Question

An electric field can just support a water droplet $1.0\times {10}^{-6}$m in diameter carrying one electron charge. The magnitude of electric field strength is

• A. $3.21\times {10}^{4}V/m$
• B. $2.31\times {10}^{4}V/m$
• C. $1.32\times {10}^{4}V/m$
• D. $6.42\times {10}^{4}V/m$

Answer 1

The correct option is A $3.21\times {10}^{4}V/m$

Answer is A.

Force in electric field
$F=qE$
Mass of water droplet
$=f_{w}V$
$=f_w\frac{4}{3}\pi r^3$
$=f_w\frac{4}{3}\pi (\frac{d}{2}{)}^3$
$=1000\times \frac{4}{3}\times \pi \times (\frac{{10}^{-6}}{2}{)}^3$
$=1000\times \frac{4}{3}\times \pi \times \frac{1}{8}\times {10}^{-18}$
$=5.23\times {10}^{-16}kg$
So, gravitational force $=mg$
$=5.23\times {10}^{-16}\times 10$
$=5.23\times {10}^{-15}N$
Now, for equilibrium
$qE=$ gravitational force
$E=\frac{5.23\times {10}^{-15}}{1.6\times {10}^{-19}}$ ($\because electroncharge=1.6\times {10}^{-19}C$)
$=3.21\times {10}^4\frac{V}{m}$