Correct Answer - Option 1 : 8 : 1

__CONCEPT:__

Linear charge density:

- It is defined as the quantity of charge per unit length.
- Its SI unit is C/m.
- If ΔQ charge is contained in the line element Δl, the linear charge density λ will be,

\(⇒ \lambda=\frac{Δ Q}{Δ l}\)

Surface charge density:

- It is defined as the quantity of charge per unit area.
- Its SI unit is C/m2.
- If ΔQ charge is contained in the elemental area Δs, the surface charge density σ will be,

\(⇒ \sigma=\frac{Δ Q}{Δ s}\)

Volume charge density:

- It is defined as the quantity of charge per unit volume.
- Its SI unit is C/m3.
- If ΔQ charge is contained in the elemental volume Δv, the volume charge density ρ will be,

\(⇒ \rho=\frac{Δ Q}{Δ v}\)

__EXPLANATION:__

Given \(\frac{R_A}{R_B}=\frac{1}{2}\) and \(\frac{Δ Q_A}{Δ Q_B}=\frac{2}{1}\)

If ΔQ charge is contained in the elemental area Δs, the surface charge density σ will be,

\(⇒ \sigma=\frac{Δ Q}{Δ s}\) ----(1)

- The area of the sphere of radius R is given as,

⇒ Δs = 4πR^{2} ----(2)

- The surface charge density for sphere A is given as,

\(⇒ \sigma_A=\frac{Δ Q_A}{Δ s_A}\)

\(⇒ \sigma_A=\frac{Δ Q_A}{4\pi R_A^2}\) ----(3)

- The surface charge density for sphere B is given as,

\(⇒ \sigma_B=\frac{Δ Q_B}{Δ s_B}\)

\(⇒ \sigma_B=\frac{Δ Q_B}{4\pi R_B^2}\) ----(4)

By equation 3 and equation 4,

\(⇒ \frac{\sigma_A}{\sigma_B}=\frac{Δ Q_A}{4\pi R_A^2}\times\frac{4\pi R_B^2}{Δ Q_B}\)

\(⇒ \frac{\sigma_A}{\sigma_B}=\frac{Δ Q_A}{Δ Q_B}\times\frac{R_B^2}{R_A^2}\)

\(⇒ \frac{\sigma_A}{\sigma_B}=\frac{2}{1}\times\left ( 2 \right )^2\)

\(⇒ \frac{\sigma_A}{\sigma_B}=\frac{8}{1}\)

- Hence, option 1 is correct.

- The charge density is a measure of how much electric charge is accumulated in a particular field. Specifically, it finds the charge density per unit volume, surface area, and length. Charge density depends on the distribution of charge and it can be positive or negative.