Five charges, ' q ' each are placed at the comers of a regular pentagon of side ' a ' as shown in figure. First, charge from ' A ' is removed with other charges intact, then charge at ' A ' is replaced with an equal opposite charge. The ratio of magnitudes of electric fields at O , without charge at A and that with equal and opposite charge at A is

Step-by-step solution

Let the electric field due to a charge q at a corner of the regular pentagon at the center O be E . The distance from each corner to the center O is r . By symmetry, the sum of the electric fields due to all five charges q at the corners A, B, C, D, E is zero, i.e., E _A + E _B + E _C + E _D + E _E = 0 . When the charge at A is removed, the net electric field at O is E _ net, 1 = E _B + E _C + E _D + E _E . From the symmetry condition, E _ net, 1 = - E _A . The magnitude is | E _ net, 1 | = E , where E = kq r^2 . W