Answer:
The question is incomplete. the complete question is giving below "negative charge of −0.550μC exerts an upward 0.600-N force on an unknown charge that is located 0.300 m directly below the first charge. What are (a) the value of the unknown charge (magnitude and sign); (b) the magnitude and direction of the force that the unknown charge exerts on the −0.550μC charge?"
answer
a 10.9μC
b.0.600N,downward
Explanation:
Since the force applied on the second charge is upward while the charge is below, we can conclude that the second charge is a positive charge since the force is attractive.
From coulombs law, the force between two charges is express as
F=(kq₁q₂)/r²
where q is the charge and r is the distance between the charges.
if we make q₂ subject of formula,we arrive at
q₂=Fr²/kq₁
q₂=(0.6N*0.3²)/(9*10⁹*0.55*10⁻⁶)
q₂=0.054/4950
q₂=1.09*10⁻⁵c
q₂=10.9μC
b.since the magnitude of the charge is constant, the magnitude of the force is constant also i.e 0.600N
the direction of the force on the first charge is downward since the charges are dislike charges.
