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

Options

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

Explanation

To find the dimension of the expression \(\frac{1}{2} \epsilon_0 E^2\), 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 \(\epsilon_0\)
The permittivity of free space \(\epsilon_0\) has the dimensions:
\[ [\epsilon_0] = M^{-1} L^{-3} T^{4} A^{2} \]

### Step 2: Identify the dimensions of the electric field \(E\)
The electric field \(E\) has the dimensions:
\[ [E] = M L T^{-2} A^{-1} \]

### Step 3: Calculate the dimensions of \(E^2\)
To find the dimensions of \(E^2\), we square the dimensions of \(E\):
\[ [E^2] = (M L T^{-2} A^{-1})^2 = M^2 L^2 T^{-4} A^{-2} \]

### Step 4: Combine the dimensions of \(\epsilon_0\) and \(E^2\)
Now, we can combine the dimensions of \(\epsilon_0\) and \(E^2\) in the expression \(\frac{1}{2} \epsilon_0 E^2\):
\[ [\frac{1}{2} \epsilon_0 E^2] = [\epsilon_0] \cdot [E^2] = (M^{-1} L^{-3} T^{4} A^{2}) \cdot (M^2 L^2 T^{-4} A^{-2}) \]

### Step 5: Perform the multiplication
Now, we multiply the dimensions:
\[ [\frac{1}{2} \epsilon_0 E^2] = M^{-1 + 2} L^{-3 + 2} T^{4 - 4} A^{2 - 2} \]
This simplifies to:
\[ [\frac{1}{2} \epsilon_0 E^2] = M^{1} L^{-1} T^{0} A^{0} \]

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

### Summary of the Solution
The dimension of \(\frac{1}{2} \epsilon_0 E^2\) is \(M^{1} L^{-1}\).

---