A copper ball of density `8.6 g//cc` amd 1 cm in diameter is immersed in oil of density `0.8 g//cc`. What is the charge on the ball, if it remains just suspended in oil in electric field of intensity `3600 V//m` acting in the upward direction ?

Answer

`1.1xx10^(-5)C`

Explanation

Here, density of copper ball, `rho_(1) = 8.6 g//c c`
`=8600 kg//m^(3)`
radius of ball, `r = (1)/(2) cm = 0.5xx10^(-2)m`
density of oil, `rho_(2) = 0.8 g//cc = 800 kg//m^(3)`
Charge, `q = ?, E = 3600 V//m`
The ball will remain first suspended in oil, when force due to electric field (F) is equal to apparent weight of the ball in the oil.

i.e., F = qE = apparent weight
= actual wt. - upthrust (wt. of oil displaced)
`= (4)/(3) pi r^(3) rho_(1) g - (4)/(3) pi r^(3) rho_(2) g = (4)/(3) pi r^(3) g (rho_(1) - rho_(2))`
`q = ((4)/(3) pi r^(3) g (rho_(1) - rho_(2)))/(E)`
`= ((4)/(3) xx (22)/(7) (0.5xx10^(-2))^(3) xx9.8 (8600 - 800 ))/(3600)`
`q = 1.1xx10^(-5)C`