Electric Field and Potential of Hexagonal Charge Configuration

To determine the correct statements, let us set up a Cartesian coordinate system with the origin at the center $O$ of the regular hexagon. Let the line $AD$ lie along the $x$-axis.

The distance from the center $O$ to any vertex of a regular hexagon with side length $L$ is equal to $L$.

The coordinates and charges at the six vertices are:

• $A: (-L, 0)$ with charge $q_A = +2q$
• $B: \left(-\frac{L}{2}, -\frac{\sqrt{3}L}{2}\right)$ with charge $q_B = +q$
• $C: \left(\frac{L}{2}, -\frac{\sqrt{3}L}{2}\right)$ with charge $q_C = -q$
• $D: (L, 0)$ with charge $q_D = -2q$
• $E: \left(\frac{L}{2}, \frac{\sqrt{3}L}{2}\right)$ with charge $q_E = -q$
• $F: \left(-\frac{L}{2}, \frac{\sqrt{3}L}{2}\right)$ with charge $q_F = +q$

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1. Evaluation of Statement (A): Electric Field at $O$

The electric field at $O(0,0)$ due to a charge $q_i$ located at $\vec{r}_i$ is given by: $\vec{E}_i = \frac{1}{4\pi\varepsilon_0} \frac{q_i}{r_i^2} (-\hat{r}_i)$

Given $K = \frac{1}{4\pi\varepsilon_0}\frac{q}{L^2}$:

• Field due to pair $A$ and $D$:

  • Charge $+2q$ at $A$ creates a field at $O$ directed along $+\hat{i}$ with magnitude $\frac{1}{4\pi\varepsilon_0}\frac{2q}{L^2} = 2K$.
  • Charge $-2q$ at $D$ creates a field at $O$ directed along $+\hat{i}$ with magnitude $\frac{1}{4\pi\varepsilon_0}\frac{2q}{L^2} = 2K$. $\vec{E}_{AD} = (2K + 2K)\hat{i} = 4K\hat{i}$

• Field due to pair $B$ and $E$:

  • Charge $+q$ at $B$ creates a field at $O$ directed towards $E$ (at an angle of $+60^\circ$ with the $+x$-axis) with magnitude $K$.
  • Charge $-q$ at $E$ creates a field at $O$ directed towards $E$ with magnitude $K$. $\vec{E}_{BE} = 2K \left(\cos 60^\circ \hat{i} + \sin 60^\circ \hat{j}\right) = 2K \left(\frac{1}{2}\hat{i} + \frac{\sqrt{3}}{2}\hat{j}\right) = K\hat{i} + \sqrt{3}K\hat{j}$

• Field due to pair $F$ and $C$:

  • Charge $+q$ at $F$ creates a field at $O$ directed towards $C$ (at an angle of $-60^\circ$ with the $+x$-axis) with magnitude $K$.
  • Charge $-q$ at $C$ creates a field at $O$ directed towards $C$ with magnitude $K$. $\vec{E}_{FC} = 2K \left(\cos(-60^\circ) \hat{i} + \sin(-60^\circ) \hat{j}\right) = 2K \left(\frac{1}{2}\hat{i} - \frac{\sqrt{3}}{2}\hat{j}\right) = K\hat{i} - \sqrt{3}K\hat{j}$

• Net Electric Field at $O$: $\vec{E}_O = \vec{E}_{AD} + \vec{E}_{BE} + \vec{E}_{FC} = 4K\hat{i} + \left(K\hat{i} + \sqrt{3}K\hat{j}\right) + \left(K\hat{i} - \sqrt{3}K\hat{j}\right) = 6K\hat{i}$

Since $\hat{i}$ is along the direction $OD$, the electric field at $O$ is $6K$ along $OD$.
Statement (A) is correct.

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2. Evaluation of Statement (B): Potential at $O$

The electrostatic potential at the center $O$ is the scalar sum of the potentials due to each point charge: $V_O = \frac{1}{4\pi\varepsilon_0 L} \sum_{i} q_i = \frac{1}{4\pi\varepsilon_0 L} \Big[(+2q) + (+q) + (-q) + (-2q) + (-q) + (+q)\Big] = 0$

Statement (B) is correct.

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3. Evaluation of Statement (C): Potential on Line $PR$

The line $PR$ is perpendicular to $AD$ and passes through $O$, coinciding with the $y$-axis ($x = 0$).
For any point $(0, y)$ on the line $PR$:

1. Contribution from $A$ and $D$:

   • Distance to $A(-L,0)$ is $r_A = \sqrt{L^2 + y^2}$.
   • Distance to $D(L,0)$ is $r_D = \sqrt{L^2 + y^2}$. $V_{AD}(0, y) = \frac{1}{4\pi\varepsilon_0} \left(\frac{2q}{\sqrt{L^2+y^2}} - \frac{2q}{\sqrt{L^2+y^2}}\right) = 0$

2. Contribution from $F$ and $E$:

   • Distance to $F\left(-\frac{L}{2}, \frac{\sqrt{3}L}{2}\right)$ is $r_F = \sqrt{\left(\frac{L}{2}\right)^2 + \left(y - \frac{\sqrt{3}L}{2}\right)^2}$.
   • Distance to $E\left(\frac{L}{2}, \frac{\sqrt{3}L}{2}\right)$ is $r_E = \sqrt{\left(\frac{L}{2}\right)^2 + \left(y - \frac{\sqrt{3}L}{2}\right)^2} = r_F$. $V_{FE}(0, y) = \frac{1}{4\pi\varepsilon_0} \left(\frac{q}{r_F} - \frac{q}{r_E}\right) = 0$

3. Contribution from $B$ and $C$:

   • Distance to $B\left(-\frac{L}{2}, -\frac{\sqrt{3}L}{2}\right)$ is $r_B = \sqrt{\left(\frac{L}{2}\right)^2 + \left(y + \frac{\sqrt{3}L}{2}\right)^2}$.
   • Distance to $C\left(\frac{L}{2}, -\frac{\sqrt{3}L}{2}\right)$ is $r_C = \sqrt{\left(\frac{L}{2}\right)^2 + \left(y + \frac{\sqrt{3}L}{2}\right)^2} = r_B$. $V_{BC}(0, y) = \frac{1}{4\pi\varepsilon_0} \left(\frac{q}{r_B} - \frac{q}{r_C}\right) = 0$

Summing all contributions, the potential at any point on line $PR$ is: $V(0, y) = V_{AD} + V_{FE} + V_{BC} = 0$

Since $V(0, y) = 0$ everywhere on $PR$, the potential remains constant (zero) across all points on line $PR$.
Statement (C) is correct.

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4. Evaluation of Statement (D): Potential on Line $ST$

The line $ST$ lies along the $x$-axis ($y = 0$). For a point $(x, 0)$ on $ST$: $V_{AD}(x, 0) = \frac{1}{4\pi\varepsilon_0} \left(\frac{2q}{|x + L|} - \frac{2q}{|x - L|}\right) \neq 0$

Since $V_{AD}(x, 0)$ varies with $x$, the electrostatic potential is not identical across all points on $ST$.
Statement (D) is incorrect.

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Conclusion

The correct statements are A, B, and C.