Match Signs of Four Point Charges to Direction of Force on a Test Charge

To determine the direction of the net electrostatic force on the positive test charge $q$ located at $(0, b)$, we calculate the force contribution from each of the four fixed point charges $Q_1, Q_2, Q_3,$ and $Q_4$ located at $(-2a, 0)$, $(-a, 0)$, $(+a, 0)$, and $(+2a, 0)$, respectively.

1. General Force Expression

The position vector of the test charge $q$ is $\vec{r}_q = b\hat{j}$. The position vectors of the four charges are: $\vec{r}_1 = -2a\hat{i}, \quad \vec{r}_2 = -a\hat{i}, \quad \vec{r}_3 = a\hat{i}, \quad \vec{r}_4 = 2a\hat{i}$

Using Coulomb's Law, the force exerted on charge $q$ by charge $Q_i$ is given by: $\vec{F}_i = \frac{k q Q_i}{|\vec{r}_q - \vec{r}_i|^3} (\vec{r}_q - \vec{r}_i)$

Writing this in component form:

• For $Q_1$ at $(-2a, 0)$: $\vec{F}_1 = \frac{k q Q_1}{(4a^2 + b^2)^{3/2}} (2a\hat{i} + b\hat{j})$

• For $Q_2$ at $(-a, 0)$: $\vec{F}_2 = \frac{k q Q_2}{(a^2 + b^2)^{3/2}} (a\hat{i} + b\hat{j})$

• For $Q_3$ at $(+a, 0)$: $\vec{F}_3 = \frac{k q Q_3}{(a^2 + b^2)^{3/2}} (-a\hat{i} + b\hat{j})$

• For $Q_4$ at $(+2a, 0)$: $\vec{F}_4 = \frac{k q Q_4}{(4a^2 + b^2)^{3/2}} (-2a\hat{i} + b\hat{j})$

Let $C_1 = \frac{k q}{(a^2 + b^2)^{3/2}} > 0$ and $C_2 = \frac{k q}{(4a^2 + b^2)^{3/2}} > 0$. Since $a^2 + b^2 < 4a^2 + b^2$, we have $C_1 > C_2$.

The net force components are: $F_x = 2a C_2 Q_1 + a C_1 Q_2 - a C_1 Q_3 - 2a C_2 Q_4$ $F_y = b C_2 Q_1 + b C_1 Q_2 + b C_1 Q_3 + b C_2 Q_4$

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2. Case-by-Case Analysis

Let $|Q_i| = Q > 0$.

Case P: $Q_1, Q_2, Q_3, Q_4$ all positive ($+Q$)

• $x$-component: $F_x = 2a C_2 Q + a C_1 Q - a C_1 Q - 2a C_2 Q = 0$
• $y$-component: $F_y = b C_2 Q + b C_1 Q + b C_1 Q + b C_2 Q = 2b Q (C_1 + C_2) > 0$

Since $F_x = 0$ and $F_y > 0$, the net force is directed along the $+y$ direction (3). $\text{P } \rightarrow \text{ 3}$

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Case Q: $Q_1, Q_2$ positive ($+Q$); $Q_3, Q_4$ negative ($-Q$)

• $x$-component: $F_x = 2a C_2 (+Q) + a C_1 (+Q) - a C_1 (-Q) - 2a C_2 (-Q) = 4a C_2 Q + 2a C_1 Q > 0$
• $y$-component: $F_y = b C_2 (+Q) + b C_1 (+Q) + b C_1 (-Q) + b C_2 (-Q) = 0$

Since $F_x > 0$ and $F_y = 0$, the net force is directed along the $+x$ direction (1). $\text{Q } \rightarrow \text{ 1}$

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Case R: $Q_1, Q_4$ positive ($+Q$); $Q_2, Q_3$ negative ($-Q$)

• $x$-component: $F_x = 2a C_2 (+Q) + a C_1 (-Q) - a C_1 (-Q) - 2a C_2 (+Q) = 0$
• $y$-component: $F_y = b C_2 (+Q) + b C_1 (-Q) + b C_1 (-Q) + b C_2 (+Q) = 2b Q (C_2 - C_1)$

Since $C_1 > C_2$, we have $C_2 - C_1 < 0$, which implies $F_y < 0$. Thus, the net force is directed along the $-y$ direction (4). $\text{R } \rightarrow \text{ 4}$

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Case S: $Q_1, Q_3$ positive ($+Q$); $Q_2, Q_4$ negative ($-Q$)

• $y$-component: $F_y = b C_2 (+Q) + b C_1 (-Q) + b C_1 (+Q) + b C_2 (-Q) = 0$
• $x$-component: $F_x = 2a C_2 (+Q) + a C_1 (-Q) - a C_1 (+Q) - 2a C_2 (-Q) = 2a Q (2C_2 - C_1) < 0$

Thus, the net force is directed along the $-x$ direction (2). $\text{S } \rightarrow \text{ 2}$

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Conclusion

Matching the results from List I to List II: $\text{P } \rightarrow \text{ 3}, \quad \text{Q } \rightarrow \text{ 1}, \quad \text{R } \rightarrow \text{ 4}, \quad \text{S } \rightarrow \text{ 2}$

This corresponds to Option A.