Question

The dimensions of a physical quantity \( \epsilon_0 \frac{d\Phi_E}{dt} \) are similar to [Symbols have their usual meanings]

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

Hint:

The dimensions of the given quantity are similar to the dimensions of electric current.

Answer 1

The Correct Option is B

We are given the quantity \( \epsilon_0 \frac{d\Phi_E}{dt} \), and we need to find its dimensions. - \(\epsilon_0\) is the permittivity of free space. Its dimensions are: \[ [\epsilon_0] = \frac{\text{C}^2}{\text{Nm}^2} = \frac{\text{A}^2 \cdot \text{s}^4}{\text{kg} \cdot \text{m}^3} \] - \( \Phi_E \) is the electric flux. The electric flux is given by \( \Phi_E = E \cdot A \), where \( E \) is the electric field and \( A \) is the area. The dimensions of \( E \) are: \[ [E] = \frac{\text{N}}{\text{C}} = \frac{\text{kg} \cdot \text{m/s}^2}{\text{A}} \] And the dimensions of area \( A \) are: \[ [A] = \text{m}^2 \] Thus, the dimensions of electric flux \( \Phi_E \) are: \[ [\Phi_E] = [E] \cdot [A] = \frac{\text{kg} \cdot \text{m/s}^2}{\text{A}} \cdot \text{m}^2 = \frac{\text{kg} \cdot \text{m}^3}{\text{A} \cdot \text{s}^2} \] Now, we calculate the dimensions of \( \epsilon_0 \frac{d\Phi_E}{dt} \): - The dimensions of \( \frac{d\Phi_E}{dt} \) are: \[ \left[\frac{d\Phi_E}{dt}\right] = \frac{\text{kg} \cdot \text{m}^3}{\text{A} \cdot \text{s}^3} \] Thus, the dimensions of \( \epsilon_0 \frac{d\Phi_E}{dt} \) are: \[ [\epsilon_0 \frac{d\Phi_E}{dt}] = [\epsilon_0] \cdot \left[\frac{d\Phi_E}{dt}\right] = \frac{\text{A}^2 \cdot \text{s}^4}{\text{kg} \cdot \text{m}^3} \cdot \frac{\text{kg} \cdot \text{m}^3}{\text{A} \cdot \text{s}^3} = \frac{\text{A}^1 \cdot \text{s}^1}{\text{s}^3} = \text{A} \cdot \text{s}^{-2} \] This is the dimension of electric current, as electric current has the dimension of \( \text{A} \). Therefore, the correct answer is (B) Electric current.