Question

If $ \epsilon_0 $ denotes the permittivity of free space and $ \Phi_E $ is the flux of the electric field through the area bounded by the closed surface, then the dimension of $ \epsilon_0 \frac{d\Phi_E}{dt} $ are that of:

• A. Electric field
• B. Electric potential
• C. Electric charge
• D. Electric current

Hint:

When solving problems involving physical quantities, ensure that you use the correct dimensional analysis to find the desired dimension. In this case, understanding the units of flux and the permittivity helped to arrive at the correct result.

Answer 1

The Correct Option is D

We are given that \( \epsilon_0 \) is the permittivity of free space and \( \Phi_E \) is the electric flux. The electric flux \( \Phi_E \) is given by: \[ \Phi_E = \int_S \mathbf{E} \cdot d\mathbf{A} \] where \( \mathbf{E} \) is the electric field and \( d\mathbf{A} \) is the area element. The dimension of \( \epsilon_0 \) (the permittivity of free space) is: \[ [\epsilon_0] = \frac{\text{C}^2}{\text{N m}^2} = \frac{\text{C}^2}{\text{kg} \cdot \text{m}^3 \cdot \text{s}^2} \] The electric flux \( \Phi_E \) has the dimension of: \[ [\Phi_E] = \text{C} \cdot \text{m}^2 \] Now, we are interested in the dimension of \( \epsilon_0 \frac{d\Phi_E}{dt} \). The dimension of \( \frac{d\Phi_E}{dt} \) is the rate of change of electric flux, which has the dimension: \[ \left[\frac{d\Phi_E}{dt}\right] = \frac{\text{C} \cdot \text{m}^2}{\text{s}} \] Now, multiplying this by \( \epsilon_0 \), we get: \[ \left[\epsilon_0 \frac{d\Phi_E}{dt}\right] = \left[\epsilon_0\right] \times \left[\frac{d\Phi_E}{dt}\right] = \frac{\text{C}^2}{\text{kg} \cdot \text{m}^3 \cdot \text{s}^2} \times \frac{\text{C} \cdot \text{m}^2}{\text{s}} \] Simplifying this: \[ \left[\epsilon_0 \frac{d\Phi_E}{dt}\right] = \frac{\text{C}^3 \cdot \text{m}^2}{\text{kg} \cdot \text{m}^3 \cdot \text{s}^3} = \frac{\text{C}}{\text{s}} \] This is the dimension of electric current (since electric current has the dimension \( \frac{\text{C}}{\text{s}} \)).
Thus, the dimension of \( \epsilon_0 \frac{d\Phi_E}{dt} \) is electric current.

Answer 2

Step 1: 
Electric flux is given by: \[ Φ_E = \int \vec{E} \cdot \vec{A} \] Hence, dimension of Φ_E = (Electric field) × (Area).

Step 2:
The dimension of electric field E is: \[ [E] = \frac{MLT^{-3}}{A} \] and the dimension of area is L².
Therefore, \[ [Φ_E] = [E] \times [A] = \frac{ML^3T^{-3}}{A} \]

Step 3:
Now differentiate w.r.t. time: \[ \left[\frac{dΦ_E}{dt}\right] = \frac{ML^3T^{-4}}{A} \]

Step 4:
Dimension of permittivity of free space (ε₀) is: \[ [ε₀] = \frac{A^2T^4}{ML^3} \]

Step 5:
Multiply both: \[ [ε₀ \frac{dΦ_E}{dt}] = \frac{A^2T^4}{ML^3} \times \frac{ML^3T^{-4}}{A} = A \] which is the dimension of electric current.

Final Answer:
\[ \boxed{\text{Electric current}} \]