Answer:
The force exerted on Q3 has a magnitude of 7.8662*^10-7 N and it goes down the Y axis.
F = -7.8662*10^-7 N
Explanation:
What you need to now?
Coulombs law :
F12 = k*(q1*q2)/ r^2 --> the force particle 1 makes to the particle 2
k = 8.99*10^9 N*m^2 / C^2 --> this is a constant
First of all you have to convert the units to th SI,
nano= 10^-9
Q1 or Charge 1 = -1.50nC = -1.50*(10^-9) C
Q2 or Charge 2 = 3.20nC = 3.20*(10^-9) C
Q3 or Charge 3 = 5nC = 5*(10^-9) C
Then you have to do your diagram as you see at the bottom, take into account :
like charges repel each other, alike charges attract each other.
Let's find the distance between the particles :
R32 = 0.4 m
R31 = 0.2 m
Now we apply Coulombs law for every charge that exerts a force to Q3 as follows:
F23 = k*(Q2*Q3)/ (R23)^2 , we replace the data we already have an we find:
F23 = 8.99*10^-7 N
F13 = k*(Q1*Q3)/ (R31)^2 , we replace the data we already have an we find:
F13 = -1.685*10^-6 N
Now you have to sum the forces and find the total force exerted on Q3
Sum F = F23 + F13
F = -7.8662*10^-7 N
The force exerted on Q3 has a magnitude of 7.8662*^10-7 N and it goes down the Y axis.
^ Y
l
l o --> Q2 = 3.2 nC at y=0
l
l
l
l o --> Q3 = 5 nC at y = -0.4m
l l l
l v l
l F13 v
l F23
l o --> Q1 = -1.50 nC at y = - 0.6m
