Question

The dimensions of a physical quantity $\epsilon_0 \frac{d\Phi_E}{dt}$ are similar to:

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

Hint:

To find the dimensions of a given physical quantity, express each term in terms of its fundamental dimensions and simplify.

Answer 1

The Correct Option is A

We are asked to find the dimensions of the quantity $\epsilon_0 \frac{d\Phi_E}{dt}$. Here, $\epsilon_0$ is the permittivity of free space, and $\Phi_E$ is the electric flux. - The dimensions of $\epsilon_0$ are given by: $$[\epsilon_0] = \frac{\text{C}^2}{\text{Nm}^2} = \frac{\text{A}^2 \cdot \text{s}^4}{\text{kg} \cdot \text{m}^3}$$ - The dimensions of electric flux $\Phi_E$ are: $$[\Phi_E] = \text{Electric field} \times \text{Area} = \left[\frac{\text{N}}{\text{C}}\right] \times \text{m}^2 = \frac{\text{kg} \cdot \text{m}^3}{\text{A} \cdot \text{s}^2}$$ Now, we calculate the dimensions of $\frac{d\Phi_E}{dt}$: - 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: $$[\epsilon_0 \frac{d\Phi_E}{dt}] = [\epsilon_0] \cdot \left[\frac{d\Phi_E}{dt}\right] = \frac{\text{A}^2 \cdot \text{s}^4}{\text{kg} \cdot \text{m}^3} \cdot \frac{\text{kg} \cdot \text{m}^3}{\text{A} \cdot \text{s}^3} = \frac{\text{A}^1 \cdot \text{s}^{-2}}{\text{s}^3} = \text{A} \cdot \text{s}^{-2}$$ This corresponds to the dimension of electric current. Therefore, the correct answer is (1) Electric current.