Question:

A charged particle is suspended in equilibrium in a uniform vertical electric field of intensity 20000  V/m 20000\,\,V/m . If mass of the particle is 9.6×10−16kg 9.6\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) :

Updated On: Aug 1, 2022
  • 4.8×10−19C,  3 4.8\times {{10}^{-19}}C,\,\,3
  • 5.8×10−19C,  4 5.8\times {{10}^{-19}}C,\,\,4
  • 3.8×10−19C,  2 3.8\times {{10}^{-19}}C,\,\,2
  • 2.8×10−19C,  1 2.8\times {{10}^{-19}}C,\,\,1
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The Correct Option is A

Solution and Explanation

Charge on the particle is given by qE=mgq E =m g or q=mgE q=\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
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Concepts Used:

Electric Field

Electric Field is the electric force experienced by a unit charge. 

The electric force is calculated using the coulomb's law, whose formula is:

F=k∣q1q2∣r2F=k\dfrac{|q_{1}q_{2}|}{r^{2}}

While substituting q2 as 1, electric field becomes:

 E=k∣q1∣r2E=k\dfrac{|q_{1}|}{r^{2}}

SI unit of Electric Field is V/m (Volt per meter).