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1. Electric Charges and Fields
easy
A body has $-\, 80$ $micro\, coulomb$ of charge. Number of additional electrons in it will be
A
$8 \times {10^{ - 5}}$
B
$80 \times {10^{ - 17}}$
C
$5 \times {10^{14}}$
D
$1.28 \times {10^{ - 17}}$
Solution
(c) By $Q = Ne$ or $N = \frac{Q}{e}\therefore N = \frac{{80 \times {{10}^{ – 6}}}}{{1.6 \times {{10}^{ – 19}}}} = 5 \times {10^{14}}$
Standard 12
Physics
