A conductor has 14.4 times 10^{-19} C of pos | vedclass

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(D) The quantization of charge is given by the formula $Q = ne$,where $Q$ is the total charge,$n$ is the number of electrons,and $e$ is the elementary

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$9$ electrons in short

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(D) The quantization of charge is given by the formula $Q = ne$,where $Q$ is the total charge,$n$ is the number of electrons,and $e$ is the elementary charge.Given $Q = 14.4 	imes 10^{-19} \ C$ and $e = 1.6 	imes 10^{-19} \ C$.We need to find $n = \frac{Q}{e}$.$n = \frac{14.4 	imes 10^{-19}}{1.6 	imes 10^{-19}} = \frac{14.4}{1.6} = 9$.Since the conductor has a positive charge,it indicates a deficiency of electrons.Therefore,the conductor has $9$ electrons in short.

$A$ conductor has $14.4 \times 10^{-19} \ C$ of positive charge. The conductor has (Charge on an electron $= 1.6 \times 10^{-19} \ C$):

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