Ask Question
9 December, 11:50

In Millikan's oil-drop experiment, one looks ata small oil drop held motionless between two plates. Take the voltage betwee the plates to be 2033 V, and the plate separation to be 2.00 cm. The oil drop (of density 0.81 g/cm3) has a diameter of 3.8x10^6m. Find the charge on the drop, in terms of electron units. You need to round your answer to the nearest integer

+4
Answers (1)
  1. 9 December, 13:58
    0
    The charge on the drop is

    q = 1.741 x 10 ⁻²¹ C

    Explanation:

    Electric field due to plates

    Ef = V/d

    Ef = 2033 V / (2.0 * 10^-2 m)

    Ef = 101650 V/m

    So, we can write

    Ef * q = m*g

    q = m*g / E f

    The mass can be equal using the density and the volume so:

    m = ρ * v

    The volume can be find as:

    v = 2.298 x 10 ⁻ ¹⁶ m³

    q = ρ * v * g / Ef

    q = 81 x 10 ³ kg / m³ * 2.2298 x 10 ⁻ ¹⁶ m³ * 9.8 m/s² / 101650 V/m

    The charge on the drop is

    q = 1.741 x 10 ⁻²¹ C
Know the Answer?
Not Sure About the Answer?
Find an answer to your question ✅ “In Millikan's oil-drop experiment, one looks ata small oil drop held motionless between two plates. Take the voltage betwee the plates to ...” in 📘 Physics if you're in doubt about the correctness of the answers or there's no answer, then try to use the smart search and find answers to the similar questions.
Search for Other Answers