Two point charges, separated by 15 cm, have charge values of +2.0 μC and -4.0 μC, respectively. a) What is the magnitude of the electric force between them? b) Is it an attractive or repulsive force? (ke = 8.99 x 10^9 Nm^2/C^2)
Added by Felipe H.
Step 1
99 x 10^9 Nm^2/C^2, q1 = 2.0 x 10^-6 C, q2 = -4.0 x 10^-6 C, and r = 15 cm = 0.15 m. Show more…
Show all steps
Your feedback will help us improve your experience
Ravindra Yadav and 99 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
Calculate the electric force between two protons at a distance of 10^-15 m apart (roughly the size of atomic nucleus). Charge of proton = 1.6 x 10^-19 C. k = 9.0 x 10^9 N m^2 / C^2. Is the electric force attractive or repulsive?
Adi S.
Two electrostatic point charges of-80 x 10^-6 C and 70 x 10 ^-6 C exert a force on each other of 15 N. a) What is the distance between the two charges? b) State whether the force attractive or repulsive?
Akancha C.
A 7.50-nC point charge is located 1.80 m from a 4.20-nC point charge. (a) Find the magnitude of the electric force that one particle exerts on the other. (b) Is the force attractive or repulsive?
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