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wonderswan
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Hello, I'm having a little trouble with this:
Coulomb Force Point Charges on Cube
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]
[tex]F = \frac{Q_1Q_2}{4\pi\epsilon_0R^2}\cdot\hat{a}\ \ (N)[/tex]
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.
Coulomb Force Point Charges on Cube
Homework Statement
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]
Homework Equations
[tex]F = \frac{Q_1Q_2}{4\pi\epsilon_0R^2}\cdot\hat{a}\ \ (N)[/tex]
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.