Question

Five equal point charges, each +q, are placed at five of the six vertices of a regular hexagon of side length a. What is the magnitude of the electric field at the center of the hexagon?

• A. kq/a\(^2\)
• B. 2kq/a\(^2\)
• C. 0
• D. kq/2a\(^2\)

Hint:

For symmetric arrangements of charges, if one charge is missing, the net field at the center is equal in magnitude and opposite in direction to the field that would have been produced by that missing charge alone. This shortcut significantly simplifies calculations.

Answer 1

The Correct Option is A

Step 1: In a regular hexagon, every vertex has a diametrically opposite vertex at distance \( 2a \), and two equal charges sitting at opposite vertices produce fields at the centre that are equal in magnitude and opposite in direction, so they cancel each other out.

Step 2: With six vertices, there are three such opposite pairs. Since only one vertex out of six is left empty, two of the three pairs are still complete (both charges present) and fully cancel; only the pair containing the empty vertex is incomplete.

Step 3: In that incomplete pair, the single charge present has no partner to cancel its field, so its full field of magnitude \( \frac{kq}{a^2} \) survives as the net field at the centre.
\[ \boxed{E = \dfrac{kq}{a^2}} \]