The quantity `X = (epsilon_(0)LV)/(t)` where `epsilon_(0)` is the permittivity of free space, L is length, V is the potential difference and t is time. The dimensions of X are the same as that of

Options

A. Resistance
B. Charge
C. Voltage
D. Current

Answer

To find the dimensions of the quantity \( X = \frac{\epsilon_0 L V}{t} \), we will break down each component in the expression and find their respective dimensions.

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

The permittivity of free space \( \epsilon_0 \) can be derived from Coulomb's law, which states:

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

From this, we can express \( \epsilon_0 \):

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

The dimensions of the quantities involved are:
- Charge \( q \) has the dimension \( [Q] \).
- Force \( F \) has the dimension \( [M][L][T^{-2}] \).
- Distance \( r \) has the dimension \( [L] \).

Thus, the dimensions of \( \epsilon_0 \) can be calculated as follows:

\[ [\epsilon_0] = \frac{[Q]^2}{[M][L][T^{-2}][L^2]} = \frac{[Q]^2}{[M][L^3][T^{-2}]} \]

### Step 2: Determine the dimensions of \( L \)

The dimension of length \( L \) is simply:

\[ [L] = [L] \]

### Step 3: Determine the dimensions of \( V \)

The potential difference \( V \) can be defined in terms of electric field \( E \):

\[ V = E \cdot L \]

Where \( E \) (electric field) has the dimension \( [M][L][T^{-3}][Q^{-1}] \). Therefore, the dimension of \( V \) is:

\[ [V] = [E][L] = [M][L][T^{-3}][Q^{-1}][L] = [M][L^2][T^{-3}][Q^{-1}] \]

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

The dimension of time \( t \) is:

\[ [t] = [T] \]

### Step 5: Combine the dimensions to find \( X \)

Now we can substitute the dimensions back into the expression for \( X \):

\[ X = \frac{\epsilon_0 L V}{t} \]

Substituting the dimensions we found:

\[ [X] = \frac{\left(\frac{[Q]^2}{[M][L^3][T^{-2}]}\right) \cdot [L] \cdot \left([M][L^2][T^{-3}][Q^{-1}]\right)}{[T]} \]

Simplifying this expression:

\[ [X] = \frac{[Q]^2 \cdot [L] \cdot [M][L^2][T^{-3}][Q^{-1}]}{[M][L^3][T^{-2}][T]} \]

This simplifies to:

\[ [X] = \frac{[Q]^{2-1} \cdot [L]^{1+2-3} \cdot [M]^{1-1} \cdot [T]^{-3-1+2}}{1} \]

Thus:

\[ [X] = [Q][L^0][M^0][T^{-2}] = [Q][T^{-2}] \]

### Conclusion

The dimensions of \( X \) are the same as that of electric current \( I \) (since \( I = \frac{Q}{T} \)), which is given by:

\[ [X] = [I] \]