Animated Solution for Physics - Electrostatics: Six charges are placed around a regular hexagon of side length a as shown in the figure. Five of them have charge q, and the remaining one has charge x. The perpendicular from each charge to the nearest hexagon side passes through the center O of the hexagon and is bisected by the side.

Let's first decode the geometry of this beautiful setup. We have a regular hexagon of side $a$. The problem states a very specific geometric condition: the perpendicular from each charge to the nearest hexagon side passes through the center $O$ and is bisected by that side.

What does this mean physically? It means the distance from the charge to the side is exactly equal to the distance from the center $O$ to the side. Since the distance from the center of a regular hexagon to any of its sides is $r=a\cos (30^{\circ })=\frac{\sqrt{3}}{2}a$, the total distance $d$ from any charge to the center $O$ is simply twice of that. Therefore, $d=2r=\sqrt{3}a$. Imagine this: all six charges are sitting perfectly on a large circle of radius $d$ centered at $O$.

The Power of Symmetry

Now, let's test Option A. What happens if we replace the unknown charge $x$ with $q$? Suddenly, we have perfect symmetry! We have six identical charges placed at equal distances around the center.

For every charge in this symmetric arrangement, there is an identical charge exactly opposite to it. Because electric field is a vector quantity, their electric fields at the center will be equal in magnitude but opposite in direction, perfectly cancelling each other out. So, the net electric field at $O$ is strictly zero. Option A is absolutely correct.

Breaking the Symmetry

Moving to Option B, let's make $x=-q$. Now the perfect symmetry is broken, but only slightly. The four charges on the sides (top-left, top-right, bottom-left, bottom-right) still have identical opposite partners, so their electric fields still cancel out completely.

We only need to worry about the top charge $+q$ and the bottom charge $-q$. The positive charge at the top pushes the electric field downwards, and the negative charge at the bottom pulls the electric field downwards. Both fields point in the exact same direction!

Let's calculate this net field. Since both fields point downwards, we just add their magnitudes. Each charge contributes $\frac{kq}{d^2}$. Adding them gives $\frac{2kq}{d^2}$. Now, remember that distance $d=\sqrt{3}a$ we found earlier? Let's substitute it here. Squaring that gives $3a^2$. So, the net field is $\frac{2kq}{3a^2}$.

To match the options, let's expand the Coulomb constant $k$ as $\frac{1}{4\pi {ϵ}_0}$. The $2$ in the numerator and the $4$ in the denominator will cancel out, leaving a $2$ in the denominator. Multiplying that $2$ with $3$ gives us $6$. So, the final magnitude is $\frac{q}{6\pi {ϵ}_0{a}^2}$. This perfectly matches Option B! So, B is correct.

The Simplicity of Scalars

Let's check Option C. Now we are dealing with electric potential. Remember, potential is a scalar quantity, which makes our lives much easier! No vectors, no directions. We just add the potentials algebraically. Since all charges are at the exact same distance $d$ from the center, the total potential is simply $k$ times the total charge divided by $d$.

For Option C, $x=2q$. So, the total charge is $5q+2q=7q$. Plugging this into our potential formula, we get $V=\frac{k(7q)}{\sqrt{3}a}$. Expanding $k$ again, we get $\frac{7q}{4\sqrt{3}\pi {ϵ}_{0}a}$. This matches Option C perfectly! So, C is also correct.

Finally, let's evaluate Option D. Here, $x=-3q$. The total charge becomes $5q-3q=2q$. The potential is $V=\frac{k(2q)}{\sqrt{3}a}$. Expanding $k$, the $2$ cancels with the $4$, leaving us with $\frac{q}{2\sqrt{3}\pi {ϵ}_{0}a}$. But Option D says it should be $\frac{3q}{4\sqrt{3}\pi {ϵ}_{0}a}$. So, Option D is incorrect.

By carefully analyzing the geometry and applying the principles of superposition for both vector fields and scalar potentials, we found that options A, B, and C are correct. Always remember, symmetry is your best friend in electrostatics!