Question

Calculate the electric field intensity required to just support a water drop let of mass ${10}^{-7}kg$ and having a charge of $+4.0\times {10}^{-13}C$.

Answer 1

Given that,

Mass $m={10}^{-7}kg$

Charge $q=4\times {10}^{-13}C$

Now, the electric field strength be applied vertically upwards to oppose gravity

Electric force = charge * electric field strength

$ma=q\times E$

${10}^{-7}\times 9.8=4\times {10}^{-13}\times E$

$E=\frac{{10}^{-7}\times 9.8}{4\times {10}^{-13}}$

$E=2.45\times {10}^{6}N/C$

Hence, the electric field intensity is $2.45\times {10}^{6}N/C$