Question

A charged oil drop is suspended in uniform field of $3 \times 10^4\, V/m$ so that it neither falls nor rises. The charge on the drop will be
(Take the mass of the charge $= 9.9 \times 10^{-15}\, kg$ and $g = 10\, m/s^2)$

• A. $3.3 \times 10^{-18}\, C$
• B. $3.2 \times 10^{-18}\, C$
• C. $1.6 \times 10^{-18}\, C$
• D. $4.8 \times 10^{-18}\, C$

Answer 1

Answer: (A)

Given : mass o f charged drop
$m=9.9\times 10^{-15}\,kg$
Electric field $E=3\times 10^{4} V/m$
In steady state.
Electric force on a drop = weight of a drop
$\therefore qE=mg$
or $q=\frac{mg}{E}$
$=\frac{9.9\times10^{-15}\times10}{3\times 10^{4}}$
$=3.3\times 10^{-18}\,C$