- #1
Blink691
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I did this problem, yet I found out that its wrong because I didn't plug the correct force in. I have no idea what to do now though. If someone can show me what I did wrong and explain it, I would greatly appreciate that. Thank you!
Two coins with identical charges are placed on a lab table 1.35 m apart.
A.If each coin experiences a force of 2.0 N because of the presence of the other coin, calculate the charge on each coin.
FE=KE (q1q2/r^2)
R^2 = 1.35m
q1q2 = 2.0N
FE=KE (2.0N*2.0N)/(1.35^2m)
FE=2.194 or 2.195 A
B.Would the force be classified as a force of attraction or repulsion? Explain your answer.
Force of attraction because it represents a weight.
Two coins with identical charges are placed on a lab table 1.35 m apart.
A.If each coin experiences a force of 2.0 N because of the presence of the other coin, calculate the charge on each coin.
FE=KE (q1q2/r^2)
R^2 = 1.35m
q1q2 = 2.0N
FE=KE (2.0N*2.0N)/(1.35^2m)
FE=2.194 or 2.195 A
B.Would the force be classified as a force of attraction or repulsion? Explain your answer.
Force of attraction because it represents a weight.