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 wonderswan Nov18-10 06:02 PM

Coulomb Force Point Charges on Cube

Hello, I'm having a little trouble with this:

Coulomb Force Point Charges on Cube

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

Identical charges of Q (C) are located at the eight corners of a cube with side L (m). Show that the coulomb force on each charge has magnitude:

$$3.29Q^2/4\pi\epsilon_0l^2$$

2. Relevant equations

$$F = \frac{Q_1Q_2}{4\pi\epsilon_0R^2}\cdot\hat{a}\ \ (N)$$

3. The attempt at a solution

The total force at the point observed (at origin) is the sum of the seven other vertices of the cube. As we are concerned with the magnitude we can neglect the vector (I'm unsure about this).

$$F = Q^2/4\pi\epsilon_0\cdot\displaystyle\sum_{n=1}^{7}\frac{1}{R^2_n}$$

$$R_1 = \sqrt{(-l)^2} \ \ \small(-1,0,0)$$
$$R_2 = \sqrt{(-l)^2} \ \ \small(0,-1,0)$$
$$R_3 = \sqrt{(-l)^2} \ \ \small(-1,0,0)$$
$$R_4 = \sqrt{(-l)^2+(-l)^2} = \sqrt{2l^2} \ \ \small(-1,-1,0)$$
$$R_5 = \sqrt{(-l)^2+(-l)^2} = \sqrt{2l^2} \ \ \small(-0,-1,-1)$$
$$R_6 = \sqrt{(-l)^2+(-l)^2} = \sqrt{2l^2} \ \ \small(-1,0,-1)$$
$$R_7 = \sqrt{(-l)^2+(-l)^2 +(-l)^2} = \sqrt{3l^2} \ \ \small(-1,-1,-1)$$

$$F = Q^2/4\pi\epsilon_0\left[3\left(\frac{1}{l^2}\right)+3\left(\frac{1}{2l^2}\right)+\left(\frac{1} {3l^2}\right)\right]$$

$$= 4.8\dot{3}Q^2/4\pi\epsilon_0l^2$$

Any help appreciated.

 berkeman Nov18-10 06:30 PM

Re: Coulomb Force Point Charges on Cube

Quote:
 Quote by wonderswan (Post 2992772) Hello, I'm having a little trouble with this: Coulomb Force Point Charges on Cube 1. The problem statement, all variables and given/known data Identical charges of Q (C) are located at the eight corners of a cube with side L (m). Show that the coulomb force on each charge has magnitude: $$3.29Q^2/4\pi\epsilon_0l^2$$ 2. Relevant equations $$F = \frac{Q_1Q_2}{4\pi\epsilon_0R^2}\cdot\hat{a}\ \ (N)$$ 3. The attempt at a solution The total force at the point observed (at origin) is the sum of the seven other vertices of the cube. As we are concerned with the magnitude we can neglect the vector (I'm unsure about this). $$F = Q^2/4\pi\epsilon_0\cdot\displaystyle\sum_{n=1}^{7}\frac{1}{R^2_n}$$ $$R_1 = \sqrt{(-l)^2} \ \ \small(-1,0,0)$$ $$R_2 = \sqrt{(-l)^2} \ \ \small(0,-1,0)$$ $$R_3 = \sqrt{(-l)^2} \ \ \small(-1,0,0)$$ $$R_4 = \sqrt{(-l)^2+(-l)^2} = \sqrt{2l^2} \ \ \small(-1,-1,0)$$ $$R_5 = \sqrt{(-l)^2+(-l)^2} = \sqrt{2l^2} \ \ \small(-0,-1,-1)$$ $$R_6 = \sqrt{(-l)^2+(-l)^2} = \sqrt{2l^2} \ \ \small(-1,0,-1)$$ $$R_7 = \sqrt{(-l)^2+(-l)^2 +(-l)^2} = \sqrt{3l^2} \ \ \small(-1,-1,-1)$$ $$F = Q^2/4\pi\epsilon_0\left[3\left(\frac{1}{l^2}\right)+3\left(\frac{1}{2l^2}\right)+\left(\frac{1} {3l^2}\right)\right]$$ $$= 4.8\dot{3}Q^2/4\pi\epsilon_0l^2$$ Any help appreciated.
Welcome to the PF!

This is the problem:
Quote:
You cannot neglect it. You need to do the vector sum first, and then you can take the magnitude of the resultant vector.

 wonderswan Nov19-10 11:04 AM

Re: Coulomb Force Point Charges on Cube

Quote:
 Quote by berkeman (Post 2992808) Welcome to the PF! This is the problem: You cannot neglect it. You need to do the vector sum first, and then you can take the magnitude of the resultant vector.
Hi, thanks for your response. I'm not sure where I can get the 3.69 value. The resultant vector is $$4l\hat{i}+4l\hat{j}+4l\hat{k}$$.

$$F = \frac{Q^2}{4\pi\epsilon_0R^2}\cdot\hat{a}$$

Is R the magnitude of the resultant vector - it would be $$\sqrt{48l^2}$$???

