Calculate the voltage needed to balance an oil drop

Asked by admin @ 17/08/2021 in Physics viewed by 284 People

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Calculate the voltage needed to balance an oil drop carrying 10 electrons when located between the plates of capacitor which are 5mm apart. ( g = 10 m/s^2) mass of oil drop = 3 * 10^-16 kg

Answered by admin @ 17/08/2021

Answer:

9.4V

Explanation:

Given,

Mass of oil drop =

$3\times{10}^{-16}kg$

Distance between the capacitors = 5mm

$5\times10^{-3} m$

number of electrons (n) = 10

Q = n e

$10\times1.6\times10^{-19}$

$= 1.6 \times 10^{-18}c$

As we know,

V = Work / Charge

Work = m * g * distance

$3 \times {10}^{-16} \times 10 \times 5 \times 10^{-3}$

$150 \times {10}^{-19} joule$

On putting the value,

V =

$\frac{150 \times {10}^{ - 19}}{1.6 \times {10}^{ - 18}}$

$93.75 \times 10$

= 9.4 V

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