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Each of magnitude q are placed at the four vertices of a regular hexagon of side a. The electric field at the centre of the:

(a) Hexagon is zero.

(b) Hexagon is non-zero and points outward.

(c) Hexagon is non-zero and points inward.

(d) Hexagon is non-zero and points tangentially.

User Buzkall
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Final answer:

The net electric field at the center of a regular hexagon with identical charges at each vertex is zero due to symmetry and the principle of superposition, where the fields from opposite charges cancel each other out.

Step-by-step explanation:

The electric field in the center of a regular hexagon with identical charges placed at each vertex can be understood by invoking the principles of symmetry and superposition. In a regular hexagon where each vertex has a point charge of magnitude q, and the side of the hexagon is 'a,' the problem is specific to electric fields and electrostatics.

Given the symmetry of the hexagon, we can say that each pair of opposite charges will generate electric fields at the center that are equal in magnitude but opposite in direction. As a result, all these contributions will cancel each other out. Therefore, the net electric field at the center of the hexagon due to these charges is zero; it doesn't point tangentially or in any other direction because the vector sum of all the fields from the individual charges is null.

This principle of vector superposition of electric fields and the use of symmetry applies to any regular polygon with an even number of sides and uniformly distributed like charges at its vertices. Hence, no matter how many charges you place at the corners of a square, hexagon, or any other polygon with these properties, the net electric field at the center will surely be zero.

User Kyle Dumovic
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