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

The given expression for electric flux is: 

\(\varphi = \alpha \sigma + \beta \lambda\)

where:

• \(\sigma\) is the surface charge density (charge per unit area), usually measured in coulombs per square meter \((\text{C/m}^2)\).
• \(\lambda\) is the linear charge density (charge per unit length), usually measured in coulombs per meter \((\text{C/m})\).
• \(\alpha\) and \(\beta\) are constants with appropriate units that relate to the flux.

The ratio \(\left( \frac{\alpha}{\beta} \right)\) is what the question asks us to interpret.

Let's analyze the units involved in the expression:

• The electric flux \(\varphi\) is measured in volts-meter \((\text{V}\cdot\text{m})\) or alternatively in newton-meter squared per coulomb \((\text{N}\cdot\text{m}^2/\text{C})\).
• Surface charge density \(\sigma\) has units \lambda has units \alpha must have units of \((\text{V}\cdot\text{m}^3/\text{C})\) and \(\beta\) must have units of \((\text{V}\cdot\text{m}^2/\text{C})\).

 

The ratio \(\left( \frac{\alpha}{\beta} \right)\) then has units of:

\(\frac{\text{V}\cdot\text{m}^3/ \text{C}}{\text{V}\cdot\text{m}^2/ \text{C}} = \text{m}\)

Therefore, \(\left( \frac{\alpha}{\beta} \right)\) represents a length or displacement.

This analysis leads to the conclusion that the correct answer is:

• displacement

Answer 2

To understand what \( \left( \frac{\alpha}{\beta} \right) \) represents in the equation for electric flux \( \varphi = \alpha \sigma + \beta \lambda \), we need to analyze the physical dimensions and properties involved.

Electric flux \( \varphi \) relates to the surface covered by an electric field and is typically calculated as the dot product of the electric field \( E \) and area \( A \), expressed as \( \varphi = E \cdot A \), implying that electric flux has dimensions of an electric field times an area.

In the given equation \( \varphi = \alpha \sigma + \beta \lambda \):

• \( \sigma \) is the surface charge density, with dimensions of charge per unit area \(([Q][L]^{-2})\).
• \( \lambda \) is the linear charge density, with dimensions of charge per unit length \(([Q][L]^{-1})\).

The coefficients \( \alpha \) and \( \beta \) must have dimensions that, when multiplied by their respective charge densities, result in the dimensions of electric flux \([E][A]\).

Let's analyze:

For the term \( \alpha \sigma \):

• \([E][A] = [\alpha][Q][L]^{-2}\)
• \([\alpha] = [E][L]^{2}/[Q]\)

For the term \( \beta \lambda \):

• \([E][A] = [\beta][Q][L]^{-1}\)
• \([\beta] = [E][L]/[Q]\)

Now, compute the ratio \( \frac{\alpha}{\beta} \):

• \(\frac{\alpha}{\beta} = \frac{[E][L]^{2}/[Q]}{[E][L]/[Q]} = [L]\)

Thus, the ratio \( \frac{\alpha}{\beta} \) represents a physical quantity with dimensions of length, typically associated with displacement in physics. Therefore, the correct interpretation of \( \left( \frac{\alpha}{\beta} \right) \) in this context is displacement.