

7. **Substituting Dimensions**:
Now substituting the dimensions into the equation:
\[ \sigma = \frac{[I^2][T]}{[L^2][M L T^{-2}]} \]
Simplifying this gives:
\[ \sigma = \frac{[I^2][T]}{[M][L^3][T^{-2}]} = \frac{[I^2][T^3]}{[M][L^3]} \]

8. **Final Dimensions of Electrical Conductivity**:
Therefore, the dimensions of electrical conductivity (σ) are:
\[ [\sigma] = [M^{-1} L^{-3} T^3 I^2] \]

### Summary:
The dimensions of electrical conductivity are:
\[ [M^{-1} L^{-3} T^3 I^2] \]