Question

eLLø Brainliacs!! ❤️❤️

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​

Answer 1

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

Answer 2

Hello!!!

Charge on oil drop =ne = 10×1.6×10^{−19}=1.6×10^{−18}C

mass of of oil drop (m) = 3 * 10^{-16} kg , E = V/r where V is the potential difference and r = radius of oil drop

Oil drop is balanced as

qE= mg

q *v/r = mg

V= mgr/q

after solving we get,

9.19V ->Ans

Thanks ❤⚠