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

[jimithing]

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)?

 

[Doc Al]

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.

 

[jimithing]

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?

 

[Doc Al]

You are right!

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.

 

[jimithing]

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?

 

[jimithing]

ok, assuming they attract.
got it.

 

[Doc Al]

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!