Calculate the voltage required to balanced an oil drop carrying 10 electrons, when located between plates of a capacitor, which are 5mm apart. Given mass of drop `= 3xx10^(-16) kg`, charge on electron `= 1.6xx10^(-19)C and g = 9.8 m//s^(2)`.

Calculate the voltage required to balanced an oil drop carrying 10 ele

1.	Calculate the voltage required to balanced an oil drop carrying 10 electrons, when located between plates of a capacitor, which are 5mm apart. Given mass of drop `= 3xx10^(-16) kg`, charge on electron `= 1.6xx10^(-19)C and g = 9.8 m//s^(2)`.
Answer» Correct Answer - 9.2V Here, `V = ?, n = 10, d = 5 mm = 5xx10^(-3) m`, `m = 3xx10^(-16) kg, e = 1.6xx10^(-19)C`, `g = 9.8m//s^(2)` To balance the oil drop `F = qE = mg or q((V)/(d)) = mg` `:. V = (mgd)/(q) = (mgd)/(n e) = (3xx10^(-16)xx9.8xx5xx10^(-3))/(10xx1.6xx10^(-19))` = 9.2 volt

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Calculate the voltage required to balanced an oil drop carrying 10 electrons, when located between plates of a capacitor, which are 5mm apart. Given mass of drop `= 3xx10^(-16) kg`, charge on electron `= 1.6xx10^(-19)C and g = 9.8 m//s^(2)`.

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