Question:
A charged particle is suspended in equilibrium in a uniform vertical electric field of intensity $ 20000\,\,V/m $ . If mass of the particle is $ 9.6\times {{10}^{-16}}kg $ , the charge on it and excess number of electrons on the particle are respectively $ (g=10\,m/s^2) $ :
A charged particle is suspended in equilibrium in a uniform vertical electric field of intensity $ 20000\,\,V/m $ . If mass of the particle is $ 9.6\times {{10}^{-16}}kg $ , the charge on it and excess number of electrons on the particle are respectively $ (g=10\,m/s^2) $ :
Updated On: Aug 1, 2022
- $ 4.8\times {{10}^{-19}}C,\,\,3 $
- $ 5.8\times {{10}^{-19}}C,\,\,4 $
- $ 3.8\times {{10}^{-19}}C,\,\,2 $
- $ 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
$q E =m g $ or $ q=\frac{m g}{E} $
$=\frac{9.6 \times 10^{-16} \times 10}{20000}=4.8 \times 10^{-19}$
When there is an excess of $n$ electrons on the particle, then
$q=n e$
So, $n=\frac{q}{e}=\frac{4.8 \times 10^{-19}}{1.6 \times 10^{-19}}=3$
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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:
\(F=k\dfrac{|q_{1}q_{2}|}{r^{2}}\)
While substituting q2 as 1, electric field becomes:
\(E=k\dfrac{|q_{1}|}{r^{2}}\)
SI unit of Electric Field is V/m (Volt per meter).