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 out in Case A?

In Case A, all charges are positive. $Q_2$ pushes the test charge to the right and up, while $Q_3$ pushes it to the left and up. Because $Q_2$ and $Q_3$ are at the same distance from the y-axis and have the same magnitude, their horizontal pushes are exactly equal and opposite, perfectly cancelling each other. The same applies to $Q_1$ and $Q_4$.

Q: In Case B, why don't the forces just cancel out completely?

While the y-components cancel out (the left charges push up, the right charges pull down), the x-components actually reinforce each other. The positive charges on the left repel the test charge towards the right ($+x$), and the negative charges on the right attract it towards the right ($+x$). They work together!

Q: How do we know the inner charges exert a stronger force in Case C?

Coulomb's Law states that the electrostatic force is inversely proportional to the square of the distance ($F \propto 1/r^2$). Since the inner charges ($Q_2, Q_3$) are located at $x = \pm a$, they are closer to the test charge at $(0, b)$ than the outer charges at $x = \pm 2a$. A smaller distance results in a larger force magnitude.

Q: For Case D, isn't it complicated to calculate the exact x-components?

We don't need to calculate the exact values! We only need the direction. The leftward forces come from the inner charges ($Q_2, Q_3$), and the rightward forces come from the outer charges ($Q_1, Q_4$). Because the inner charges are closer, their combined leftward force will always be greater than the combined rightward force from the outer charges, regardless of the exact value of $b$.

Q: Does the magnitude of the test charge $+q$ affect the direction of the net force?

No, the magnitude of $+q$ only affects the magnitude of the net force, not its direction. The direction is entirely determined by the signs and positions of the fixed charges $Q_1, Q_2, Q_3, Q_4$ relative to the test charge.

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.

$\begin{array}{ll}
	ext{	extbf{Column I}} & 	ext{	extbf{Column II}} \
	ext{A. } Q_1, Q_2, Q_3, Q_4 	ext{ all positive} & 	ext{p. } +x \
	ext{B. } Q_1, Q_2 	ext{ positive; } Q_3, Q_4 	ext{ negative} & 	ext{q. } -x \
	ext{C. } Q_1, Q_4 	ext{ positive; } Q_2, Q_3 	ext{ negative} & 	ext{r. } +y \
	ext{D. } Q_1, Q_3 	ext{ positive; } Q_2, Q_4 	ext{ negative} & 	ext{s. } -y
\end{array}$

Options:

(A) A-r, B-p, C-s, D-q

(B) A-s, B-q, C-r, D-p

(C) A-r, B-p, C-q, D-s

(D) A-s, B-q, C-p, D-r

Sigma Percentile · Accepted Answer

Welcome to a beautiful exploration of electrostatics! This problem is a masterclass in using **symmetry** and **Coulomb's Law** to simplify what looks like a complicated vector addition nightmare. We have four charges fixed along the x-axis and a test charge sitting on the y-axis. Our mission is to determine the direction of the net force on this test charge for four different sign configurations. Let's dive in!

### Analyzing the Setup

Imagine the x-y plane. We have four charges, $Q_1, Q_2, Q_3$, and $Q_4$, placed symmetrically along the x-axis at $x = -2a, -a, +a$, and $+2a$. A positive test charge, $+q$, is positioned on the positive y-axis at $(0, b)$.

Because the setup is perfectly symmetric about the y-axis, we can pair the charges up: the inner pair ($Q_2$ and $Q_3$) and the outer pair ($Q_1$ and $Q_4$). This symmetry is our secret weapon. It means that for any symmetric charge distribution, the horizontal ($x$) components of the forces will often perfectly cancel out or perfectly add up.

### Case A: The Power of Pure Repulsion

In our first scenario, all four fixed charges are positive. Since our test charge $+q$ is also positive, every single force acting on it will be **repulsive**.

Let's look at the inner pair, $Q_2$ and $Q_3$. They push the test charge symmetrically upwards and outwards. Because they are equidistant from the y-axis and have the same magnitude, their horizontal components ($x$-components) are equal and opposite. They perfectly cancel each other out! We are left with only their vertical components, which add together to push the test charge straight up.

