A cube of edge length 'a' has a point charge +Q at every vertex except one vertex (taken at the origin) where the charge is -Q instead. What is the electric field at the centre of the cube?

Question

Accepted Answer

-Q/(3*sqrt(3)*pi*e0*a²) (x_hat + y_hat + z_hat). By symmetry, 8 equal +Q at all vertices give zero field at the centre. The actual configuration = (all +Q) + (an extra -2Q at the origin vertex). Only the -2Q matters. Distance from a vertex to centre is (sqrt(3)/2)a. Field magnitude from -2Q: E = k*2Q/((sqrt(3)/2 a)²) = k*2Q/(3a²/4) = 8kQ/(3a²), directed from centre toward the origin vertex (since charge is negative, field points toward it). Writing along the unit vector and using k = 1/(4*pi*e0), the field works out to -Q/(3*sqrt(3)*pi*e0*a²)(x_hat+y_hat+z_hat).

A cube of edge length 'a' has a point charge +Q at every vertex except one vertex (taken at the origin) where the charge is -Q instead. What is the electric field at the centre of the cube?

Solution

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