Question

A charged water drop of radii $0.1\mu m$ under equilibrium in some electric field. The charge on the drop is equivalent to electronic charge. The intensity of electric field is:

• A. $1.61N/C$
• B. $25.2N/C$
• C. $262N/C$
• D. $1610N/C$

Answer 1

The correct option is C $262N/C$

According to question

$mg=cE$

Where $m=$ Mass of drop.

$E=$ Intensity of elecric field.

$\rho \times \frac{4}{3}\pi ({10}^{-7}{)}^3\times 10=1.6\times {10}^{-19}E$

[Take $g=10m/s^2$; $\rho =$Density of water.]

$E=\frac{4\pi \times {10}^{-21}\times 10\times 1000}{3\times 1.6\times {10}^{-19}}$

$=261.8N/C$

$E\approx 262N/C$