Number of electrons present in a negative charge of $8$ coulomb is?
$5 \times 10^{19}$
$2.5 \times 10^{19}$
$12.8 \times 10^{19}$
$1.6 \times 10^{19}$

Question

Number of electrons present in a negative charge of 8 coulomb is-5times 10-19-2-5times 10-19-12-8times 10-19-1-6times 10-19-

Toppr Admin · Accepted Answer

The correct option is A 5-xD7-1019Magnitude of charge-xA0- Q-8-xA0-CMagnitude of one electronic charge-xA0- e-1-6-xD7-10-x2212-19-xA0-CLet the number of electron present be N-Thus- number of electrons -xA0- N-Qe-x27F9-xA0-N-81-6-xD7-10-x2212-19-5-xD7-1019 electrons

Number of electrons present in a negative charge of 8 coulomb is?5×10192.5×101912.8×10191.6×1019

The correct option is A 5×1019Magnitude of charge Q=8 CMagnitude of one electronic charge e=1.6×10−19 CLet the number of electron present be N.Thus, number of electrons N=Qe⟹ N=81.6×10−19=5×1019 electrons

Number of electrons present in a negative charge of $8$ coulomb is?
$5 \times 10^{19}$
$2.5 \times 10^{19}$
$12.8 \times 10^{19}$
$1.6 \times 10^{19}$

The correct option is A $5 \times 10^{19}$
Magnitude of charge $Q = 8 C$
Magnitude of one electronic charge $e = 1.6 \times 10^{- 19} C$
Let the number of electron present be $N$ .
Thus, number of electrons $N = \frac{Q}{e}$
$⟹ N = \frac{8}{1.6 \times 10^{- 19}} = 5 \times 10^{19}$ electrons