When 10^{19} electrons are removed from a neu | vedclass

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Question

(B) The correct option is $B$.According to the quantization of charge,the total charge $Q$ on a body is given by the formula $Q = n 	imes e$,where $n

Vedclass · Accepted Answer

+$1.6$ $C$

Vedclass · Answer

(B) The correct option is $B$.According to the quantization of charge,the total charge $Q$ on a body is given by the formula $Q = n 	imes e$,where $n$ is the number of electrons transferred and $e$ is the elementary charge $(e \approx 1.6 	imes 10^{-19} \ C)$.Given:$n = 10^{19}$$e = 1.6 	imes 10^{-19} \ C$Calculation:$Q = 10^{19} 	imes 1.6 	imes 10^{-19} \ C$$Q = 1.6 	imes 10^{(19-19)} \ C$$Q = 1.6 	imes 10^0 \ C$$Q = 1.6 \ C$Since electrons (which are negatively charged) are removed from the neutral plate,the plate develops a deficiency of electrons,resulting in a net positive charge.Therefore,the charge on the plate becomes $+1.6 \ C$.

When $10^{19}$ electrons are removed from a neutral metal plate through some process,the charge on it becomes . . . . . . .

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