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.
Added by Brenda C.
Close
Step 1
The weight can be calculated using the formula: Weight = mass * acceleration due to gravity The acceleration due to gravity is approximately 9.8 m/s^2. Weight = 8.2 g * 9.8 m/s^2 = 80.36 g*m/s^2 Show more…
Show all steps
Your feedback will help us improve your experience
Madhur L and 101 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 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?
Prabhat T.
In Millikan's experiment, an oil drop of radius 1.64$\mu \mathrm{m}$ and density 0.851 $\mathrm{g} / \mathrm{cm}^{3}$ is suspended in chamber $\mathrm{C}($ Fig. $22-16)$ when a downward electric field of $1.92 \times 10^{5} \mathrm{N} / \mathrm{C}$ is applied. Find the charge on the drop, in terms of $e .$
Pritesh R.
In Millikan's experiment, an oil drop of radius $1.64 \mu \mathrm{m}$ and density $0.851 \mathrm{~g} / \mathrm{cm}^{3}$ is suspended in chamber $\mathrm{C}$ (Fig. $22-16$ ) when a downward electric field of $1.92 \times 10^{5} \mathrm{~N} / \mathrm{C}$ is applied. Find the charge on the drop, in terms of $e$.
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