# Homework Help: Coulomb's Forces

1. Apr 15, 2014

### pumpernickel

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

Consider the three charges in the figure below, in which d = 4.7 cm, q = 11 nC, and the positive x-axis points to the right. What is the force Fvec on the 5 nC charge? Give your answer as a magnitude and a direction.

2. Relevant equations

F= (K*Q*q)/r^2

3. The attempt at a solution

First I found the distance between +5 and 11.
(.047^2)+(.03^2)=.055758 m

arctan(.047/.03)= 57.45

Then the magnitude: .000164

The the magnitude of +5, -5 --> .00025 (Horizontal component)

cos 57.45 = x/.000164 = 8.823X10^-5 (Vertical component of line)

sin 57.45 = x/.000164= 1.3824x10^-4 (Horizontal component of line)

Add horizontal components and use that and vertical to make a triangle and find the third side.

So my answer is incorrect, can someone help me out please. Thank you.

#### Attached Files:

• ###### help 3.gif
File size:
3.7 KB
Views:
106
2. Apr 15, 2014

### SammyS

Staff Emeritus
Is the force Between the two positive charges attractive or is it repulsive ?

3. Apr 15, 2014

### pumpernickel

Repulsive. I was unclear, but if the question is whether I took it into account when adding the horizontal components, I did. -->.00025 + (-1.3824x10^-4)

4. Apr 16, 2014

### dauto

Many of the equations you wrote are incorrect even if the solution is correct. That makes your solution very hard to understand. For instance, you wrote
cos 57.45 = 8.823X10^-5 which is completely wrong. There are at least 3 incorrect things with that line and I see at least 3 other lines with similar mistakes.

5. Apr 16, 2014

### SammyS

Staff Emeritus
dauto makes an excellent point.

cos 57.45 = x/.000164 = 8.823X10^-5 (Vertical component of line)​
Is x (in this case) supposed to be the vertical component of the force that the 11 nC charge exerts on the 5 nC charge at the origin ?

If so you should use the sine, not the cosine. Even then, it's only the absolute value of that component.

As dauto points out, what you have written says that cos(57.45°) = 8.823X10^-5 . I doubt that you mean that.

Write something like:
cos(57.45°) = x/.000164 → x = 8.823X10^-5​
(Of course this is not the vertical component.)

6. Apr 16, 2014

### BiGyElLoWhAt

ok, you're given the distance, and you have the correct formula for your force, why not start with a N2L equation?
$\Sigma \vec{F}=m_{5}\vec{a_{5}}=K[\frac{q_{-5}q_{5}}{.03^2}\hat{i}+\frac{q_{11}q_{5}}{r_{11→5}^2}\hat{r}]$
where:
$\hat{r}=cos(tan^{-1}(\frac{d}{.03}))\hat{i} + sin(tan^{-1}(\frac{d}{.03}))\hat{j}$

7. Apr 16, 2014

### BiGyElLoWhAt

I think I made a mistake, the i hat's going to give you a force in - i, that should be rhat_{-5->5} and the second should be rhat_{11->5}

Last edited: Apr 16, 2014