A charged oil drop of mass `2.5xx10^(-7)`kg is in space between the two plates, each of area `2xx10^(-2) m^(2)` of a parallel plate capacitor. When the upper plate has a charge of `5 xx10^(-7) C` and the lower plate has an equal negative charge, then the oil remains stationary. the charge of the oil drop is (take, `g=10 m//s^(2)`)

Options

A. `9xx10^(-1) C`
B. `9xx10^(-6) C`
C. `8.85xx10^(-13) C`
D. `1.8xx10^(-14) C`

Answer

C

Explanation

Since, qE = mg
`implies(qQ)/(epsilon_0A)=mg impliesq =(epsilon_0Amg)/(Q)`
`=(8.85 xx10^(-12) xx2xx10^(-2) xx2.5 xx10^(-7) xx10)/(5xx10^(-7))`
`= 8.85 xx10^(-13) C`