A negative charge of - 0.510 μC 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?

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.