Cube with charges at each vertex

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SUMMARY

The discussion centers on calculating the electric field (E) at the center of the upper face of a cube with charges at each vertex. The derived formula is $$E=4k\frac{q}{r^2}\sin\theta$$, where ##\theta=\arcsin\left(\frac{s}{r}\right)=\arcsin{\sqrt\frac{2}{3}}## and ##r=\sqrt{\frac{3}{2}}s##. The final expression for the electric field is confirmed as $$E=\frac{8\sqrt6}{9}k\frac{q}{s^2}$$, although a participant noted a potential omission of a factor of ##\sqrt{2}## in the final answer, which was later corrected to ##\sqrt6##.

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archaic
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Homework Statement
Determine an expression for the magnitude of the electric field at the center of any face of the cube.
Relevant Equations
$$E=k\frac{q}{r^2}$$
22-c-op-027-alt.png
For convenience, I take the center of the upper face.
The charges at the top cancel each other's effects, and those at the bottom cancel each other's horizontal effects, so I get$$E=4k\frac{q}{r^2}\sin\theta$$I have found that ##\theta=\arcsin\left(\frac{s}{r}\right)=\arcsin{\sqrt\frac{2}{3}}##, with ##r=\sqrt{\frac{3}{2}}s##:$$E=4\frac{2}{3}\sqrt\frac23k\frac{q}{s^2}=\frac{8\sqrt6}{9}k\frac{q}{s^2}$$
Correct, right?
 
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The method looks right, but I think you have dropped a factor of ##\sqrt{2}## in your very final answer.
 
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etotheipi said:
The method looks right, but I think you have dropped a factor of ##\sqrt{2}## in your very final answer.
should have been ##\sqrt6## instead of ##\sqrt3##. thanks, corrected!
 
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