Question

A dust particle of mass $ {{10}^{-13}}g $ is stationary between plates of a horizontal parallel plate capacitor of 0.016 m separation which is connected to a voltage of 100 V. How many fundamental charges $ (e=1.6\times {{10}^{-19}}C) $ the dust particle carries : $ (g=10\text{ }m/{{s}^{2}}) $

• A. 1
• B. 10
• C. 100
• D. 1000

Answer 1

Answer: (A)

From the relation, $ q\,E=mg $ $ \left( \text{Since,}\,E=\frac{V}{d} \right) $ $ \left( E=\frac{100}{0.016} \right) $ $ \therefore $ $ q=\frac{mg}{E} $ ?(i) $ \left( \begin{align} & \text{Given: m =1}{{\text{0}}^{-13}}\,g \\ & ={{10}^{-16}}\,\text{kg} \\ & d=0.016 \\ & V=100\,\text{volt} \\ & e\text{ }=\text{ }1.6\times {{10}^{-19}}\text{coulomb} \\ & g\text{ }=\text{ }10\text{ }m/{{s}^{2}} \\ \end{align} \right) $ Now, putting the given value and value of E then form (ii) in Eq. (i) $ q=\frac{{{10}^{-6}}\times 10\times 0.016}{100}\,\text{coulomb} $ Number of charges are given as $ n=\frac{q}{e} $ $ =\frac{{{10}^{-16}}\times 10\times 0.016}{100\times 1.6\times {{10}^{-19}}} $ = 1