The dimension of `((1)/(2))epsilon_(0)E^(2)` (`epsilon_(0)` : permittivity of free space, E electric field

Options

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

Explanation

To find the dimension of the term \(\frac{1}{2} \epsilon_0 E^2\), where \(\epsilon_0\) is the permittivity of free space and \(E\) is the electric field, we can follow these steps:

### Step 1: Identify the dimensions of the electric field \(E\).
The electric field \(E\) is defined as force per unit charge. The dimension of force is given by:
\[ [F] = M L T^{-2} \]
The dimension of charge \(Q\) is simply \(Q\). Therefore, the dimension of the electric field \(E\) can be expressed as:
\[ [E] = \frac{[F]}{[Q]} = \frac{M L T^{-2}}{Q} = M L T^{-2} Q^{-1} \]

### Step 2: Calculate the dimension of \(E^2\).
Now, we need to find the dimension of \(E^2\):
\[ [E^2] = (M L T^{-2} Q^{-1})^2 = M^2 L^2 T^{-4} Q^{-2} \]

### Step 3: Identify the dimensions of \(\epsilon_0\).
The permittivity of free space \(\epsilon_0\) can be related to the electric field and force. It is defined in terms of the electric field and charge as follows:
\[ \epsilon_0 = \frac{1}{4\pi k} \quad \text{(where \(k\) is Coulomb's constant)} \]
The dimension of \(\epsilon_0\) can also be derived from the equation of capacitance:
\[ [C] = \frac{Q^2}{[F]} \quad \Rightarrow \quad [\epsilon_0] = \frac{[Q]^2}{[F][L]} = \frac{Q^2}{M L T^{-2} L} = \frac{Q^2}{M L^2 T^{-2}} = M^{-1} L^{-3} T^4 Q^2 \]

### Step 4: Combine the dimensions of \(\epsilon_0\) and \(E^2\).
Now we can find the dimension of \(\epsilon_0 E^2\):
\[ [\epsilon_0 E^2] = [\epsilon_0] \cdot [E^2] = (M^{-1} L^{-3} T^4 Q^2) \cdot (M^2 L^2 T^{-4} Q^{-2}) \]

### Step 5: Simplify the expression.
Now, we simplify the expression:
\[ [\epsilon_0 E^2] = M^{-1} L^{-3} T^4 Q^2 \cdot M^2 L^2 T^{-4} Q^{-2} = M^{2-1} L^{2-3} T^{4-4} Q^{2-2} = M^1 L^{-1} T^0 Q^0 = M L^{-1} \]

### Step 6: Final dimension.
Thus, the dimension of \(\frac{1}{2} \epsilon_0 E^2\) is:
\[ [M L^{-1}] \]

### Conclusion:
The final answer is:
\[ \text{Dimension of } \frac{1}{2} \epsilon_0 E^2 = M L^{-1} \]