Dimensions of `epsi_(0) (dphi_(E))/(dt)` are same as that of

Options

A. charge
B. potential
C. capacitance
D. current

Explanation

To find the dimensions of the expression \( \epsilon_0 \frac{d\Phi_E}{dt} \), we will break it down step by step.

### Step 1: Understand the terms involved
- \( \epsilon_0 \) is the permittivity of free space.
- \( \Phi_E \) is the electric flux.
- \( \frac{d\Phi_E}{dt} \) represents the rate of change of electric flux with respect to time.

### Step 2: Find the dimensions of \( \epsilon_0 \)
The permittivity of free space \( \epsilon_0 \) has dimensions given by:
\[ [\epsilon_0] = \frac{[C^2]}{[N \cdot m^2]} \]
Where:
- \( [C] \) is the dimension of charge.
- \( [N] \) (Newton) can be expressed as \( [kg \cdot m/s^2] \).

Thus, we can rewrite the dimensions of \( \epsilon_0 \):
\[ [\epsilon_0] = \frac{[C^2]}{[kg \cdot m/s^2] \cdot [m^2]} = \frac{[C^2]}{[kg \cdot m^3/s^2]} = [C^2 \cdot s^2/(kg \cdot m^3)] \]

### Step 3: Find the dimensions of electric flux \( \Phi_E \)
Electric flux \( \Phi_E \) is defined as:
\[ \Phi_E = E \cdot A \]
Where:
- \( E \) is the electric field with dimensions \( [E] = \frac{[V]}{[m]} = \frac{[J/C]}{[m]} = \frac{[kg \cdot m^2/(s^2 \cdot C)]}{[m]} = \frac{[kg \cdot m]}{[s^2 \cdot C]} \).
- \( A \) is the area with dimensions \( [A] = [m^2] \).

Thus, the dimensions of electric flux \( \Phi_E \) are:
\[ [\Phi_E] = [E] \cdot [A] = \left(\frac{[kg \cdot m]}{[s^2 \cdot C]}\right) \cdot [m^2] = \frac{[kg \cdot m^3]}{[s^2 \cdot C]} \]

### Step 4: Find the dimensions of \( \frac{d\Phi_E}{dt} \)
The rate of change of electric flux with respect to time has dimensions:
\[ \left[\frac{d\Phi_E}{dt}\right] = \frac{[\Phi_E]}{[T]} = \frac{[kg \cdot m^3/(s^2 \cdot C)]}{[s]} = \frac{[kg \cdot m^3]}{[s^3 \cdot C]} \]

### Step 5: Combine the dimensions
Now we can find the dimensions of the entire expression \( \epsilon_0 \frac{d\Phi_E}{dt} \):
\[ \left[\epsilon_0 \frac{d\Phi_E}{dt}\right] = [\epsilon_0] \cdot \left[\frac{d\Phi_E}{dt}\right] \]
Substituting the dimensions we found:
\[ = \left[\frac{C^2 \cdot s^2}{kg \cdot m^3}\right] \cdot \left[\frac{kg \cdot m^3}{s^3 \cdot C}\right] \]
When we multiply these, we get:
\[ = \frac{C^2 \cdot s^2 \cdot kg \cdot m^3}{kg \cdot m^3 \cdot s^3 \cdot C} = \frac{C \cdot s^2}{s^3} = \frac{C}{s} = [I] \]
Where \( [I] \) is the dimension of electric current (Ampere).

### Final Answer
The dimensions of \( \epsilon_0 \frac{d\Phi_E}{dt} \) are the same as that of electric current, which is \( [I] \).