MHTCET 18 Apr 2026 Shift 1 Paper

Concept:In a regular hexagon, equal charges at opposite vertices produce equal and opposite electric fields at the centre, so they cancel each other.Explanation:Let the distance of each vertex from the centre $O$ be $r$.The field due to any charge $q$ at a vertex is $E = \frac{1}{4\pi\epsilon_0}\frac{q}{r^2}$.The opposite pairs are $(A,D)$, $(B,E)$, and $(C,F)$.Charge at $A$ is $+q$ and charge at $D$ is also $+q$, so they cancel at $O$.Charge at $C$ is $-q$ and charge at $F$ is also $-q$, so they also cancel at $O$.Thus, among the five charges at $A, B, C, D, F$, only the charge $-q$ at $B$ remains effective.So, the field due to these five charges is $E_1 = \frac{1}{4\pi\epsilon_0}\frac{q}{r^2}$.The field due to charge $+Q$ at $E$ alone is $E_2 = \frac{1}{4\pi\epsilon_0}\frac{Q}{r^2}$.Given $E_1 = 3E_2$:$$\frac{1}{4\pi\epsilon_0}\frac{q}{r^2} = 3\left(\frac{1}{4\pi\epsilon_0}\frac{Q}{r^2}\right)$$Cancelling common terms gives $q = 3Q$.Hence, $Q = \frac{q}{3}$.Answer:$Q = \frac{q}{3}$, which corresponds to option A.