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An oil drop 'B' has charge 1.6 xx 10^(-1...

An oil drop 'B' has charge `1.6 xx 10^(-19)C and " mass " 1.6 xx 10^(-14)kg`. If the drop is in equilibrium position, then what will be potential difference between the plates. [The distance between the plates is 100mm]

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To solve the problem of finding the potential difference between the plates given an oil drop in equilibrium, we can follow these steps: ### Step-by-Step Solution 1. **Identify the Forces Acting on the Oil Drop**: The oil drop has a charge \( Q = 1.6 \times 10^{-19} \, \text{C} \) and a mass \( m = 1.6 \times 10^{-14} \, \text{kg} \). When the drop is in equilibrium, the electric force acting on it due to the electric field between the plates balances the gravitational force acting on it. 2. **Calculate the Gravitational Force**: ...
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Knowledge Check

  • A proton has a mass 1.67 xx 10^(-27) kg and charge + 1.6 xx 10^(-19) C. If the proton is accelerated through a potential difference of million volts, then the kinetic energy is

    A
    `1.6 xx 10^(-15)` J
    B
    `1.6 xx 10^(-13)` J
    C
    `1.6 xx 10^(-21)` J
    D
    `3.2 xx 10^(-13)`J

  • A charged oil drop of mass 9.75 xx 10^(-15) kg and charge 30 xx 10^(-16) C is suspended in a uniform electric field existing between two parallel plates. The field between the plates is (take g = 10 ms^(-2) )

    A
    `3.25 V m^(-1)`
    B
    `300 V m^(-1)`
    C
    `325 V m^(-1)`
    D
    `32.5 V m^(-1)`

  • The de-Broglie wavelength of proton (" charge" = 1.6 xx 10^(-19)C, "mass" = 1.6 xx 10^(-27) Kg) accelerated through a potential difference of 1kV is

    A
    600 Å
    B
    `0.9 xx 10^(-12) m`
    C
    7 Å
    D
    0.9 mm

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