# Electrostatic Problem

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)?

#### Attachments

• 3 KB Views: 250
Last edited:

Related Introductory Physics Homework Help News on Phys.org
Doc Al
Mentor
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.

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
Mentor
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. Last edited:
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?

ok, assuming they attract.
got it.

Doc Al
Mentor
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!