- #1

- 6

- 0

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.