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: [tex]3.29Q^2/4\pi\epsilon_0l^2[/tex] 2. Relevant equations [tex]F = \frac{Q_1Q_2}{4\pi\epsilon_0R^2}\cdot\hat{a}\ \ (N)[/tex] 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). [tex]F = Q^2/4\pi\epsilon_0\cdot\displaystyle\sum_{n=1}^{7}\frac{1}{R^2_n}[/tex] [tex]R_1 = \sqrt{(l)^2} \ \ \small(1,0,0)[/tex] [tex]R_2 = \sqrt{(l)^2} \ \ \small(0,1,0)[/tex] [tex]R_3 = \sqrt{(l)^2} \ \ \small(1,0,0)[/tex] [tex]R_4 = \sqrt{(l)^2+(l)^2} = \sqrt{2l^2} \ \ \small(1,1,0)[/tex] [tex]R_5 = \sqrt{(l)^2+(l)^2} = \sqrt{2l^2} \ \ \small(0,1,1)[/tex] [tex]R_6 = \sqrt{(l)^2+(l)^2} = \sqrt{2l^2} \ \ \small(1,0,1)[/tex] [tex]R_7 = \sqrt{(l)^2+(l)^2 +(l)^2} = \sqrt{3l^2} \ \ \small(1,1,1)[/tex] [tex]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][/tex] [tex]= 4.8\dot{3}Q^2/4\pi\epsilon_0l^2[/tex] Any help appreciated. 
Re: Coulomb Force Point Charges on Cube
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This is the problem: Quote:

Re: Coulomb Force Point Charges on Cube
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[tex]F = \frac{Q^2}{4\pi\epsilon_0R^2}\cdot\hat{a}[/tex] Is R the magnitude of the resultant vector  it would be [tex]\sqrt{48l^2}[/tex]??? I think I'm getting confused. 
Re: Coulomb Force Point Charges on Cube
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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). 
Re: Coulomb Force Point Charges on Cube
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[tex](1,0,0) \ \ a_1 = l\hat{i}[/tex] [tex](0,1,0) \ \ a_2 = l\hat{j}[/tex] [tex](0,0,1) \ \ a_3 = l\hat{k}[/tex] [tex](1,1,0) \ \ a_4 = l\hat{i}l\hat{j}[/tex] [tex](0,1,1) \ \ a_5 = l\hat{j}l\hat{k}[/tex] [tex](1,0,1) \ \ a_6 = l\hat{i}l\hat{k}[/tex] [tex](1,1,1) \ \ a_7 = l\hat{i}l\hat{k}l\hat{j}[/tex] [tex]a_{res} = 4l\hat{i}4\hat{j}4\hat{k}[/tex] 
Re: Coulomb Force Point Charges on Cube
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Re: Coulomb Force Point Charges on Cube
Is this the sum of the individual force components (the vectors are in brackets)?
[tex]\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][/tex] [tex]+\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][/tex] [tex]+\frac{kQ^2}{3l^2}\left[\frac{l\hat{i}l\hat{j}l\hat{k}}{\sqrt{3}l}\right][/tex] 
Re: Coulomb Force Point Charges on Cube
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Re: Coulomb Force Point Charges on Cube
Okay here's the next step.
[tex]\frac{kQ^2}{l^2}\left[\frac{l\hat{i}l\hat{j}l\hat{k}}{l}\right][/tex] [tex]+\frac{kQ^2}{2l^2}\left[\frac{2l\hat{i}2l\hat{j}2l\hat{k}}{\sqrt{2}l}\right][/tex] [tex]+\frac{kQ^2}{3l^2}\left[\frac{l\hat{i}l\hat{j}l\hat{k}}{\sqrt{3}l}\right][/tex] I don't know how to add them to find the resultant magnitude? 
Re: Coulomb Force Point Charges on Cube
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Why are you dividing the vectors components by it's magnitude? 
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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: [tex]\Vert{a}\Vert = \sqrt{x^2+y^2+z^2}[/tex] To find the vector [tex]a[/tex], we must add the individual force components. Here are the individual components: [tex]F_1 = \frac {kQ^2} {l^2} [\hat{i}][/tex] [tex]F_2 = \frac {kQ^2} {l^2} [\hat{j}][/tex] [tex]F_3 = \frac {kQ^2} {l^2} [\hat{k}][/tex] [tex]F_4 = \frac 1 {2\sqrt{2}} \left(\frac{kQ^2}{l^2}\right)[\hat{i}\hat{j}][/tex] [tex]F_5 = \frac 1 {2\sqrt{2}} \left(\frac{kQ^2}{l^2}\right)[\hat{j}\hat{k}][/tex] [tex]F_6 = \frac 1 {2\sqrt{2}} \left(\frac{kQ^2}{l^2}\right)[\hat{i}\hat{k}][/tex] [tex]F_7 = \frac 1 {3\sqrt{3}} \left(\frac{kQ^2}{l^2}\right)[\hat{i}\hat{j}\hat{k}][/tex] And the total gives us: [tex]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}][/tex] The magnitude of this vector, from above, is: [tex]\Vert{a}\Vert = \sqrt{x^2+y^2+z^2}[/tex] where the individual components x, y and z (or of i, j and k) are the same. [tex]\Vert{a}\Vert = \sqrt{3x^2}[/tex] So [tex]\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)[/tex] [tex]\Vert{a}\Vert \simeq \frac {3.29Q^2} {4\pi\epsilon_0 l^2}[/tex] Hope this helps. 
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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