An electrically neutral insulating body loses $1$ electron in the first step and gains $6$ electrons in the second step of charge transfer. The final charge on its surface can be accurately calculated as

- 1.6 \times 10^{- 19} C

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- 9.6 \times 10^{- 19} C

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8 \times 10^{- 19} C

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- 8 \times 10^{- 19} C

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Solution

The correct option is D $- 8 \times 10^{- 19} C$
One electron carries a charge of $- e = - 1.6 \times 10^{- 19} C$

Now, total electron gained= $6 - 1 = 5$

Hence, charge on surface= $5 \times$ charge on each electron
$= 5 \times (- e)$

$= - 5 \times 1.6 \times 10^{- 19} C$

$= - 8 \times 10^{- 19} C$

Answer-(D)

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An electrically neutral insulating body loses 1 electron in the first step and gains 6 electrons in the second step of charge transfer. The final charge on its surface can be accurately calculated as

The correct option is D −8×10−19COne electron carries a charge of −e=−1.6×10−19CNow, total electron gained=6−1=5Hence, charge on surface=5× charge on each electron=5×(−e)=−5×1.6×10−19C=−8×10−19CAnswer-(D)

An electrically neutral insulating body loses $1$ electron in the first step and gains $6$ electrons in the second step of charge transfer. The final charge on its surface can be accurately calculated as