What is the force between two point charges at a 45 degree angle?

In summary, we discussed the placement of point charges q1 and q2 in space, with q1 at the origin and q2 a distance r from q1 making a 45 degree angle with the horizontal. We found the force between q1 and q2 using unit vectors i and j, with the force of q1 on q2 being positive in the given coordinate system. We also discussed how the magnitude of the force changes when q1 and q2 have the same value and when the distance between them is 2.0 m.
  • #1
jimithing
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Point charges q1 and q2 are placed in space, with q1 at the origin and q2 a distance r from q1 making a 45 degree angle with the horizontal.
a) Find the force using unit vectors i and j from q1 to q2
b) " " from q2 to q1
c) If q1=q2, what is the magnitude of the force?

so far i have:

q2:
Fx = F cos (theta)
Fy = F sin (theta) - so F(1on2) = F cos(theta) i + F sin (theta) j

am I on the right track? and would F(2on1) be -F(1on2)?
 

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  • #2
jimithing said:
q2:
Fx = F cos (theta)
Fy = F sin (theta) - so F(1on2) = F cos(theta) i + F sin (theta) j
Why use F to represent the force? Use Coulomb's law and write "F" in terms of q1, q2, and r. Remember: signs matter.

and would F(2on1) be -F(1on2)?
Yes.
 
  • #3
so for 1 on 2:
Fx = k(q1)(q2)cos(theta)/r^2 i
Fy = k(q1)(q2)sin(theta)/r^2 j
2 on 1
Fx = -k(q1)(q2)cos(theta)/r^2 i
Fy = -k(q1)(q2)sin(theta)/r^2 j

Part (c) when q1=q2=5 x 10^-6 C and r = 2.0 m

sub values into:
F = k (q1)(q2)/r^2 or F = kq^2/r^2

am i correct?
 
  • #4
You are right!

jimithing said:
so for 1 on 2:
Fx = k(q1)(q2)cos(theta)/r^2 i
Fy = k(q1)(q2)sin(theta)/r^2 j
2 on 1
Fx = -k(q1)(q2)cos(theta)/r^2 i
Fy = -k(q1)(q2)sin(theta)/r^2 j
You are correct!
Part (c) when q1=q2=5 x 10^-6 C and r = 2.0 m

sub values into:
F = k (q1)(q2)/r^2 or F = kq^2/r^2

am i correct?
Sounds good to me.

Edit: I messed up the signs before! You are correct. :smile:
 
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  • #5
Doc Al said:
If the magnitude of the total force is F, where F = k(q1)(q2)/r^2,
then the components of the force on q1 are positive:
Fx = F cos(theta); Fy = F sin(theta)
and the components of the force on q2 are negative:
Fx = -F cos(theta); Fy = -F sin(theta)

Sounds good to me.
wouldn't the force of q1 on q2 be positive on the coordinate system used?
 
  • #6
ok, assuming they attract.
got it.
 
  • #7
jimithing said:
wouldn't the force of q1 on q2 be positive on the coordinate system used?
Right. I messed up the signs before. (Funny... I was telling you to be careful of signs and I goofed up! :blush: )

Good work!
 

FAQ: What is the force between two point charges at a 45 degree angle?

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