Electrostatic – A regular hexagon has charges placed at all six vertices

Alternate vertices carry charges +q and −q.

Find the magnitude and direction of the electric field at the center of the hexagon.

Given: A regular hexagon has charges at all six vertices.

Alternate vertices carry +q and −q.

Distance from center to each vertex = a

Electric field due to one charge at center:

E_0=\frac{kq}{a^2}

Now consider one opposite pair (+q,−q).

The field due to +q is away from charge while field due to −q is towards charge.

Hence both fields act in the same direction.

E_{pair}=2E_0

E_{pair}=2\frac{kq}{a^2}

Three equal vectors are separated by 120°.

The resultant of three equal vectors separated by 120° is zero.

\therefore \vec{E}_{net}=0

\boxed{\vec{E}_{net}=0}