Six charges, three positive and three negative of equal magnitude are to be placed at the vertices o

Q: Why does 'double the electric field' imply the same direction?

Electric field is a vector quantity. When a problem states the field is 'double' another field without specifying 'magnitude only', it implies scalar multiplication of the vector, which preserves its direction.

Q: Why do we check opposite vertices together?

In a regular hexagon, opposite vertices lie on a straight line passing through the center. Charges on these vertices produce electric fields along the same line, making it very easy to see if they add up or cancel out.

Q: What if we placed a negative charge at $R$ initially?

If a negative charge was at $R$, the initial field would point towards $R$. The target net field would then be double that, pointing towards $R$. The logic remains the same, just the target vector flips.

Q: Could there be multiple correct arrangements?

Yes, depending on the options provided. For instance, if $P$ and $S$ were both $+q$, they would still cancel out. But among the given options, only (d) satisfies all conditions.

Q: Is it necessary to calculate the exact magnitude of the field?

Not at all! This is a classic JEE trick. By focusing on symmetry and vector directions, you can solve the problem purely conceptually without writing a single equation.

Question

Six charges, three positive and three negative of equal magnitude are to be placed at the vertices of a regular hexagon such that the electric field at $O$ is double the electric field when only one positive charge of same magnitude is placed at $R$. Which of the following arrangements of charge is possible for, $P, Q, R, S, T$ and $U$ respectively?

Options:

(A) $+,-,+,-,-,+$

(B) $+,-,+,-,+,-$

(C) $+,+,-,+,-,-$

(D) $-,+,+,-,+,-$

Sigma Percentile · Accepted Answer

The problem of finding the correct arrangement of charges on a regular hexagon might seem daunting at first glance. With six vertices and multiple combinations of positive and negative charges, calculating the net electric field for each option could take forever. But what if I told you there's a highly elegant, visual way to solve this using pure symmetry?

Let's embark on this journey and decode the physics behind the problem!

### The Target: Setting the Benchmark

Before we dive into the options, we must clearly understand our target. The question states that the net electric field at the center $O$ must be **double** the electric field produced when *only one positive charge* is placed at vertex $R$.

Imagine a single positive charge $+q$ sitting at $R$. The electric field it produces at the center $O$ will point radially away from $R$. Since $U$ is the vertex directly opposite to $R$, this electric field vector, let's call it $\vec{E}_R$, points exactly towards $U$.

Our target is a net electric field $\vec{E}_{net} = 2\vec{E}_R$. Notice that the problem says "double the electric field," which implies a vector relationship. The net field must not only have twice the magnitude but must also point in the **exact same direction**—towards $U$.

### The Power of Superposition and Symmetry

In a regular hexagon, opposite vertices lie on straight lines passing through the center. This geometric property is our secret weapon. By pairing up opposite vertices, we can easily see if their electric fields add up or cancel out.

Let's test the winning arrangement, **Option (d)**, which proposes the sequence: $P(-), Q(+), R(+), S(-), T(+), U(-)$.

#### 1. The $P$ and $S$ Pair
Vertices $P$ and $S$ are diagonally opposite. In this arrangement, both hold a negative charge $-q$. 
- The field due to $P$ points towards $P$.
- The field due to $S$ points towards $S$.
Since the charges are identical and equidistant from the center, these two vectors are equal in magnitude but perfectly opposite in direction. They annihilate each other! $\vec{E}_P + \vec{E}_S = 0$.

#### 2. The $Q$ and $T$ Pair
Next, let's look at the diagonally opposite vertices $Q$ and $T$. Both hold a positive charge $+q$.
- The field due to $Q$ points away from $Q$ (towards $T$).
- The field due to $T$ points away from $T$ (towards $Q$).
Once again, we have a perfect tug-of-war resulting in a tie. The fields cancel out completely. $\vec{E}_Q + \vec{E}_T = 0$.

#### 3. The $R$ and $U$ Pair
Finally, we arrive at the crucial pair: $R$ and $U$. Here, $R$ holds a positive charge $+q$, and $U$ holds a negative charge $-q$.
- The field due to the positive charge at $R$ points away from $R$, towards $U$. This is our original vector $\vec{E}_R$.
- The field due to the negative charge at $U$ points towards $U$. This vector is also equal to $\vec{E}_R$ in magnitude and direction!

When we superimpose these two fields, they don't cancel; they reinforce each other! 
$$ \vec{E}_{net} = \vec{E}_R + \vec{E}_U = \vec{E}_R + \vec{E}_R = 2\vec{E}_R $$

### The Grand Conclusion

By systematically pairing opposite vertices, we discovered that the charges at $P, S, Q,$ and $T$ create a perfectly balanced, zero net field at the center. The entire net field is solely generated by the charges at $R$ and $U$, which perfectly align to give us exactly $2\vec{E}_R$.

This confirms that **Option (d)** is the only arrangement that satisfies the problem's strict condition. Always remember: in physics, symmetry isn't just a visual treat; it's a powerful mathematical tool!

Select Answer:

Visualized Solution

$The Hexagon Setup$

$Field due to + q at R$

$The Target Field$

$Analyzing Option (d): Charges at P and S$

$Analyzing Option (d): Charges at Q and T$

$Analyzing Option (d): Charges at R and U$

$Conclusion$

The Sigma Insight: Electric Field

The Target

The Power of Superposition and Symmetry

# 1

# 2

# 3

The Grand Conclusion

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Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Hexagon Setup

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Field due to +q at R

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Target Field

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Analyzing Option (d): Charges at P and S

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Analyzing Option (d): Charges at Q and T

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Analyzing Option (d): Charges at R and U

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Conclusion