Question

A charged oil drop is suspended in a uniform field of $3\times {10}^{4}V/m$ so that it neither falls nor rises. The charge on the drop will be,

(Take $massofcharge=9.9\times {10}^{-15}kg$ and $g=10m/s^2$)

• 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

The correct option is A

$3.3\times {10}^{-18}C$

Step 1: Given

Electric force: $E=3\times {10}^{4}V/m$

Mass of charge: $m=9.9\times {10}^{-15}kg$

Acceleration due to gravity: $g=10m/s^2$

Step 2: Formula Used

At equilibrium, the electric force of a drop balances the weight of the drop.

So, it is expressed as,

$qE=mg$

(Where $q$ is the charge of the drop, $E$ is the electric force, $m$ is mass of charge and $g$ is the acceleration due to gravity)

This can be further rewritten as,

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

Step 3: Calculation

Substitute the values to find the charge on drop,

$\begin{array}{rcl}q& =& \frac{9.9\times {10}^{-15}\times 10}{3\times {10}^4}\\ & =& 3.3\times {10}^{-18}C\end{array}$

The correct option is option (A).