The exact same logic applies to the outer pair, $Q_1$ and $Q_4$. Their $x$-components cancel, and their $y$-components push upwards.

Therefore, the total net force is strictly along the **positive y-axis** ($+y$). This matches Case A with option **r**.

### Case B: A Coordinated Push and Pull

Now, let's mix things up. $Q_1$ and $Q_2$ (on the left) are positive, while $Q_3$ and $Q_4$ (on the right) are negative.

The positive charges on the left will **repel** the test charge, pushing it towards the right and upwards. The negative charges on the right will **attract** the test charge, pulling it towards the right and downwards.

Notice what happens to the vertical components. The left side pushes *up*, while the right side pulls *down*. Because of the perfect symmetry in their magnitudes and distances, these upward and downward $y$-components completely cancel out!

However, both the repulsive push from the left and the attractive pull from the right have horizontal components pointing in the $+x$ direction. They work together! The net force is purely along the **positive x-axis** ($+x$). This matches Case B with option **p**.

### Case C: The Proximity Advantage

In Case C, the outer charges ($Q_1, Q_4$) are positive, and the inner charges ($Q_2, Q_3$) are negative.

The positive outer charges push the test charge inwards and upwards. The negative inner charges pull it inwards and downwards. Once again, the left-right symmetry guarantees that all $x$-components cancel out. The battle is entirely vertical.

So, who wins? The upward push or the downward pull? This is where **Coulomb's Law** comes into play. The force between two charges is inversely proportional to the square of the distance between them ($F \propto \frac{1}{r^2}$).

The inner charges ($Q_2, Q_3$) are physically closer to the test charge than the outer charges ($Q_1, Q_4$). Because they are closer, their attractive downward force is significantly stronger than the repulsive upward force from the outer charges. The downward pull dominates!

The net force is along the **negative y-axis** ($-y$). This matches Case C with option **s**.

### Case D: The Horizontal Tug of War

Finally, we have Case D: $Q_1$ and $Q_3$ are positive, while $Q_2$ and $Q_4$ are negative. This creates a horizontal tug of war.

Let's break down the horizontal forces acting on $+q$:
- $Q_1$ (positive, far left) pushes to the right ($+x$).
- $Q_4$ (negative, far right) pulls to the right ($+x$).
- $Q_2$ (negative, inner left) pulls to the left ($-x$).
- $Q_3$ (positive, inner right) pushes to the left ($-x$).

The forces pushing/pulling to the right come from the outer charges ($Q_1, Q_4$). The forces pushing/pulling to the left come from the inner charges ($Q_2, Q_3$).

Just like in Case C, proximity is the deciding factor. The inner charges are much closer to the test charge. Therefore, their combined leftward force ($F_{2x} + F_{3x}$) is much stronger than the combined rightward force ($F_{1x} + F_{4x}$) from the outer charges.

The net force points along the **negative x-axis** ($-x$). This matches Case D with option **q**.

### Final Conclusion

Let's summarize our elegant deductions:
- **A $\rightarrow$ r** ($+y$)
- **B $\rightarrow$ p** ($+x$)
- **C $\rightarrow$ s** ($-y$)
- **D $\rightarrow$ q** ($-x$)

Looking at our options, this perfectly aligns with option **(a)**. By leveraging symmetry and the distance dependence of Coulomb's law, we completely bypassed complex vector math and arrived at the solution through pure physical intuition!

Select Answer:

Visualized Solution

$V i s u a l i z in g t h e S e t u p$

$C a se A : A l l P os i t i v e$

$C a se B : C oor d ina t e d P u s h an d P u l l$

$C a se C : T h e P r o x imi t y A d v an t a g e$

$C a se D : H or i z o n t a l T ug o f W a r$

$F ina l C o n c l u s i o n$

The Sigma Insight: Coulomb's Law

Analyzing the Setup

Case A

Case B

Case C

Case D

Final Conclusion

Similar Questions

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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; } Visualizing the Setup

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Case B: Coordinated Push and Pull

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Case C: The Proximity Advantage

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Case D: Horizontal Tug of War

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