Q. 24.3( 24 Votes )
Let O be the center of the hexagon. It contains the charges at all its 6 vertices, each charge = + 5 μC = 5 × 10-6 C. The side of the hexagon is 10 cm = 0.1 m
It follows that the point O, when joined to the two ends of a side of the hexagon forms an equilateral triangle
Electric potential at O due to one charge,
Here, q = amount of charge = 5 × 10-6 C
r = distance between charge and O = 0.1 m
ε0 = absolute permittivity of free space
Since at the each of the hexagon, a charge of 5 μC (5 × 10-6 C) is placed, total electric potential at the point O due to the charges at the six corners,
⇒ V1 = 2.7 × 106 V
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