# Simple Coulomb's Law Problem

1. Sep 8, 2008

### godmoktail

1. The problem statement, all variables and given/known data

2. Relevant equations
F = | kqQ / r*r |

3. The attempt at a solution
Magnitude:
F_net = F_1on3 + F_2on3
F_1on3 = (9E9)(70E-6)(15E-6) / ( (sqrt(5))^2 ) = 1.89N
F_2on3 = (9E9)(-36E-6)(15E-6) / ( (2)^2 ) = -1.215N ~ -1.22N
F_net = 1.89 - 1.22 = 0.67N

Direction = arctan(F_y/F_x)
In component form the vectors for r_13 and r_23 are:
r_13 = 2i + 1j && r_23 = 2j

Therefore the force in component notation would be:
Fvec_13 = 1.89(2i + 1j) = 3.78i + 1.89j
Fvec_23 = -1.22(2j) = -2.44j
Fvec_net = 3.78i - 0.55j

Thus angle must be:
theta = arctan(-0.55/3.78) = -8.13 degrees

I entered 0.675 for the magnitude and was told it was incorrect, although I'm positive I did the work correctly. Am I wrong? (I am a little unsure about the angle part)

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Sep 8, 2008

### edziura

One mistake I see is that, for example, r_13 must be converted to a unit vector before you can use it to write Fvec_13. r_23 as a unit vector is just j.

3. Sep 8, 2008

### godmoktail

So r_13 as a unit vector would be squareroot(5), and r_23 would be squareroot(2), no?

4. Sep 8, 2008

### edziura

I'm not sure what you mean.

r_13 as a unit vector: (2i + 1j) / sqrt(5)

r_23 as a unit vector = j

5. Sep 8, 2008

### LowlyPion

Are you sure that's the magnitude?

6. Sep 8, 2008

### godmoktail

Are you saying that I should have entered 0.67N? Either way, I would like to ensure that my magnitude calculations are correct, regardless of what I entered on my homework software.

If they asked for the Force vector, then I would have to multiply the magnitude by r^ (r hat), no?

7. Sep 8, 2008

### LowlyPion

What I'm asking is given your statement of the force vector is that it's magnitude?

8. Sep 8, 2008

### godmoktail

I take Fvec_net as the Force vector 3.78i - 0.55j (with magnitude 0.67N) which I wrote in component notation

9. Sep 8, 2008

### LowlyPion

The problem is that this is not the net magnitude. To properly calculate the magnitude of the result you need to have done a vector addition as these magnitudes are not acting along the same line.

10. Sep 8, 2008

### godmoktail

Ok, here is my retry.

r_vec31 = -2i - j
r_vec32 = -2j

r_hat31 = (-2i - j) / sqrt(5)
r_hat32 = (-2j) / 2 = -j

Fvec_31 = 4.23 * r_hat31 N = -3.78i - 1.89j
Fvec_32 = -1.22 * r_hat32 N = 1.22j
Fvec_net = (-3.78i - 0.67j)
Fvec_mag = 3.84 N

So the magnitude would be 3.84N, NO?

11. Sep 9, 2008

### godmoktail

And since the components are -3.78 and 0.67, then the direction is:
arctan(-.67/-3.78) = 10 deg, no?

12. Sep 9, 2008

### LowlyPion

Yes, that looks good.

13. Sep 9, 2008

### LowlyPion

10 degrees with respect to what? x-axis?

14. Sep 9, 2008

### godmoktail

yes, the x-axis

15. Sep 9, 2008

### LowlyPion

OK. Looks good then.

16. Mar 4, 2009

### r kumar

hello sir ..........can u help me i have a problem.........two point charges placed at d distance .two mediums of dielectric constants k1 and k2 are filled between them.one upto distance d1 and other upto d2 distance.what is the force acting between the charge.d1+d2=d.

17. Mar 4, 2009

### LowlyPion

I think I would use the equivalent εeq in place of εo

ε1 = k1*εo
ε2 = k2*εo

which I think intuitively can be determined similarly to capacitors in series by

d/εeq = d1/(k1*εo) + d2/(k2*εo)

You may want to satisfy yourself that this is correct.

18. Mar 4, 2009

### r kumar

sir thanks for replying. sir i have another question also if these mediums are placed one above the other between two point charges .the lengh of mediums is same as the distance between charges .what will be force between the charges .
_____________________________________
q1|________________k2_____________________| q2
|________________k1_____________________|

19. Mar 4, 2009

### LowlyPion

That seems to be more complicated because the thicknesses are unknown. If the thicknesses are large and the boundary bisects the point charges then by symmetry I would suppose that

εeq = 1/2*(k1 + k2)εo

BUT for thicknesses less than d between the charges, ... that looks more complicated than I want to consider.

20. Mar 4, 2009

### r kumar

sir both the mediums have thickness same as that of d distance between the chages.