The dimensions of `1/2 epsilon_(0)E^(2) (epsilon_(0)=` permittivity of free space, `E=` electric field) is

Options

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

Answer

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

### Step 1: Identify the dimensions of \( \epsilon_0 \) (Permittivity of free space)
The dimensions of \( \epsilon_0 \) are given as:
\[ [\epsilon_0] = M^{-1} L^{-3} T^{4} I^{2} \]

### Step 2: Identify the dimensions of \( E \) (Electric field)
The dimensions of the electric field \( E \) are given as:
\[ [E] = M^{1} L^{1} I^{-1} T^{-3} \]

### Step 3: Calculate the dimensions of \( E^2 \)
To find the dimensions of \( E^2 \), we square the dimensions of \( E \):
\[ [E^2] = (M^{1} L^{1} I^{-1} T^{-3})^2 = M^{2} L^{2} I^{-2} T^{-6} \]

### Step 4: Combine the dimensions of \( \epsilon_0 \) and \( E^2 \)
Now, we combine the dimensions of \( \epsilon_0 \) and \( E^2 \) in the expression \( \frac{1}{2} \epsilon_0 E^2 \):
\[ [\frac{1}{2} \epsilon_0 E^2] = [\epsilon_0] \cdot [E^2] = (M^{-1} L^{-3} T^{4} I^{2}) \cdot (M^{2} L^{2} I^{-2} T^{-6}) \]

### Step 5: Perform the multiplication of dimensions
Now, we multiply the dimensions:
\[ M^{-1} \cdot M^{2} = M^{1} \]
\[ L^{-3} \cdot L^{2} = L^{-1} \]
\[ T^{4} \cdot T^{-6} = T^{-2} \]
\[ I^{2} \cdot I^{-2} = I^{0} \quad (\text{dimensionless}) \]

### Step 6: Combine the results
Combining all the results, we get:
\[ [\frac{1}{2} \epsilon_0 E^2] = M^{1} L^{-1} T^{-2} \]

### Final Answer
Thus, the dimensions of \( \frac{1}{2} \epsilon_0 E^2 \) are:
\[ M L^{-1} T^{-2} \]

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