In Millikan’s oil drop experiment, an oil drop of mass kg 16 *10^-6 kg is balanced by an electric field of 10^6 V/m .What is the charge in coulomb on the drop?
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Step 1
Step 1: Write the expression for the force acting on the oil drop in the electric field: \[ F = qE \] where: \( F \) = force (in Newtons) \( q \) = charge on the oil drop (in Coulombs) \( E \) = electric field strength (in N/C) Show more…
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Kamlesh G.
In Millikan's oil drop experiment on oil drop of mass $16 \times 10^{-6} \mathrm{~kg}$ is balanced by a electric field of $10^{6} \mathrm{Vm}^{-1}$. The charge in coulomb on the drop will be $\left(g=10 \mathrm{~ms}^{-2}\right)$ (a) $16 \times 10^{-13} \mathrm{C}$ (b) $16 \times 10^{-11} \mathrm{C}$ (c) $6.2 \times 10^{-11} \mathrm{C}$ (d) $16 \times 10^{-9} \mathrm{C}$
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Round 2
A positively charged oil drop of mass $1.0 \times 10^{-15} \mathrm{~kg}$ is placed in the region of a uniform electric field between two oppositely charged, horizontal plates. The drop is found to remain stationary under the influence of the Earth's gravitational field and the uniform electric field of $6.1 \times 10^{4} \mathrm{~N} / \mathrm{C}$. What is the magnitude of the charge on the drop? (Ignore the small buoyant force on the drop.)
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