PLZ help! In a particular Millikan oil-drop apparatus, the plates are 2.25 cm ap

Question

PLZ help!

In a particular Millikan oil-drop apparatus, the plates are 2.25 cm apart. The oil used has a density of 0.800 g/cm^3, and the atomizer that sprays the oil drops produces drops of diameter 1.00 times 10^-3 mm What strength of electric field is needed to hold such a drop stationary against gravity if the drop contains five excess electrons? What should be the potential difference across the plates to produce this electric field? If another drop of the same oil requires a plate potential of 64.1 V to hold it stationary, how many excess electrons did it contain? Express your answer as an integer.

Explanation / Answer

(a)
mg = qE
(800 kg/m^3)(4(5e-7 m)^3/3)(9.81 m/s^2) = (5)(1.6e-19 C)(E)
E = 5.13e^3 N/C

(b)
For parallel plates:
V = E*x
V = (5.13e^3 N/C)(0.0225 m)
V = 115.42 V

(c)
mg = (ne)(V/x)
n = mgx/(eV)
n = (800 kg/m^3)(4(5e-7 m)^3/3)(9.81 m/s^2)(0.0225 m) / [(1.6e-19 C)(64.1 V)]
n = 9

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.