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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 center of the cube. |
Answer» We know that the length of diagonal of the cube side b is `sqrt(3b^(2)) = b sqrt(3)` `:.` Distance between center of cube and each vertex, `r = (b sqrt(3))/(2)` AS `V = (1)/(4pi in_(0)) (q)/(r )` and 8 charges each of value q are present at the eight verticies of the cube, therefore, `V = (1)/(4pi in_(0)) (8q)/(b sqrt(3)//2) or V = (4q)/(sqrt(3) pi in_(0) b)` Further electric field intensity at the center due to all the eight charges is zero, because the fileds due to individual charges cancel in pairs. |
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