99 Percentile Qs Bank for JEE MainPhysicsElectrostatics
A neutral sphere of radius r and density ρ is placed in a uniform electric field E that exists on the earth's surface in the vertically upward direction. If the atomic number and the mass number of the material of the sphere are Z and A respectively, then the fraction of electrons that should be removed from the sphere for it to remain in equilibrium is [Assume that the sphere is able to hold the necessary charge wit
Options
- An n total = ρ g A e E N A Z
- Bn n total = 4 g A π e E N A Z
- Cn n total = g A e E N A Z
- Dn n total = π ρ g A 3 e E N A Z
Correct answer
C. n n total = g A e E N A Z
Step-by-step solution
Let us assume that we remove n electrons from the sphere due to which the net charge on the sphere is q = n e For the sphere to be in equilibrium m g = q E = n e E 4 3 π r 3 ρ g = n e E ⇒ n = 4 π r 3 ρ g 3 e E The number of moles of the metal in the sphere is n mol = m A = 4 π r 3 ρ 3 A So the number of metal atoms will be n metal atoms = 4 π r 3 ρ 3 A × N A This means the total number of electrons in the sphere is n total = 4 π r 3 ρ N A 3 A × Z Hence, the fraction is n n total = 4 π r 3 ρ g 3 e E × 3 A 4 π r 3 ρ