A small oil drop of mass `10^(-6)kg` is hanging in at rest between two plates separated by1 mm having a potential difference of 500V. The charge on the drop is ____ `(g = 10 ms^(-2))`

Options

A. `2 xx 10^(-9)C`
B. `2 xx 10^(-11)C`
C. `2 xx 10^(-6)C`
D. `2 xx 10^(-8)C`

Answer

B

Explanation

For the drop remains at rest, Force on the drop due to electric field = Weight of the drop
`qE= mg`
`q((V)/(d)) = mg " " ( :' E+ (V)/(d))`
`q= (mgd)/(V)`
Here, `m= 10^(-6) kg, d=1mm = 10^(-3)m, V= 500V, g= 10 ms^(-2)`
`:. q= ((10^(-6)kg) (10 ms^(-2)) (10^(-3)m))/(500V) = 2 xx 10^(-11)C`