A cube of side b has a charge q at each of its vertices. Determine the potential and electric field due to this charge array at the centre of the cube.

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Answer 1

Diagonal of the cube face, $d = (b^{2} + b^{2}) = b 2$
Diagonal of cube $l = (d^{2} + b^{2}) = b 3$
Distance between center of cube and vertices, $r = l / 2 = b 3 / 2$
Potential, $v = 8 q / 4 π ϵ_{o} r = 4 q / 3 π ϵ_{o} b$ (8 due to presence of 8 charges at the vertices)
by symmetry of charge electric field, E=0.

Answer 2

Diagonal of the cube face,

d = (b^{2} + b^{2}) = b 2

Diagonal of cube

l = (d^{2} + b^{2}) = b 3

Distance between center of cube and vertices,

r = l / 2 = b 3 / 2

Potential,

v = 8 q / 4 π ϵ_{o} r = 4 q / 3 π ϵ_{o} b

(8 due to presence of 8 charges at the vertices)
by symmetry of charge electric field, E=0.

A cube of side b has a charge q at each of its vertices. Determine the potential and electric field due to this charge array at the centre of the cube.

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A cube of side b has a charge q at each of its vertices. Determine the potential and electric field due to this charge array at the centre of the cube.

Diagonal of the cube face, d=(​b2+b2)=b2​ Diagonal of cube l=(​d2+b2)=b3​ Distance between center of cube and vertices, r=l/2=b3​/2 Potential, v=8q/4πϵo​r=4q/3​πϵo​b (8 due to presence of 8 charges at the vertices) by symmetry of charge electric field, E=0.

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