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(D) Let the charges at corners $A, B, C, D, E, F$ be $q_A = -q, q_B = 3q, q_C = -2q, q_D = 2q, q_E = q, q_F = -2q$.At the center $O$,the electric fiel

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None of these

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(D) Let the charges at corners $A, B, C, D, E, F$ be $q_A = -q, q_B = 3q, q_C = -2q, q_D = 2q, q_E = q, q_F = -2q$.At the center $O$,the electric field $\vec{E}$ due to a charge $q_i$ at a distance $r$ is $\vec{E}_i = \frac{k q_i}{r^2} \hat{r}_i$.Since the electron has charge $-e$,the force on it is $\vec{F} = -e \vec{E}_{net}$.We can pair opposite charges: $1$. Pair $(A, D)$: $q_A = -q, q_D = 2q$. Net charge effect at $O$ is equivalent to a charge of $q$ at $D$.$2$. Pair $(B, E)$: $q_B = 3q, q_E = q$. Net charge effect at $O$ is equivalent to a charge of $2q$ at $B$.$3$. Pair $(C, F)$: $q_C = -2q, q_F = -2q$. Net charge effect at $O$ is $0$.Summing these,the resultant electric field at $O$ is directed such that the force on the electron is non-zero and does not align simply with $OF$ or $OC$. Thus,the correct answer is None of these.

Six charges are placed at the corners of a regular hexagon as shown. If an electron is placed at its centre $O$,the force on it will be:

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