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Physics Question on Electric Field

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 are respectively (g=10 m/s2)(g=10\,m/s^2) :

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