Physics - Electrostatics Question with Solution | TestHub

Let the side length of the regular hexagon be $r$. The distance from each vertex to the centre $O$ is $r$.

The electric field at the centre $O$ due to a point charge $q$ at a distance $r$ has a magnitude $E=\frac{1}{4\pi {ϵ}_0}\frac{q}{r^2}$.

In the given regular hexagon, the diametrically opposite pairs of vertices are $(A,D)$, $(C,F)$, and $(B,E)$.

1. For the pair $(A,D)$:
The charge at $A$ is $+q$ and the charge at $D$ is $+q$. Since both charges are equal and of the same sign, their electric fields at the centre $O$ are equal in magnitude and opposite in direction. Thus, they cancel each other out: ${\vec{E}}_A+{\vec{E}}_D=0$.

2. For the pair $(C,F)$:
The charge at $C$ is $-q$ and the charge at $F$ is $-q$. Similarly, their electric fields at the centre $O$ are equal in magnitude and opposite in direction, so they also cancel each other out: ${\vec{E}}_C+{\vec{E}}_F=0$.

3. Total electric field due to the five charges ($A,B,C,D,F$):
The net electric field at $O$ due to these five charges is simply the electric field due to the charge at $B$ alone, because the contributions from $A,D,C,$ and $F$ sum to zero.
${\vec{E}}_5={\vec{E}}_A+{\vec{E}}_B+{\vec{E}}_C+{\vec{E}}_D+{\vec{E}}_F={\vec{E}}_B$
Since the charge at $B$ is $-q$, the electric field ${\vec{E}}_B$ is directed towards $B$ with magnitude $E_5=\frac{1}{4\pi {ϵ}_0}\frac{q}{r^2}$.

4. Electric field due to charge at $E$ alone:
The charge at $E$ is $+Q$. Its electric field ${\vec{E}}_E$ at $O$ is directed away from $E$. Since $E$ is diametrically opposite to $B$, the direction away from $E$ is towards $B$.
The magnitude is $E_E=\frac{1}{4\pi {ϵ}_0}\frac{Q}{r^2}$.

According to the given condition, the electric field due to the five charges is thrice the electric field due to the charge at $E$ alone:
$E_5=3E_E$

Substituting the magnitudes:
$\frac{1}{4\pi {ϵ}_0}\frac{q}{r^2}=3\left(\frac{1}{4\pi {ϵ}_0}\frac{Q}{r^2}\right)$

$q=3Q⟹Q=\frac{+q}{3}$

Answer: $\frac{+q}{3}$