Coulomb's Forces Homework: Answer 5 nC Force

[pumpernickel]

Homework Statement

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.

Homework Equations

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

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.

 

[SammyS]

Homework Statement

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.

Homework Equations

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

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.

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

 

[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)

 

[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.

 

[SammyS]

...

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)
...

dauto makes an excellent point.

Look at your line:

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.)

It wouldn't hurt to include units in your answer.

 

[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}

 

[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}