`epsilon_(0)E^(2)` has the dimensions of
(`epsilon_(0)=` permittivity of free space, `E=` electric field)
Here `k=` Boltzmann consant
`T=` absolute temperature
`R=` universal gas constant.

Options

A. Pressure
B. `kT`
C. `R//T`
D. `RT`

Answer

To find the dimensions of \( \epsilon_0 E^2 \), where \( \epsilon_0 \) is the permittivity of free space and \( E \) is the electric field, we will follow these steps:

### Step 1: Determine the dimensions of \( \epsilon_0 \)

The permittivity of free space \( \epsilon_0 \) is defined in terms of the electric force between two charges. The formula for the electric force \( F \) between two point charges \( q_1 \) and \( q_2 \) separated by a distance \( r \) is given by:

\[ F = \frac{1}{4 \pi \epsilon_0} \frac{q_1 q_2}{r^2} \]

From this equation, we can rearrange it to express \( \epsilon_0 \):

\[ \epsilon_0 = \frac{q_1 q_2}{4 \pi F r^2} \]

### Step 2: Identify the units of \( F \), \( q \), and \( r \)

- The unit of force \( F \) is Newton (N).
- The unit of charge \( q \) is Coulomb (C).
- The unit of distance \( r \) is meter (m).

### Step 3: Substitute the units into the expression for \( \epsilon_0 \)

Substituting the units into the expression for \( \epsilon_0 \):

\[ \epsilon_0 = \frac{C^2}{N \cdot m^2} \]

### Step 4: Determine the dimensions of \( E \)

The electric field \( E \) is defined as the force per unit charge:

\[ E = \frac{F}{q} \]

Substituting the units:

\[ E = \frac{N}{C} \]

### Step 5: Calculate \( E^2 \)

Now, we square the electric field:

\[ E^2 = \left(\frac{N}{C}\right)^2 = \frac{N^2}{C^2} \]

### Step 6: Calculate \( \epsilon_0 E^2 \)

Now we can find the dimensions of \( \epsilon_0 E^2 \):

\[ \epsilon_0 E^2 = \left(\frac{C^2}{N \cdot m^2}\right) \left(\frac{N^2}{C^2}\right) \]

### Step 7: Simplify the expression

When we multiply these two expressions, we get:

\[ \epsilon_0 E^2 = \frac{C^2 \cdot N^2}{N \cdot m^2 \cdot C^2} = \frac{N^2}{N \cdot m^2} = \frac{N}{m^2} \]

### Step 8: Identify the final dimensions

The final expression \( \frac{N}{m^2} \) is the same as pressure, which has the dimensions of force per unit area.

### Conclusion

Thus, the dimensions of \( \epsilon_0 E^2 \) are the same as that of pressure.