An object A is charged by rubbing with an object B. In the process \( 10^{9} \) electrons are transferred from A to B. Charge developed on A will be :
(Take electronic charge \( e = 1.602\times10^{-19}\,\mathrm{C} \))
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( 1.6\times10^{-10}\,\mathrm{C} \)
Charge is quantised: an object that gains or loses \( n \) electrons has charge \( q = ne \). A loses electrons, so it becomes positively charged.
\( \displaystyle q = ne = 10^{9}\times1.602\times10^{-19}\,\mathrm{C} = 1.602\times10^{-10}\,\mathrm{C} \approx 1.6\times10^{-10}\,\mathrm{C} \)
The charge on A is \( +1.6\times10^{-10}\,\mathrm{C} \) (and B gets \( -1.6\times10^{-10}\,\mathrm{C} \)).