Explore Westonci.ca, the leading Q&A site where experts provide accurate and helpful answers to all your questions. Discover in-depth solutions to your questions from a wide range of experts on our user-friendly Q&A platform. Explore comprehensive solutions to your questions from a wide range of professionals on our user-friendly platform.

What must the charge (sign and magnitude) of a particle of mass 1.44 g be for it to remain stationary when placed in a downward-directed electric field of magnitude 680 N/C

Sagot :

Answer:

Q E = m g      electric and gravitational forces balance - both E and g are taken as positive in the downward direction

Q = m g / E = .00144 kg * 9.8 m/s^2 / 680 N/C = 2.08E-5 coul

This is the charge required for the specified force

Therefore Q = -2.08-5 C      for the charge to remain stationary

Thank you for your visit. We are dedicated to helping you find the information you need, whenever you need it. We hope this was helpful. Please come back whenever you need more information or answers to your queries. Thank you for choosing Westonci.ca as your information source. We look forward to your next visit.