Four charges Q1, Q2, Q3 and Q4 of same magnitude are fixed along the x-axis at x = -2a, -a, +a and +

Q: Why do the x-components cancel in Case A and Case C?

Because the charge distribution is perfectly symmetric about the y-axis. For every charge on the left pushing/pulling in the x-direction, there is an identical charge on the right pushing/pulling with the exact same magnitude in the opposite x-direction.

Q: In Case C, how do we know the downward force is stronger without calculating?

Coulomb's law states that force is inversely proportional to the square of the distance. The inner charges $Q_2$ and $Q_3$ are physically closer to the test charge $q$ than the outer charges $Q_1$ and $Q_4$. Thus, their attractive downward force easily overpowers the repulsive upward force.

Q: Why don't the y-components cancel in Case A?

In Case A, all charges are positive. They all repel the positive test charge $q$ away from the x-axis. Since they all push 'upwards', their y-components add together rather than cancel.

Q: In Case D, both pairs of charges create forces along the x-axis. How do we determine the net direction?

Just like in Case C, proximity is key. The inner pair ($Q_2, Q_3$) creates a force in the $-x$ direction, while the outer pair ($Q_1, Q_4$) creates a force in the $+x$ direction. Since the inner pair is closer, their $-x$ force dominates.

Q: Does the value of distance 'b' affect the final answer?

For the given options, no. The symmetry arguments hold true for any $b > 0$. The cancellation of components is exact, and the dominance of the closer charges remains valid for the net direction.

Question

Four charges $Q_1, Q_2, Q_3$ and $Q_4$ of same magnitude are fixed along the $x$-axis at $x = -2a, -a, +a$ and $+2a$ respectively. A positive charge $q$ is placed on the positive $y$-axis at a distance $b > 0$. Four options of the signs of these charges are given in Column I. The direction of the forces on the charge $q$ is given in Column II. Match Column I with Column II and select the correct answer using the code given below the lists.

List-I:

P. A. $Q_1, Q_2, Q_3, Q_4$ all positive

Q. B. $Q_1, Q_2$ positive; $Q_3, Q_4$ negative

R. C. $Q_1, Q_4$ positive; $Q_2, Q_3$ negative

S. D. $Q_1, Q_3$ positive; $Q_2, Q_4$ negative

List-II:

1. p. $+x$

2. q. $-x$

3. r. $+y$

4. s. $-y$

Sigma Percentile · Accepted Answer

### Mastering Electrostatic Symmetry: A Dance of Vectors

Welcome to a beautiful exploration of electrostatic forces and symmetry. When faced with multiple charges, our first instinct is often to dive into heavy algebra and Coulomb's law calculations. However, this problem is a masterclass in using symmetry to bypass the math entirely. Let's break down the forces into their $x$ and $y$ components and see how they interact.

### Analyzing the Setup

We have four charges placed symmetrically on the $x$-axis at $\pm a$ and $\pm 2a$. A positive test charge $+q$ sits on the $y$-axis at $(0,b)$.

The force magnitude from any charge $Q$ at $(x_0, 0)$ is given by $F = \frac{kQq}{x_0^2+b^2}$.

By resolving this force, the $x$ and $y$ components are $F_x = F \frac{-x_0}{\sqrt{x_0^2+b^2}}$ and $F_y = F \frac{b}{\sqrt{x_0^2+b^2}}$. The beauty of this setup is that for every charge on the left, there is a corresponding charge on the right at the exact same distance.

### The Power of Symmetry

**Case A: All charges positive**
Imagine all four charges are positive. Since like charges repel, every single charge pushes our test charge $q$ upwards and away. The $y$-components of all these forces point straight up in the $+y$ direction.

What about the $x$-components? The push to the right from the left charges is perfectly balanced by the push to the left from the right charges. They cancel out completely! So, the net force points purely in the $+y$ direction.

**Case B: $Q_1, Q_2$ positive; $Q_3, Q_4$ negative**
The two charges on the left are positive, repelling $q$ up and to the right. The two charges on the right are negative, attracting $q$ down and to the right.

Notice what happens to the $y$-components! The upward push from the left perfectly cancels the downward pull from the right. However, all four charges are working together to move $q$ to the right. Thus, the net force is strictly in the $+x$ direction.

### Proximity and Dominance

**Case C: $Q_1, Q_4$ positive; $Q_2, Q_3$ negative**
The outer charges are positive, pushing $q$ upwards. The inner charges are negative, pulling $q$ downwards. Because the setup is symmetric left-to-right, the $x$-components cancel out again.

Now, it's a tug-of-war along the $y$-axis. Who wins? The inner charges are closer to $q$, so their downward pull is much stronger than the upward push from the outer charges. Therefore, the net force points downwards, in the $-y$ direction.

**Case D: $Q_1, Q_3$ positive; $Q_2, Q_4$ negative**
The charges alternate in sign. $Q_1$ is positive and $Q_4$ is negative, both contributing to a force in the $+x$ direction. Meanwhile, $Q_2$ is negative and $Q_3$ is positive, both contributing to a force in the $-x$ direction.

The $y$-components cancel out. So, it's a battle along the $x$-axis. Since the inner charges $Q_2$ and $Q_3$ are closer to $q$, their influence is stronger. They win the tug-of-war, making the net force point in the $-x$ direction.

### Final Matching

By mastering symmetry, we solved a complex vector problem without a single heavy calculation.

- **A** matches with **r** ($+y$)
- **B** matches with **p** ($+x$)
- **C** matches with **s** ($-y$)
- **D** matches with **q** ($-x$)

Keep visualizing the physics, and you'll never get lost!

List-I

List-II

Select Matching Pairs:

Visualized Solution

$S y mm e t r y o f t h e S e t u p$

$C a se A : A l l c ha r g es p os i t i v e$

$C a se B : Q_{1}, Q_{2} > 0 and Q_{3}, Q_{4} < 0$

$C a se C : Q_{1}, Q_{4} > 0 and Q_{2}, Q_{3} < 0$

$C a se D : Q_{1}, Q_{3} > 0 and Q_{2}, Q_{4} < 0$

$F ina l M a t c hin g$

The Sigma Insight: Coulomb's Law

Mastering Electrostatic Symmetry

Analyzing the Setup

The Power of Symmetry

Proximity and Dominance

Final Matching

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List-I

List-II

Select Matching Pairs:

Visualized Solution

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Case A: All charges positive

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Case B: Q1​,Q2​>0 and Q3​,Q4​<0

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Case C: Q1​,Q4​>0 and Q2​,Q3​<0

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Case D: Q1​,Q3​>0 and Q2​,Q4​<0

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