Question

Two charged spherical conductors of radii $ {{R}_{1}} $ and $ {{R}_{2}} $ are connected by a wire. Then the ratio of surface charge densities of the spheres $ {{\sigma }_{1}}/{{\sigma }_{2}} $ is

• A. $ {{R}_{1}}/{{R}_{2}} $
• B. $ {{R}_{2}}/{{R}_{1}} $
• C. $ \sqrt{({{R}_{1}}/{{R}_{2}})} $
• D. $ R_{1}^{2}/R_{2}^{2} $

Answer 1

The Correct Option is B

$ \frac{{{q}_{1}}}{{{q}_{2}}}=\frac{{{C}_{1}}}{{{C}_{2}}}=\frac{{{R}_{1}}}{{{R}_{2}}} $ and $ \frac{{{q}_{1}}}{{{q}_{2}}}=\frac{4\pi R_{1}^{2}{{\sigma }_{1}}}{4\pi R_{2}^{2}{{\sigma }_{2}}} $ $ \frac{{{R}_{1}}}{{{R}_{2}}}=\frac{4\pi R_{1}^{2}{{\sigma }_{1}}}{4\pi R_{2}^{2}{{\sigma }_{2}}} $ $ \therefore $ $ \frac{{{\sigma }_{1}}}{{{\sigma }_{2}}}=\frac{{{R}_{2}}}{{{R}_{1}}} $

Answer 2

Surface potential of a spherical conductor of radius R, carrying charge Q, \(V_{S}=\frac{Q}{4\pi \varepsilon _{0}R}\)

Let Q1 and Q2 are the charges of two spherical conductors having radii R₁ and R₂ respectively.

 When these two charged spherical conductors are connected with the wire, the potential at their surfaces become equal

 Therefore, Vs₁ = Vs₂

\(V_{S}=\frac{Q_{1}}{4\pi \varepsilon _{0}R_{1}}\) = \(V_{S}=\frac{Q_{1}}{4\pi \varepsilon _{0}R_{1}}\)⇒ \(\frac{Q_{1}}{Q_{2}}\)=\(\frac{R_{1}}{R_{2}}\)……(i)

Surface charge density, σ = \(\frac{Charge}{Area}\)

\(\frac{\sigma _{1}}{\sigma _{2}}\) =\(\frac{(\frac{Q_{1}}{4\pi \varepsilon _{0}R_{1}^{2}})}{(\frac{Q_{1}}{4\pi \varepsilon _{0}R_{2}^{2}})}\)= \(\frac{Q_{1}}{Q_{2}}\) × \(\frac{R_{2}^{2}}{R_{1}^{2}}\)

= \(\frac{R_{1}}{R_{2}}\) ×\(\frac{R_{2}^{2}}{R_{1}^{2}}\) = \(\frac{R_{2}}{R_{1}}\)[Using(i)]