Question

A copper ball of density 8.6 $\frac{g}{c{m}^3}$, 1 cm in diameter is immersed in oil of density 0.8 $\frac{g}{c{m}^3}$. What is the charge in $\mu$C on the ball, if it remains suspended in an electric field of intensity 3600 $\frac{V}{m}$ acting in upward direction. (Answer the nearest integer)

Answer 1

Given
density of copper ball ${\rho }_{Cu}=8.6\frac{g}{c{m}^3}=8.6\times {10}^3\frac{kg}{m^3}$
${\rho }_{oil}=0.8\frac{g}{c{m}^3}\Rightarrow 0.8\times {10}^3\frac{kg}{m^3}$
E = $3600\frac{V}{m},r=\frac{1}{2}cm=0.5\times {10}^{-2}m$
$F_{net}$= qE
From equilibrium
qE +$\frac{4}{3}\pi R^3\times {\rho }_{oil}\times g=\frac{4}{3}\pi R^3\times {\rho }_{Cu}\times g$
$q=\frac{\frac{4}{3}\times 3.14\times (0.5\times {10}^{-2}{)}^3(7.8)\times {10}^3\times 10}{E}$
$q=\frac{\frac{4}{3}\times 3.14\times (0.5\times {10}^{-2}{)}^3(7.8)\times {10}^3\times 10}{3600}$
$q=\frac{\frac{4}{3}\times 3.14\times 7.8\times 0.125\times {10}^{-2}}{3600}$
q = $1.1\times {10}^{-5}$ C
= $11\times {10}^{-6}$ C
= 11 $\mu$ C