Electric Flux Dimensional Analysis JEE Main PYQ

 

❓ Question

The electric flux is given by

$$\phi=\alpha\sigma+\beta\lambda$$

where $\lambda$ and $\sigma$ are linear charge density and surface charge density, respectively.

What does

$$\frac{\alpha}{\beta}$$

represent?

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✍️ Solution

Step 1: Dimension of electric flux

Electric flux is

$$\phi=\vec E\cdot \vec A$$

Therefore,

$$[\phi]=[E][L^2]$$

Now,

$$[E]=\frac{[F]}{[q]}$$
$$[E] = \frac{[MLT^{-2}]}{[AT]} = [MLT^{-3}A^{-1}]$$

Hence,

$$[\phi] = [MLT^{-3}A^{-1}][L^2]$$
$$\boxed{[\phi]=[ML^3T^{-3}A^{-1}]}$$

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Step 2: Find dimension of $\alpha$

Since

$$\phi=\alpha\sigma+\beta\lambda$$

we must have

$$[\phi]=[\alpha\sigma]$$

Surface charge density is

$$\sigma=\frac{q}{L^2}$$

Therefore,

$$[\sigma]=[ATL^{-2}]$$

Hence,

$$[\alpha] = \frac{[\phi]}{[\sigma]}$$
$$= \frac{[ML^3T^{-3}A^{-1}]} {[ATL^{-2}]}$$
$$\boxed{[\alpha]=[ML^5T^{-4}A^{-2}]}$$

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Step 3: Find dimension of $\beta$

Similarly,

$$[\phi]=[\beta\lambda]$$

Linear charge density is

$$\lambda=\frac{q}{L}$$

Therefore,

$$[\lambda]=[ATL^{-1}]$$

Hence,

$$[\beta] = \frac{[\phi]}{[\lambda]}$$
$$= \frac{[ML^3T^{-3}A^{-1}]} {[ATL^{-1}]}$$
$$\boxed{[\beta]=[ML^4T^{-4}A^{-2}]}$$

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Step 4: Dimension of $\alpha/\beta$

$$\left[\frac{\alpha}{\beta}\right] = \frac{[ML^5T^{-4}A^{-2}]} {[ML^4T^{-4}A^{-2}]}$$
$$= [L]$$

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✅ Final Answer

$$\boxed{\left[\frac{\alpha}{\beta}\right]=[L]}$$

Therefore,

$$\boxed{\frac{\alpha}{\beta}\text{ represents displacement or length}}$$

Key Idea

Because $\sigma$ is charge per unit area while $\lambda$ is charge per unit length, their dimensional difference directly introduces one power of length.