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Physics Question on Electrostatic potential

A charged particle is suspended in equilibrium in a uniform vertical electric field of intensity 20000 V/m20000\, V / m. If mass of the particle is 9.6×10−16kg9.6 \times 10^{-16} kg, the charge on it and excess number of electrons on the-particle respectively are (g=10 m/s2)\left( g =10\, m / s ^{2}\right)

A

4.8×10−19C, 34.8 \times 10^{-19} C,\, 3

B

5.8×10−19C, 45.8 \times 10^{-19} C,\, 4

C

3.8×10−19C, 23.8 \times 10^{-19} C,\, 2

D

2.8×10−19C, 12.8 \times 10^{-19} C,\, 1

Answer

4.8×10−19C, 34.8 \times 10^{-19} C,\, 3

Explanation

Solution

Charge on the particle is given by
qE=mgq E =m g or q=mgEq=\frac{m g}{E}
=9.6×10−16×1020000=4.8×10−19=\frac{9.6 \times 10^{-16} \times 10}{20000}=4.8 \times 10^{-19}
When there is an excess of nn electrons on the particle, then
q=neq=n e
So, n=qe=4.8×10−191.6×10−19=3n=\frac{q}{e}=\frac{4.8 \times 10^{-19}}{1.6 \times 10^{-19}}=3