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 can break it down into its components and analyze each part step by step.

### Step 1: Understanding the components
The expression consists of two parts:
1. \( \epsilon_0 \) (the permittivity of free space)
2. \( \frac{d\Phi_E}{dt} \) (the rate of change of electric flux)

### Step 2: Finding the dimensions of \( \epsilon_0 \)
The permittivity of free space \( \epsilon_0 \) has dimensions that can be derived from the relationship between capacitance, charge, and potential difference. The dimension of capacitance \( C \) is given by:
\[ C = \frac{Q}{V} \]
Where:
- \( Q \) is charge with dimensions \( [Q] = A \cdot T \)
- \( V \) is potential difference with dimensions \( [V] = \frac{ML^2}{AT^3} \)

Rearranging gives:
\[ C = \frac{A \cdot T}{\frac{ML^2}{AT^3}} = \frac{A^2 T^4}{ML^2} \]
From the definition of capacitance, we also know:
\[ C = \epsilon_0 \cdot \frac{L}{L} = \epsilon_0 \]
Thus, the dimensions of \( \epsilon_0 \) can be expressed as:
\[ [\epsilon_0] = \frac{M^{-1} L^{-2} A^2 T^4} \]

### Step 3: Finding the dimensions of \( \frac{d\Phi_E}{dt} \)
The electric flux \( \Phi_E \) is given by:
\[ \Phi_E = E \cdot A \]
Where:
- \( E \) (electric field) has dimensions \( [E] = \frac{ML}{AT^3} \)
- \( A \) (area) has dimensions \( [A] = L^2 \)

Thus, the dimensions of electric flux are:
\[ [\Phi_E] = [E] \cdot [A] = \frac{ML}{AT^3} \cdot L^2 = \frac{ML^3}{AT^3} \]

Now, finding the dimensions of \( \frac{d\Phi_E}{dt} \):
\[ \frac{d\Phi_E}{dt} \text{ has dimensions } \frac{ML^3}{AT^3} \cdot \frac{1}{T} = \frac{ML^3}{AT^4} \]

### Step 4: Combining the dimensions
Now, we can combine the dimensions of \( \epsilon_0 \) and \( \frac{d\Phi_E}{dt} \):
\[ [\epsilon_0 \frac{d\Phi_E}{dt}] = [\epsilon_0] \cdot \left[\frac{d\Phi_E}{dt}\right] = \left(M^{-1} L^{-2} A^2 T^4\right) \cdot \left(\frac{ML^3}{AT^4}\right) \]

Calculating this gives:
\[ = M^{-1} L^{-2} A^2 T^4 \cdot \frac{ML^3}{AT^4} = L^{3-2} A^{2-1} = L^1 A^1 = LA \]

### Conclusion
The dimensions of \( \epsilon_0 \frac{d\Phi_E}{dt} \) are the same as that of current \( I \), which has dimensions \( [I] = A \).

### Final Answer
Thus, the dimensions of \( \epsilon_0 \frac{d\Phi_E}{dt} \) are the same as that of current, which is \( A \).

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