# A wire is bent in the form of a regular hexagon and a total charge q is distributed uniformly on it. What is the electric field at the center? You many answer this part without making any numerical calculations.

Given:
Uniformly distributed Charge on the regular hexagon : q

When a wire is bent in the form of a regular hexagon having charge q uniformly distributed, each point on the hexagon will contribute same magnitude of electric field.
Say, 6 vertices of the hexagon have same charge q.
Thus they will produce same electric field E at the center.
This electric field gets nullified as same magnitude is acting from both sides.
Formula used:
Electric field at a point due to a point charge is given as:
Where k is a constant and k= =9× 109 Nm2C-2. q is the charge and r is the distance between the point and the charge.
Mathematically:
E1=E2
From the figure E1 is the electric field due to vertex 1 at the center and E2 is the electric field due to vertex 2 at the center diametrically opposite to vertex 1.
Thus net electric field at center due to 1 and 2 is
Enet = E1-E2

Enet = 0

Eventually electric field due to all 6 vertices cancel out each other at the center.
Hence, electric field at the center of the regular hexagon is zero.

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