Which of the following represents correct dimensional formula of ` 1/2 epsi_0 E^2` where symols have usual meanings ?

Options

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

Explanation

To find the correct dimensional formula of \( \frac{1}{2} \epsilon_0 E^2 \), we will follow these steps:

### Step 1: Understand the Terms
- \( \epsilon_0 \) is the permittivity of free space.
- \( E \) is the electric field strength.

### Step 2: Identify the Dimensional Formula of Each Term
1. **Electric Field \( E \)**:
The electric field \( E \) is defined as force per unit charge. The dimensional formula for force is \( [F] = [M L T^{-2}] \) and for charge \( [Q] \).
Therefore, the dimensional formula for electric field \( E \) is:
\[ [E] = \frac{[F]}{[Q]} = \frac{[M L T^{-2}]}{[Q]} = [M L T^{-2} Q^{-1}] \]

2. **Permittivity \( \epsilon_0 \)**:
The permittivity \( \epsilon_0 \) has the dimensional formula:
\[ [\epsilon_0] = \frac{[Q^2]}{[M L^{-3} T^4]} \]
This can be derived from the relationship in electrostatics, specifically from Coulomb's law.

### Step 3: Calculate the Dimensional Formula of \( \epsilon_0 E^2 \)
Now we need to find the dimensional formula of \( \epsilon_0 E^2 \):
\[ [\epsilon_0 E^2] = [\epsilon_0] \cdot [E]^2 \]
Substituting the dimensional formulas:
\[ [\epsilon_0 E^2] = \left(\frac{[Q^2]}{[M L^{-3} T^4]}\right) \cdot \left([M L T^{-2} Q^{-1}]\right)^2 \]
Calculating \( [E]^2 \):
\[ [E]^2 = [M^2 L^2 T^{-4} Q^{-2}] \]
Now substituting back:
\[ [\epsilon_0 E^2] = \frac{[Q^2]}{[M L^{-3} T^4]} \cdot [M^2 L^2 T^{-4} Q^{-2}] \]
Combining the terms:
\[ [\epsilon_0 E^2] = \frac{[Q^2] \cdot [M^2 L^2 T^{-4} Q^{-2}]}{[M L^{-3} T^4]} = \frac{[M^2 L^2 T^{-4}]}{[M L^{-3} T^4]} \]
This simplifies to:
\[ [\epsilon_0 E^2] = [M^{2-1} L^{2+3} T^{-4-4}] = [M^1 L^5 T^{-8}] \]

### Step 4: Include the Factor \( \frac{1}{2} \)
The factor \( \frac{1}{2} \) is dimensionless and does not affect the dimensional formula. Thus, the dimensional formula for \( \frac{1}{2} \epsilon_0 E^2 \) remains:
\[ [M^1 L^5 T^{-8}] \]

### Final Answer
The correct dimensional formula of \( \frac{1}{2} \epsilon_0 E^2 \) is:
\[ [M^1 L^5 T^{-8}] \]