Question

An oil drop of mass 0.2 mg has lost 100 electrons. The density of oil is 0.8 g/cc. Find the electrostatic pressure it has.

• A. $4.42\times {10}^{-8}Pa$
• B. $4.42\times {10}^{-10}Pa$
• C. $4.42\times {10}^{-6}Pa$
• D. $4.42\times {10}^{-4}Pa$

Answer 1

Answer: (A)

$4.42\times {10}^{-8}Pa$

$r={\left(\frac{m}{\frac{4}{3}\pi 9}\right)}^{\frac{1}{3}}={\left(\frac{{10}^{-9}\times .2}{8\times \frac{4}{3}\times \pi }\right)}^{\frac{1}{3}}$

$=\frac{{10}^{-3}}{2.6}cm=\frac{{10}^{-5}}{2.6}m$.

$P=\frac{1}{2}{ϵ}_0{E}^2$

$=\frac{1}{2}\times 8.85\times {10}^{-12}\frac{{10}^{-19}\times 100\times 9\times {10}^9\times {2.6}^2}{({10}^{-5}{)}^2}$

$=4.42\times {10}^{-12}\times 1440\times 6.89$

$=4.42\times {10}^{-8}Pa$