Logo
Jeeify
MCQEasyJEE 2025Dimensions & Dimensional Analysis

JEE Physics 2025 Question with Solution

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

  • A

    charge

  • B

    electric field

  • C

    displacement

  • D

    area

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given: Electric flux is given by φ=ασ+βλ\varphi = \alpha \sigma + \beta \lambda where σ\sigma is surface charge density and λ\lambda is linear charge density.

Find: What physical quantity is represented by (αβ)\left( \frac{\alpha}{\beta} \right).

Use dimensional analysis.

Electric flux has units of electric field into area, so its units are V⋅m\text{V} \cdot \text{m} or equivalently N⋅m2/C\text{N} \cdot \text{m}^2/\text{C}.

For the term involving surface charge density:

[α][σ]=[φ][\alpha][\sigma] = [\varphi]

Since surface charge density σ\sigma has units C/m2\text{C}/\text{m}^2,

[α]=[φ][σ]=V⋅mC/m2=V⋅m3C[\alpha] = \frac{[\varphi]}{[\sigma]} = \frac{\text{V} \cdot \text{m}}{\text{C}/\text{m}^2} = \frac{\text{V} \cdot \text{m}^3}{\text{C}}

For the term involving linear charge density:

[β][λ]=[φ][\beta][\lambda] = [\varphi]

Since linear charge density λ\lambda has units C/m\text{C}/\text{m},

[β]=[φ][λ]=V⋅mC/m=V⋅m2C[\beta] = \frac{[\varphi]}{[\lambda]} = \frac{\text{V} \cdot \text{m}}{\text{C}/\text{m}} = \frac{\text{V} \cdot \text{m}^2}{\text{C}}

Therefore,

[αβ]=V⋅m3/CV⋅m2/C=m\left[\frac{\alpha}{\beta}\right] = \frac{\text{V} \cdot \text{m}^3/\text{C}}{\text{V} \cdot \text{m}^2/\text{C}} = \text{m}

So (αβ)\left( \frac{\alpha}{\beta} \right) has the dimension of length, which here corresponds to displacement.

Therefore, the correct option is C.

Dimension Comparison

Given: φ=ασ+βλ\varphi = \alpha \sigma + \beta \lambda.

Find: The interpretation of (αβ)\left( \frac{\alpha}{\beta} \right).

Both terms on the right-hand side must have the same dimensions as electric flux.

Surface charge density has dimensions

[σ]=[Q][L]−2[\sigma] = [Q][L]^{-2}

and linear charge density has dimensions

[λ]=[Q][L]−1[\lambda] = [Q][L]^{-1}

If electric flux is represented by [φ][\varphi], then

[α]=[φ][Q][L]−2[\alpha] = \frac{[\varphi]}{[Q][L]^{-2}}

and

[β]=[φ][Q][L]−1[\beta] = \frac{[\varphi]}{[Q][L]^{-1}}

Taking the ratio,

[α][β]=[φ]/([Q][L]−2)[φ]/([Q][L]−1)=[L]\frac{[\alpha]}{[\beta]} = \frac{[\varphi]/([Q][L]^{-2})}{[\varphi]/([Q][L]^{-1})} = [L]

Hence, αβ\frac{\alpha}{\beta} has the dimensions of length.

Thus it represents displacement. The correct option is C.

Common mistakes

  • Confusing dimension of length with area. The ratio αβ\frac{\alpha}{\beta} leaves only one power of LL, not L2L^2. Compare units carefully after cancellation.

  • Using only the dimensions of σ\sigma and λ\lambda without matching them to electric flux. The coefficients α\alpha and β\beta must be found by making each term have the same units as φ\varphi.

  • Treating electric flux as charge directly. Electric flux is related to electric field through area, so its dimensions are not the same as pure charge. Start from φ=E⃗⋅A⃗\varphi = \vec{E} \cdot \vec{A} or equivalent units.

Practice more Dimensions & Dimensional Analysis questions

Get unlimited AI-adaptive practice, mastery tracking, and an AI tutor that explains every step — free to start.

Start free trialMore on Dimensions & Dimensional Analysis

Related questions