I think I'm getting confused.

 berkeman Nov19-10 01:07 PM

Re: Coulomb Force Point Charges on Cube

Quote:
 Quote by wonderswan (Post 2993900) Hi, thanks for your response. I'm not sure where I can get the 3.69 value. The resultant vector is $$4l\hat{i}+4l\hat{j}+4l\hat{k}$$. $$F = \frac{Q^2}{4\pi\epsilon_0R^2}\cdot\hat{a}$$ Is R the magnitude of the resultant vector - it would be $$\sqrt{48l^2}$$??? I think I'm getting confused.
Can you show how you got that resultant vector?

And BTW, it doesn't matter for the answer, but for me the math is easier if you choose the charge at the origin for the test charge, and put the other charges along the 3 axes (plus the one at the far corner of the cube at L,L,L).

 wonderswan Nov19-10 04:10 PM

Re: Coulomb Force Point Charges on Cube

Quote:
 Quote by berkeman (Post 2994055) Can you show how you got that resultant vector? And BTW, it doesn't matter for the answer, but for me the math is easier if you choose the charge at the origin for the test charge, and put the other charges along the 3 axes (plus the one at the far corner of the cube at L,L,L).
I have also put the charges along the axis as you have described. So the vector caused by each charge:

$$(1,0,0) \ \ a_1 = -l\hat{i}$$
$$(0,1,0) \ \ a_2 = -l\hat{j}$$
$$(0,0,1) \ \ a_3 = -l\hat{k}$$
$$(1,1,0) \ \ a_4 = -l\hat{i}-l\hat{j}$$
$$(0,1,1) \ \ a_5 = -l\hat{j}-l\hat{k}$$
$$(1,0,1) \ \ a_6 = -l\hat{i}-l\hat{k}$$
$$(1,1,1) \ \ a_7 = -l\hat{i}-l\hat{k}-l\hat{j}$$

$$a_{res} = -4l\hat{i}-4\hat{j}-4\hat{k}$$

 berkeman Nov19-10 04:40 PM

Re: Coulomb Force Point Charges on Cube

Quote:
 Quote by wonderswan (Post 2994277) I have also put the charges along the axis as you have described. So the vector caused by each charge: $$(1,0,0) \ \ a_1 = -l\hat{i}$$ $$(0,1,0) \ \ a_2 = -l\hat{j}$$ $$(0,0,1) \ \ a_3 = -l\hat{k}$$ $$(1,1,0) \ \ a_4 = -l\hat{i}-l\hat{j}$$ $$(0,1,1) \ \ a_5 = -l\hat{j}-l\hat{k}$$ $$(1,0,1) \ \ a_6 = -l\hat{i}-l\hat{k}$$ $$(1,1,1) \ \ a_7 = -l\hat{i}-l\hat{k}-l\hat{j}$$ $$a_{res} = -4l\hat{i}-4\hat{j}-4\hat{k}$$
I don't think that's how you should sum things up. For each charge, you need to find the force vector components on the test charge at the origin. You have directions in the above, but not the magnitudes of the forces, which vary inversely with distance^2...

 wonderswan Nov20-10 10:54 AM

Re: Coulomb Force Point Charges on Cube

Is this the sum of the individual force components (the vectors are in brackets)?

$$\frac{kQ^2}{l^2}\left[\frac{-l\hat{i}}{l}\right]+\frac{kQ^2}{l^2}\left[\frac{-l\hat{j}}{l}\right]+\frac{kQ^2}{l^2}\left[\frac{-l\hat{k}}{l}\right]$$

$$+\frac{kQ^2}{2l^2}\left[\frac{-l\hat{i}-l\hat{j}}{\sqrt{2}l}\right]+\frac{kQ^2}{2l^2}\left[\frac{-l\hat{i}-l\hat{k}}{\sqrt{2}l}\right]+\frac{kQ^2}{2l^2}\left[\frac{-l\hat{i}-l\hat{k}}{\sqrt{2}l}\right]$$

$$+\frac{kQ^2}{3l^2}\left[\frac{-l\hat{i}-l\hat{j}-l\hat{k}}{\sqrt{3}l}\right]$$

 berkeman Nov20-10 04:51 PM

Re: Coulomb Force Point Charges on Cube

Quote:
 Quote by wonderswan (Post 2995437) Is this the sum of the individual force components (the vectors are in brackets)? $$\frac{kQ^2}{l^2}\left[\frac{-l\hat{i}}{l}\right]+\frac{kQ^2}{l^2}\left[\frac{-l\hat{j}}{l}\right]+\frac{kQ^2}{l^2}\left[\frac{-l\hat{k}}{l}\right]$$ $$+\frac{kQ^2}{2l^2}\left[\frac{-l\hat{i}-l\hat{j}}{\sqrt{2}l}\right]+\frac{kQ^2}{2l^2}\left[\frac{-l\hat{i}-l\hat{k}}{\sqrt{2}l}\right]+\frac{kQ^2}{2l^2}\left[\frac{-l\hat{i}-l\hat{k}}{\sqrt{2}l}\right]$$ $$+\frac{kQ^2}{3l^2}\left[\frac{-l\hat{i}-l\hat{j}-l\hat{k}}{\sqrt{3}l}\right]$$
I think that looks right. What do you get for the final magnitude of the force sum?

 wonderswan Nov20-10 07:43 PM

Re: Coulomb Force Point Charges on Cube

Okay here's the next step.

$$\frac{kQ^2}{l^2}\left[\frac{-l\hat{i}-l\hat{j}-l\hat{k}}{l}\right]$$

$$+\frac{kQ^2}{2l^2}\left[\frac{-2l\hat{i}-2l\hat{j}-2l\hat{k}}{\sqrt{2}l}\right]$$

$$+\frac{kQ^2}{3l^2}\left[\frac{-l\hat{i}-l\hat{j}-l\hat{k}}{\sqrt{3}l}\right]$$

I don't know how to add them to find the resultant magnitude?

