Question

Two point charges are located on the y-axis as follows: charge ????1 = −1.50 nC at y = −0.60 m, and charge ????2 = +3.20 nC at the origin (y = 0). What is the total force (magnitude and direction) exerted by these two charges on a third charge ????3 = +5.00 nC located at y = −0.40 m.

Answer 1

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