Question

A water drop of radius ${10}^{-6}$ m is charged with one electron. The electric field required to keep it stationary is (given density of water $\rho =1000kg/m^3;g=9.8m/s^2$)

• A. $2.566\times {10}^{5}V/m$
• B. $1.283\times {10}^{5}V/m$
• C. $3.849\times {10}^{5}V/m$
• D. $5.132\times {10}^{5}V/m$

Answer 1

Answer: (A)

$2.566\times {10}^{5}V/m$

Let the electric field required be $E$.
According to equilibrium condition,
$\frac{4}{3}\pi {\left({10}^{-6}\right)}^3\times 1000\times 9.8=1.6\times {10}^{-19}\times E$
$\Rightarrow E\approx 2.566\times {10}^{5}V/m$

So, the answer is option (A).