1. Set up the geometry

The positive test charge $q$ is placed on the positive $y$-axis at point

$$(0,b), \qquad b>0.$$

The four source charges of equal magnitude are at

$$Q_1: (-2a,0),\quad Q_2: (-a,0),\quad Q_3: (a,0),\quad Q_4: (2a,0).$$

We only need the direction of the net force on $q$.

Since $q>0$:

• a positive source charge repels $q$,
• a negative source charge attracts $q$.

Also, by symmetry, for suitable sign patterns many components cancel.

---

2. Useful directional facts

Let us note the force directions on $q(0,b)$ due to charges at symmetric points.

• From a positive charge at $(-x,0)$, force on $q$ is along $(+x,+y)$.
• From a positive charge at $(+x,0)$, force on $q$ is along $(-x,+y)$.
• From a negative charge at $(-x,0)$, force on $q$ is along $(-x,-y)$.
• From a negative charge at $(+x,0)$, force on $q$ is along $(+x,-y)$.

Now evaluate each case.

---

3. Case P: $Q_1,Q_2,Q_3,Q_4$ all positive

All four charges repel $q$.

• Charges at $\pm a$ give equal and opposite $x$-components, so they cancel.
• Charges at $\pm 2a$ also give equal and opposite $x$-components, so they cancel.
• All $y$-components are upward.

Hence net force is along

$$+y.$$

So,

$$P \to 3.$$

---

4. Case Q: $Q_1,Q_2$ positive; $Q_3,Q_4$ negative

Left-side charges are positive, right-side charges are negative.

• $Q_1$ at $-2a$ (positive): force on $q$ toward $(+x,+y)$.
• $Q_2$ at $-a$ (positive): force toward $(+x,+y)$.
• $Q_3$ at $+a$ (negative): force toward $(+x,-y)$.
• $Q_4$ at $+2a$ (negative): force toward $(+x,-y)$.

Thus all four $x$-components are positive, so net force has $+x$ direction.

For $y$-components:

• symmetric pair at $\pm a$: one gives $+y$, the other gives $-y$ with equal magnitude, so cancel.
• symmetric pair at $\pm 2a$: similarly cancel.

Hence net force is along

$$+x.$$

So,

$$Q \to 1.$$

---

5. Case R: $Q_1,Q_4$ positive; $Q_2,Q_3$ negative

So outer pair ($\pm 2a$) are positive and inner pair ($\pm a$) are negative.

Outer pair: $Q_1,Q_4$ positive

By symmetry, their $x$-components cancel and their $y$-components add upward. So outer pair gives net force along $+y$.

Inner pair: $Q_2,Q_3$ negative

Again by symmetry, their $x$-components cancel and their $y$-components add downward. So inner pair gives net force along $-y$.

Now compare magnitudes.

For a pair at distance $x$ from origin, vertical component from one charge is proportional to

$$\frac{b}{(x^2+b^2)^{3/2}}.$$

So the pair at $x=a$ produces stronger vertical force than the pair at $x=2a$ because

$$\frac{b}{(a^2+b^2)^{3/2}} > \frac{b}{(4a^2+b^2)^{3/2}}.$$

Thus the downward contribution from the inner negative pair dominates.

Hence net force is along

$$-y.$$

So,

$$R \to 4.$$

---

6. Case S: $Q_1,Q_3$ positive; $Q_2,Q_4$ negative

That is:

• $Q_1$ at $-2a$: positive $\Rightarrow$ force toward $(+x,+y)$
• $Q_2$ at $-a$: negative $\Rightarrow$ force toward $(-x,-y)$
• $Q_3$ at $+a$: positive $\Rightarrow$ force toward $(-x,+y)$
• $Q_4$ at $+2a$: negative $\Rightarrow$ force toward $(+x,-y)$

Now examine components.

$y$-components

• Pair $Q_1$ and $Q_4$ at $\pm 2a$: one contributes $+y$, the other $-y$ with equal magnitude $\Rightarrow$ cancel.
• Pair $Q_2$ and $Q_3$ at $\pm a$: one contributes $-y$, the other $+y$ with equal magnitude $\Rightarrow$ cancel.

So net $y$-component is zero.

$x$-components

• From $Q_1$ and $Q_4$ (charges at $\pm 2a$), both $x$-components are toward $+x$.
• From $Q_2$ and $Q_3$ (charges at $\pm a$), both $x$-components are toward $-x$.

Now compare magnitudes. The pair at $x=a$ is closer, so its horizontal effect is stronger than the pair at $x=2a$. Therefore the negative $x$ contribution dominates.

Hence net force is along

$$-x.$$

So,

$$S \to 2.$$

---

7. Final matching

We get

$$P\to 3,\quad Q\to 1,\quad R\to 4,\quad S\to 2.$$

This corresponds to Option A.

---

8. Comparison with stored answer

Stored correct answer: A

Our derived answer: A

So they agree.