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 : $(g=10m/s^2)$

• A. 1
• B. 10
• C. 100
• D. 1000

Answer 1

Answer: (A)

From the relation, $qE=mg$ $\left(\text{Since,}E=\frac{V}{d}\right)$ $\left(E=\frac{100}{0.016}\right)$ $\therefore$ $q=\frac{mg}{E}$ ?(i) $\left(\begin{array}{rl}& \text{Given: m =1}{\text{0}}^{-13}g\\ & ={10}^{-16}\text{kg}\\ & d=0.016\\ & V=100\text{volt}\\ & e=1.6\times {10}^{-19}\text{coulomb}\\ & g=10m/s^2\end{array}\right)$ 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