Two thin discs carry charges of \( +4\ \mu\mathrm{C} \) and \( -4\ \mu\mathrm{C} \). They are attached to the two ends of a metal cylinder. The free electrons in the cylinder neutralise these charges in \( 2\ \mu\mathrm{s} \). Take \( e=1.6\times10^{-19}\ \mathrm{C} \). On average, the number of electrons that pass through a cross-section of the cylinder per second is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(1.25\times10^{19}\ \mathrm{s^{-1}}\)
The field of the two charges pushes the free electrons towards \( +Q \). While they move, they form a current. The electrons that leave the \( -Q \) end are the same electrons that reach the \( +Q \) end. So only \( 4\ \mu\mathrm{C} \) of charge crosses any section of the cylinder. Each electron carries a charge \( e \), so the number per second is \( I/e \).
Given:
\( Q=4\ \mu\mathrm{C}=4\times10^{-6}\ \mathrm{C} \)
\( t=2\ \mu\mathrm{s}=2\times10^{-6}\ \mathrm{s} \)
\( \begin{aligned} I &= \frac{Q}{t} \\ &= \frac{4\times10^{-6}}{2\times10^{-6}} \\ &= 2\ \mathrm{A} \end{aligned} \)
\( \begin{aligned} \frac{N}{t} &= \frac{I}{e} \\ &= \frac{2}{1.6\times10^{-19}} \\ &= 1.25\times10^{19}\ \mathrm{s^{-1}} \end{aligned} \)
NCERT (2026-27):
Electric Currents in Conductors – “… The electrons will be accelerated due to this field towards +Q. They will thus move to neutralise the charges. The electrons, as long as they are moving, will constitute an electric current. Hence in the situation considered, there will be a current for a very short while and no current thereafter.”
Summary – “1. Current through a given area of a conductor is the net charge passing per unit time through the area.”