Five point charges, each of value +q coulomb, are placed on five vertices of a regular hexagon of si

Q: Why is the distance from the center to a vertex equal to the side length $L$?

A regular hexagon can be divided into six equilateral triangles by drawing lines from the center to each vertex. Since all sides of an equilateral triangle are equal, the distance from the center to any vertex is exactly equal to the side length of the hexagon, $L$.

Q: What if all six vertices had a $+q$ charge?

If all six vertices had a $+q$ charge, the arrangement would be perfectly symmetric. Every charge would have a diagonally opposite partner cancelling its force. The net force on the central charge would be exactly zero.

Q: Can I solve this by adding the $x$ and $y$ components of all five forces?

Yes, you absolutely can! If you set up a coordinate system and calculate the $x$ and $y$ components of the force from each of the five charges, you will find that all the $y$ components sum to zero, and all the $x$ components sum to the force of a single charge. However, using symmetry is much faster and less prone to calculation errors.

Q: Does the sign of the central charge matter for the magnitude of the net force?

No, the sign of the central charge only affects the *direction* of the net force. If the central charge were $+q$, the unbalanced force from the rightmost vertex would be repulsive (pushing it to the left). Since it is $-q$, the force is attractive (pulling it to the right). The magnitude remains the same.

Q: What is the 'missing charge' concept?

You can think of this problem as a complete hexagon (6 charges) PLUS a negative charge $-q$ placed at the empty vertex. The complete hexagon exerts zero net force. So, the net force is simply the force exerted by the 'missing' $-q$ charge at the empty vertex, which repels the central $-q$ charge, pushing it towards the right. This gives the exact same result!

Question

Five point charges, each of value $+q$ coulomb, are placed on five vertices of a regular hexagon of side $L$ metre. The magnitude of the force on the point charge of value $-q$ coulomb placed at the centre of the hexagon is ………newton.

Sigma Percentile · Accepted Answer

### The Power of Symmetry in Electrostatics

Imagine you are faced with a problem that looks like a nightmare of vector additions: five charges arranged on the vertices of a regular hexagon, all pulling on a central charge. Calculating the force from each charge and adding them up using vector components would take a lot of time and effort. But what if there's a shortcut?

This is where the elegance of symmetry comes into play. In physics, symmetry is not just a visual property; it is a powerful mathematical tool that can turn a complex calculation into a trivial observation.

### Analyzing the Setup

We have a regular hexagon of side length $L$. At five of its six vertices, we place a positive charge $+q$. Right at the geometric center of the hexagon, we place a negative charge $-q$. Our goal is to find the net electrostatic force acting on this central charge.

According to the principle of superposition, the net force on the central charge is simply the vector sum of the individual forces exerted by each of the five positive charges.

$$ \vec{F}_{net} = \vec{F}_1 + \vec{F}_2 + \vec{F}_3 + \vec{F}_4 + \vec{F}_5 $$

### The Magic of Cancelling Pairs

Let's look closely at the arrangement. A regular hexagon has a high degree of symmetry. Specifically, its vertices can be grouped into diagonally opposite pairs.

Consider the charge at the top-left vertex and the charge at the bottom-right vertex. Both are positive charges of magnitude $+q$, and both are at the exact same distance $L$ from the central $-q$ charge. The top-left charge pulls the central charge towards it with a force $\vec{F}_1$. The bottom-right charge pulls the central charge towards it with a force $\vec{F}_4$.

Because the charges are equal and the distances are equal, the magnitudes of these forces are identical. However, because they are diagonally opposite, their directions are exactly $180^\circ$ apart.

$$ \vec{F}_1 = -\vec{F}_4 \implies \vec{F}_1 + \vec{F}_4 = 0 $$

These two forces perfectly cancel each other out!

We can apply the exact same logic to the top-right and bottom-left charges. They also form a diagonally opposite pair, and their forces on the central charge will also sum to zero.

$$ \vec{F}_2 = -\vec{F}_5 \implies \vec{F}_2 + \vec{F}_5 = 0 $$

### The Final Calculation

Out of the five original charges, four have completely cancelled each other out. We are left with only one charge: the $+q$ charge sitting at the rightmost vertex.

Because the sixth vertex (the leftmost one) is empty, there is no charge to cancel out the pull from the rightmost charge. The net force on the central charge is simply the force exerted by this single unbalanced charge.

Using Coulomb's Law, the magnitude of this force is:

$$ F_{net} = \frac{1}{4\pi\varepsilon_0} \frac{q \cdot q}{L^2} $$

$$ F_{net} = 9 	imes 10^9 \frac{q^2}{L^2} $$

**This is our final answer.** By leveraging symmetry, we bypassed all the complex vector math and arrived at the solution in seconds. Always keep an eye out for symmetry—it is a physicist's best friend!

Animated Solution for Physics - Electrostatics: Five point charges, each of value $+ q$ coulomb, are placed on five vertices of a regular hexagon of side $L$ metre. The magnitude of the force on the point charge of value $- q$ coulomb placed at the centre of the hexagon is ………newton.

Visualized Solution

Visualizing the Setup

Principle of Superposition

Symmetry Pair 1

Symmetry Pair 2

Calculating the Unbalanced Force

Conclusion and Insights

The Sigma Insight: Coulomb's Law

The Power of Symmetry in Electrostatics

Analyzing the Setup

The Magic of Cancelling Pairs

The Final Calculation

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