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NEET PHYSICSMedium

Six charges, three positive and three negative of equal magnitude are to be placed at the vertices of a regular hexagon (P, Q, R, S, T, U) such that the electric field at the center O is double the electric field when only one positive charge of same magnitude is placed at R. Which of the following arrangements of charges is possible for P, Q, R, S, T and U respectively?

A

+, −, +, −, −, +

B

+, −, +, −, +, −

C

+, +, −, +, −, −

D

−, +, +, −, +, −

Step-by-Step Solution

Let the magnitude of the electric field due to a single positive charge qq at a vertex be EE. The direction of the field due to a positive charge at R is directed away from R, towards the opposite vertex U (vector E⃗R→U\vec{E}_{R \to U}). The problem requires the net field at O to be double this value, i.e., 2E2E directed towards U.

Analyzing Option (D) [P(-), Q(+), R(+), S(-), T(+), U(-)]:

  1. Pair (P, S): Charges are (-q, -q). The field from P (towards P) cancels the field from S (towards S). Net = 0.
  2. Pair (Q, T): Charges are (+q, +q). The field from Q (away from Q) cancels the field from T (away from T). Net = 0.
  3. Pair (R, U): Charges are (+q, -q). The field from R (+q) points towards U. The field from U (-q) also points towards U (attractive). The fields add up: E+E=2EE + E = 2E directed towards U.

This arrangement results in a net field of 2E2E in the direction of U, satisfying the condition.

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