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

Options

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

Answer

To solve the question regarding the dimensions of the expression \( \epsilon_0 \frac{d\phi_E}{dt} \), we will follow these steps:

### Step 1: Understand the Terms
- \( \epsilon_0 \) is the permittivity of free space, which has specific dimensions.
- \( \phi_E \) represents the electric flux, which is defined as the product of the electric field \( E \) and the area \( A \) through which the field lines pass.

### Step 2: Find the Dimensions of \( \epsilon_0 \)
The permittivity of free space \( \epsilon_0 \) has the dimensions:
\[ [\epsilon_0] = \frac{C^2}{N \cdot m^2} \]
Where:
- \( C \) is the unit of charge (Coulombs).
- \( N \) is the unit of force (Newtons).
- \( m \) is the unit of length (meters).

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

### Step 3: Find the Dimensions of \( \frac{d\phi_E}{dt} \)
Electric flux \( \phi_E \) has the dimensions:
\[ [\phi_E] = [E] \cdot [A] \]
Where:
- The electric field \( E \) has dimensions \( [E] = \frac{V}{m} = \frac{J}{C \cdot m} = \frac{kg \cdot m^2}{s^3 \cdot C} \).
- Area \( A \) has dimensions \( [A] = m^2 \).

Thus, the dimensions of electric flux are:
\[ [\phi_E] = \left(\frac{kg \cdot m^2}{s^3 \cdot C}\right) \cdot m^2 = \frac{kg \cdot m^4}{s^3 \cdot C} \]

Now, we can find the dimensions of \( \frac{d\phi_E}{dt} \):
\[ \left[\frac{d\phi_E}{dt}\right] = \frac{kg \cdot m^4}{s^3 \cdot C} \cdot \frac{1}{s} = \frac{kg \cdot m^4}{s^4 \cdot C} \]

### Step 4: Combine the Dimensions
Now, we can find the dimensions of \( \epsilon_0 \frac{d\phi_E}{dt} \):
\[ \left[\epsilon_0 \frac{d\phi_E}{dt}\right] = \left[\epsilon_0\right] \cdot \left[\frac{d\phi_E}{dt}\right] \]
Substituting the dimensions we found:
\[ \left[\epsilon_0 \frac{d\phi_E}{dt}\right] = \left(\frac{C^2 \cdot s^2}{kg \cdot m^3}\right) \cdot \left(\frac{kg \cdot m^4}{s^4 \cdot C}\right) \]

### Step 5: Simplify the Expression
Now we simplify:
\[ = \frac{C^2 \cdot s^2 \cdot kg \cdot m^4}{kg \cdot m^3 \cdot s^4 \cdot C} = \frac{C \cdot s^2 \cdot m}{s^4} = \frac{C \cdot m}{s^2} \]

### Step 6: Identify the Result
The dimensions \( \frac{C \cdot m}{s^2} \) correspond to the dimensions of electric current \( I \) (Ampere), since:
\[ [A] = \frac{C}{s} \]

Thus, the dimensions of \( \epsilon_0 \frac{d\phi_E}{dt} \) are the same as that of electric current.

### Final Answer:
The dimensions of \( \epsilon_0 \frac{d\phi_E}{dt} \) are the same as that of electric current (Ampere).

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