Electric Fields in Millikan's experiment

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SUMMARY

In Millikan's experiment, an oil drop with a radius of 0.9493 µm and a density of 0.876 g/cm³ is suspended in an electric field of 1.92 x 105 N/C. The calculation of the charge on the drop involves determining its volume using the formula (4/3)πr³, followed by calculating the mass from the volume and density, and then finding the weight by multiplying the mass by gravity (9.81 m/s²). The weight of the drop is essential, but the next step requires understanding the forces acting on the drop in the electric field.

PREREQUISITES
  • Understanding of basic physics concepts, particularly forces and electric fields.
  • Familiarity with the formula for the volume of a sphere: (4/3)πr³.
  • Knowledge of density calculations and unit conversions.
  • Ability to apply Newton's second law in the context of charged particles in electric fields.
NEXT STEPS
  • Calculate the charge on the oil drop using the balance of forces in the electric field.
  • Explore the relationship between electric field strength and force on charged objects.
  • Review the principles of Millikan's oil drop experiment for historical context and significance.
  • Investigate the concept of quantization of charge and its implications in physics.
USEFUL FOR

Students studying physics, particularly those focusing on electromagnetism and experimental methods in physics, as well as educators looking for practical applications of theoretical concepts.

Forceflow
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In Millikan's experiment, an oil drop of radius 0.9493 µm and density 0.876 g/cm3 is suspended in chamber C (Figure 22-14) when a downward-pointing electric field of 1.92 105 N/C is applied. Find the charge on the drop, in terms of e. (Include the sign.)

So, to do this problem, i know that you have to use (4/3)pi*r^3 to find the volume. Then you times the volume by the density (in meters). That finds the mass, which you then times by gravity. That should be the answer but its wrong.

My work- Volume=3.5834e-18
(3.5834e-18)(8.74e-7)= 3.139e-24
(8.139e-24)*9.81=3.07944e-23
 
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Forceflow said:
So, to do this problem, i know that you have to use (4/3)pi*r^3 to find the volume. Then you times the volume by the density (in meters). That finds the mass, which you then times by gravity. That should be the answer but its wrong.
What this will give you is the weight of the drop. That's a needed step in the solution, but what's next? What role does the electric field play in this problem? Hint: What forces act on the suspended oil drop?
 

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