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Eight charges, each of magnitude q, are placed at the vertices of a cube placed in vacuum. Electric potential at the centre of the cube due to this system of charges is ______. Here ε₀ is the permittivity of vacuum and a is the length of each side of the cube.

Asked in GUJCET 2008 · Potential of point charges

• (1) zero
• (2) (√3 q)/(πε₀ a)
• (3) (2q)/(πε₀ a)
• (4) (4q)/(√3 πε₀ a)

Step-by-step solution

Why the other options are wrong

• (1) Potential is a scalar, so the eight contributions add as numbers. Eight equal charges of the same sign each give a potential of the same sign at the centre, so they reinforce rather than cancel. What does cancel at the centre, by symmetry, is the electric field, which is a vector.
• (2) This adds up only six contributions, the number of *faces* of a cube rather than its eight *vertices*: with r = (√3 a)/2, 6×q/(4πε₀ r) = (√3 q)/(πε₀ a), three quarters of the right answer.
• (3) This takes each vertex to be a distance a from the centre, the side of the cube, instead of half the body diagonal: 8×q/(4πε₀ a) = (2q)/(πε₀ a). The centre is closer than a full side away, so the true potential is larger than this.

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