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 = 4πR2 ----(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.
