View Full Version : Electrostatic Problem
jimithing
Aug24-04, 05:34 PM
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)?
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
Aug24-04, 06:56 PM
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?
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:
jimithing
Aug24-04, 09:00 PM
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
Aug24-04, 09:23 PM
ok, assuming they attract.
got it.
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!
vBulletin® v3.8.7, Copyright ©2000-2012, vBulletin Solutions, Inc.