Question

Electric flux \( \varphi \) is related with linear charge density \( \lambda \) and surface charge density \( \sigma \) as \( \varphi = \alpha \lambda + \beta \sigma \). Where \( \alpha \) and \( \beta \) are of appropriate dimension. The dimension of \( \frac{\beta}{\alpha} \) is:

• A. \( M L^2 T^{-2} \)
• B. \( M L^3 T^{-2} \)
• C. \( M L^2 T^{-3} \)
• D. \( M L^3 T^{-3} \)

Hint:

To solve for dimensions in equations, match the dimensions on both sides, considering that electric flux has the dimension of charge.

Answer 1

The Correct Option is B

Electric flux \( \varphi \) has the dimension of charge, and from the equation \( \varphi = \alpha \lambda + \beta \sigma \), the dimensions of \( \alpha \) and \( \beta \) can be found by equating the dimensions of each term. The dimensions of \( \lambda \) (linear charge density) are \( M L^{-1} T^{-1} \) and of \( \sigma \) (surface charge density) are \( M L^{-2} T^{-1} \). After solving for \( \frac{\beta}{\alpha} \), we find the dimensions are \( M L^3 T^{-2} \). Thus, the correct answer is option (2).