 MindZiper Mar12-11 11:42 PM

Re: Coulomb Force Point Charges on Cube

Quote:
 Quote by wonderswan (Post 2995437) Is this the sum of the individual force components (the vectors are in brackets)? $$\frac{kQ^2}{l^2}\left[\frac{-l\hat{i}}{l}\right]+\frac{kQ^2}{l^2}\left[\frac{-l\hat{j}}{l}\right]+\frac{kQ^2}{l^2}\left[\frac{-l\hat{k}}{l}\right]$$ $$+\frac{kQ^2}{2l^2}\left[\frac{-l\hat{i}-l\hat{j}}{\sqrt{2}l}\right]+\frac{kQ^2}{2l^2}\left[\frac{-l\hat{i}-l\hat{k}}{\sqrt{2}l}\right]+\frac{kQ^2}{2l^2}\left[\frac{-l\hat{i}-l\hat{k}}{\sqrt{2}l}\right]$$ $$+\frac{kQ^2}{3l^2}\left[\frac{-l\hat{i}-l\hat{j}-l\hat{k}}{\sqrt{3}l}\right]$$
I just joined this forum to ask this:

Why are you dividing the vectors components by it's magnitude?

 MindZiper Mar13-11 12:10 AM

Re: Coulomb Force Point Charges on Cube

Quote:
 Quote by MindZiper (Post 3185001) I just joined this forum to ask this: Why are you dividing the vectors components by it's magnitude?
NVM. I got it. But I'm stuck at adding the components, just like wonderswan.

 wonderswan Mar13-11 06:32 PM

Re: Coulomb Force Point Charges on Cube

Hello MindZiper,

This post was made months ago so I thank you for reminding me of it.

I think I can provide the correct answer - in fact I've noticed some mistakes I made previously so to clarify I will write it step by step. Firstly something important:

We must find the magnitude of the resultant force vector. This is very important, the magnitude is:
$$\Vert{a}\Vert = \sqrt{x^2+y^2+z^2}$$

To find the vector $$a$$, we must add the individual force components. Here are the individual components:

$$F_1 = \frac {kQ^2} {l^2} [-\hat{i}]$$
$$F_2 = \frac {kQ^2} {l^2} [-\hat{j}]$$
$$F_3 = \frac {kQ^2} {l^2} [-\hat{k}]$$
$$F_4 = \frac 1 {2\sqrt{2}} \left(\frac{kQ^2}{l^2}\right)[-\hat{i}-\hat{j}]$$
$$F_5 = \frac 1 {2\sqrt{2}} \left(\frac{kQ^2}{l^2}\right)[-\hat{j}-\hat{k}]$$
$$F_6 = \frac 1 {2\sqrt{2}} \left(\frac{kQ^2}{l^2}\right)[-\hat{i}-\hat{k}]$$
$$F_7 = \frac 1 {3\sqrt{3}} \left(\frac{kQ^2}{l^2}\right)[-\hat{i}-\hat{j}-\hat{k}]$$

And the total gives us:

$$F_{RES} = \left(\frac{kQ^2}{l^2}\right)\left(1+\frac 1 {\sqrt{2}} + \frac 1 {3\sqrt{3}}\right)[-\hat{i}-\hat{j}-\hat{k}]$$

The magnitude of this vector, from above, is:

$$\Vert{a}\Vert = \sqrt{x^2+y^2+z^2}$$

where the individual components x, y and z (or of i, j and k) are the same.

$$\Vert{a}\Vert = \sqrt{3x^2}$$

So

$$\Vert{a}\Vert = \sqrt{3\left(1+\frac 1 {\sqrt{2}} + \frac 1 {3\sqrt{3}}\right)^2} \left(\frac {kQ^2} {l^2}\right)$$
$$\Vert{a}\Vert \simeq \frac {3.29Q^2} {4\pi\epsilon_0 l^2}$$

Hope this helps.

 MindZiper Mar13-11 08:40 PM

Re: Coulomb Force Point Charges on Cube

wonderswan,

I figured how to do it on my own. It was tedious, but I got it. I was so happy that I forgot to post it on here lol

Thanks for the help! :biggrin:

Edit: I figured how to do the whole vector stuff from this: http://www.physicsforums.com/showpos...5&postcount=11

So I simply added everything together, then multiplied the scalars by the vectors to get 3 vectors and then add the 3 together. Then square them, add them and square root them. And BAM.

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