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Five equal positive charges are fixed at five of the six vertices of a regular hexagon, and the net electric field at the centre is E1. A negative charge of the same magnitude is then placed at the sixth (vacant) vertex, after which the net field at the centre becomes E2. Find the ratio E2/E1.

1. 2
2. 1
3. 3
4. None of these

Correct answer: 2

Solution

For six identical charges at the hexagon vertices the fields cancel by symmetry. With one vertex empty, the net field of the five positive charges equals the magnitude of the field a single positive charge would produce at the centre (pointing as if that one charge were absent), so E1 = kq/a² (a = circumradius). Placing a negative charge of equal magnitude at the sixth vertex adds another kq/a² in the same direction (toward that vertex), so E2 = 2kq/a². Thus E2/E1 = 2.

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