alleryStart learning →Two objects rubbed together acquire a charge of $50\ nC$ each. The number of electrons transferred is about
A$3.1\times10^{8}$
B$8.0\times10^{9}$
C$3.1\times10^{11}$
D$5.0\times10^{-8}$
Answer & Solution
Correct answer: C. $3.1\times10^{11}$
1. Charge quantisation: $n = q/e$.
2. $q = 50\times10^{-9}$ C, $e = 1.6\times10^{-19}$ C.
3. $n = \dfrac{50\times10^{-9}}{1.6\times10^{-19}} \approx 3.1\times10^{11}$ electrons.
_Source: Samacheer Kalvi Class 12 Physics Vol.1, Unit 1 'Electrostatics', Exercise Q.1_
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