In a regular polygon of n sides, each corner is at a distance r from the centre. Identical charges of magnitude Q are placed at (n-1) corners. The filed at the centre is :

Problem : In a regular polygon of n sides, each corner is at a distance r from the centre. Identical charges of magnitude Q are placed at (n-1) corners. The filed at the centre is :
a) kQ/r² ✅ 
b) (n-1)kQ/r²
c) nkQ/(n-1)r²
d) (n-1)kQ/nr²

Doubt by Rimjhim

Solution : A regular Polygon has even number of vertex. So we can say that n is even and (n-1) is odd. So by considering the concept of symmetry we can say that Electric field due to the charge Q at one of the vertex of the regular polygon will be cancelled by the the another charge Q on the opposite vertex of the regular polygon [because Electric Field is Vector quantity]. There would be possible pair of charges except one charge. So the net magnetic field due to this single charge Q would be given by, E = kQ/r²

∴ Total Magnetic Field at the centre will be a) KQ/r²

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Similar Question : In a regular polygon of n sides, each corner is at a distance r from the centre. Identical charges of magnitude Q are placed at (n-1) corners. At the centre intensity is E and the potential is V. The ratio V/E has magnitude
a) rn
b) r(n-1) ✅
c) r
d) r(n-1)/n

Ans : We know 
E = KQ/r²

Potential is scalar. 
Hence, total potential at the centre will be equal to the sum of the individual potential. 
∴ V = (n-1)kQ/r

Now, V/E = [(n-1)kQ/r]/[KQ/r^2]
⇒ b) V/E = r(n-1)