A 2.0 x 10-4 kg oil drop is held in place by a downward electric field of 3.0 x 102 N/C. What is the charge on the oil drop?
Added by Sadie D.
Step 1
This means the downward gravitational force on the oil drop is balanced by the upward electric force. The gravitational force (Fg) can be calculated using the equation Fg = mg, where m is the mass of the oil drop and g is the acceleration due to gravity Show more…
Show all steps
Your feedback will help us improve your experience
Prabhat Tyagi and 97 other Physics 101 Mechanics educators are ready to help you.
Ask a new question
Labs
Want to see this concept in action?
Explore this concept interactively to see how it behaves as you change inputs.
Key Concepts
Recommended Videos
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.)
A charged oil drop with a mass of 8.2 g is held suspended by a downward electric field of 394 N/C. What is the charge on the drop? Answer in μC.
Madhur L.
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?
Krishna J.
Recommended Textbooks
University Physics with Modern Physics
Physics: Principles with Applications
Fundamentals of Physics
Transcript
Watch the video solution with this free unlock.
EMAIL
PASSWORD