Question

Calculate the voltage needed to balance an oil drop carrying $10$ electrons between the plates of a capacitor $5mm$ apart. The mass of the drop is $3\times {10}^{-16}kg$ and $g=10m{s}^{-2}.$

Answer 1

$F_G=3\times {10}^{-6}\times 10=3\times {10}^{-15}N$

$E=\left(\frac{mg}{10e}\right)=\left(\frac{3\times {10}^{-15}}{10\times 1.6\times {10}^{-19}}\right)=0.1875\times {10}^{4}V/m$

$V=Ed$

$=0.1875\times {10}^4\times 5\times {10}^{-3}$

$=9.375volt$