Q. 434.0( 3 Votes )

# 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.

Answer :

**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× 10

^{9}Nm

^{2}C

^{-2}. q is the charge and r is the distance between the point and the charge.

Mathematically:

E

_{1}=E

_{2}

From the figure E

_{1}is the electric field due to vertex 1 at the center and E

_{2}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

E

_{net}= E

_{1}-E

_{2}

∴ E

_{net}= 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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