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 $=9.9\times {10}^{-15}andg=10$)

• 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

The correct option is A $3.3\times {10}^{-18}C$

In steady state,

Electric force on drop $=$ Weight of drop

That is,

$qE=mg$

Then,

$q=\frac{mg}{E}$

$=\frac{9.9\times {10}^{-15}\times 10}{3\times {10}^4}=3.3\times {10}^{-18}C$