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An oil drop of radius 10^(-6)m carries a...

An oil drop of radius `10^(-6)m` carries a charge equal to that on an electron. If the density of the oil is `2xx10^(3)kg m^(3)`, find the electric field required to keep it stationary. Take `e=1.6xx10^(-19)C` and `g=9.8 ms^(-2)`.

Text Solution

Verified by Experts

The correct Answer is:
`5.13 xx 10^(5) Vm^(-1)`
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Knowledge Check

  • An oil drop of radius r carrying a charge q remain stationary in the presence of electric field of intensity E. If the density of oil rho , then

    A
    `E=4/3pi r^(3)rhogq`
    B
    `E=4/3pir^(3)rho g`
    C
    `E=4/3pir^(3)rho g//q`
    D
    `E=4/3pir^(3)rho//g^(3)`

  • The electric field that can balance a charged particle of mass 3.2xx10^(-27) kg is (Given that the charge on the particle is 1.6xx10^(-19)C )

    A
    `19.6xx10^(-8)NC^(-1)`
    B
    `20xx10^(-6)NC^(-1)`
    C
    `19.6xx10^(8)NC^(-1)`
    D
    `20xx10^(6)NC^(-1)`

  • A charged water drop of weight 4 xx 10^(-18) kg and charge 1.6 xx 10^(-19) C is stable in an electric field . What is the intensity of electric field

    A
    `150 V m^(-1)`
    B
    `200 V m^(-1)`
    C
    `250 V m^(-1)`
    D
    `50 V m^(-1)`

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    A water droplet of radius 1 micron in Milikan oil drop appartus in first held stationary under the influence of an electric field of intensity 5.1xx10^(4) NC^(-1) . How many excess electrons does it carry ? Take e = 1.6xx10^(-19) C, g = 9.8 ms^(-2) and density of water of = 10^(3) kg m^(-3) .

    An oil drop has 8.0 xx 10^(-19)C charge. How many electrons does this oil drop has?

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    If an oil drop of weight 3.2xx10^(-13) N is balanced in an electric field of 5xx10^(5) Vm^(-1) , find the charge on the oil drop.

    What is the drift velcity of electrons in silver wire of length 1 m having cross-sectional area 3.14xx10^(-6)m^(2) and carrying a current of 10 A . Given atomic mass of silver =108, density of silver =10.5xx10^(3) kg m^(-3) , charge on electron =1.6xx10^(-19) C and Avogadro's number =6.023xx10^(26) per kt-atom.

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