Question

The electric flux is \( \varphi = \alpha \sigma + \beta \lambda \) where \( \lambda \) and \( \sigma \) are linear and surface charge density, respectively, and \( \left( \frac{\alpha}{\beta} \right) \) represents

• A. charge
• B. electric field
• C. displacement
• D. area

Hint:

Electric flux is a measure of the total electric field passing through a surface. The key idea is to understand how the surface charge density and linear charge density relate to the dimensions of the variables involved.

Answer 1

The Correct Option is C

We are given that the electric flux is: \[ \varphi = \alpha \sigma + \beta \lambda \] Here, \( \alpha \) and \( \beta \) are constants, \( \sigma \) is the surface charge density, and \( \lambda \) is the linear charge density. Now, let's take the dimensions of both sides of the equation. For electric flux \( \varphi \), the dimension is: \[ [\varphi] = [\alpha \sigma] = [\beta \lambda] \] \[ [\alpha] = \left[ \frac{\varphi}{\sigma} \right] = \left[ \frac{[Q/L]}{[Q/Area]} \right] = \left[ \frac{Area}{Length} \right] \] So, the dimensions of \( \alpha \) are \( \left[ \frac{L^2}{L} \right] = L \). For \( \beta \), we get: \[ [\beta] = \left[ \frac{\varphi}{\lambda} \right] = \left[ \frac{[Q/L]}{[Q/Area]} \right] = [L] \] Thus, the quantity \( \frac{\alpha}{\beta} \) represents a length, which corresponds to a displacement. Therefore, the correct answer is (3) displacement.