The number of electrons that must be removed from an electrically neutral silver dollar to give it a charge of $+ 2.4 C$ is

2.5 \times 10^{19}

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1.5 \times 10^{19}

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1.5 \times 10^{- 19}

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2.5 \times 10^{- 19}

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Solution

The correct option is D $1.5 \times 10^{19}$
Total charge, $q = + 2.4 C$
Then by quantization of charge $q = n e$
$∴$ number of electrons, $n = \frac{q}{e} = \frac{2.4 C}{1.6 \times 10^{- 19}} = 1.5 \times 10^{19}$

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The number of electrons that must be removed from an electrically neutral silver dollar to give it a charge of +2.4C is

The correct option is D 1.5×1019Total charge, q=+2.4CThen by quantization of charge q=ne∴ number of electrons, n=qe=2.4C1.6×10−19=1.5×1019

The number of electrons that must be removed from an electrically neutral silver dollar to give it a charge of $+ 2.4 C$ is

The correct option is D $1.5 \times 10^{19}$
Total charge, $q = + 2.4 C$
Then by quantization of charge $q = n e$
$∴$ number of electrons, $n = \frac{q}{e} = \frac{2.4 C}{1.6 \times 10^{- 19}} = 1.5 \times 10^{19}$