Three charges -q1, +q2 and -q3 are placed as shown in the figure. The x-component of the force on -q

Q: How do we know the angle of $F_{13}$ is $\theta$ with the negative y-axis?

The line connecting $-q_1$ and $-q_3$ makes an angle $\theta$ with the positive y-axis. The force vector is a straight continuation of this line through the origin. By vertically opposite angles, the force vector makes the same angle $\theta$ with the negative y-axis.

Q: Why is the x-component of $F_{13}$ using sine instead of cosine?

The angle $\theta$ is given with respect to the y-axis, not the x-axis. In a right triangle where the angle is with the vertical, the horizontal side (opposite to $\theta$) is proportional to $\sin\theta$.

Q: Do we need to worry about the signs of the charges when substituting into Coulomb's law?

It's usually safer to use the magnitudes of the charges to find the magnitude of the force, and use the physical attraction/repulsion to determine the direction. That's what we did here!

Q: What about the y-component of the force?

The question specifically asks only for the x-component, so we can save time by ignoring the y-component entirely.

Question

Three charges $-q_1, +q_2$ and $-q_3$ are placed as shown in the figure. The x-component of the force on $-q_1$ is proportional to

Options:

(A) $\frac{q_2}{b^2} - \frac{q_3}{a^2} \cos	heta$

(B) $\frac{q_2}{b^2} + \frac{q_3}{a^2} \sin	heta$

(C) $\frac{q_2}{b^2} + \frac{q_3}{a^2} \cos	heta$

(D) $\frac{q_2}{b^2} - \frac{q_3}{a^2} \sin	heta$

Sigma Percentile · Accepted Answer

### Analyzing the Setup
Imagine you are standing at the origin of a coordinate system. Right where you stand, there is a negative charge, $-q_1$. Looking straight ahead along the positive x-axis, at a distance $b$, you spot a positive charge, $+q_2$. Now, look up and slightly to your left into the second quadrant. At a distance $a$, there is another negative charge, $-q_3$. The line connecting you to $-q_3$ makes an angle $	heta$ with the positive y-axis.

Our mission is to find the net electrostatic force acting on you (the charge $-q_1$), but specifically, we only care about the **x-component** of this force.

### The Pull of the Positive Charge
Let's break this down using the principle of superposition. First, consider the interaction between $-q_1$ and $+q_2$. Since opposite charges attract, $+q_2$ is pulling $-q_1$ towards it. This means the force $\vec{F}_{12}$ points directly to the right, along the positive x-axis.

According to Coulomb's Law, the magnitude of this force is:
$$F_{12} = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{b^2}$$
Because it points entirely along the x-axis, its x-component is simply its full magnitude.

### The Push of the Negative Charge
Now, let's look at $-q_3$. Since both $-q_1$ and $-q_3$ are negative, they repel each other. The force $\vec{F}_{13}$ acting on $-q_1$ will point directly away from $-q_3$. Because $-q_3$ is in the upper-left, the repulsive force pushes $-q_1$ towards the lower-right, into the fourth quadrant.

By vertically opposite angles, the force vector $\vec{F}_{13}$ makes the exact same angle $	heta$ with the negative y-axis. The magnitude of this repulsive force is:
$$F_{13} = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_3}{a^2}$$

### Resolving the Vectors
We only need the x-component of $\vec{F}_{13}$. Looking at the geometry, the x-component is opposite to the angle $	heta$. Therefore, we use the sine function to resolve it:
$$F_{13x} = F_{13} \sin	heta = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_3}{a^2} \sin	heta$$

### The Final Calculation
Finally, we sum the x-components. Both $\vec{F}_{12}$ and the x-component of $\vec{F}_{13}$ point in the positive x-direction, so they add together constructively:
$$F_x = F_{12} + F_{13x}$$
$$F_x = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{b^2} + \frac{1}{4\pi\epsilon_0} \frac{q_1 q_3}{a^2} \sin	heta$$

Factoring out the common terms, we get:
$$F_x = \frac{q_1}{4\pi\epsilon_0} \left( \frac{q_2}{b^2} + \frac{q_3}{a^2} \sin	heta ight)$$

The question asks what the x-component is proportional to. By dropping the constant multiplier $\frac{q_1}{4\pi\epsilon_0}$, we find that the force is proportional to:
$$\frac{q_2}{b^2} + \frac{q_3}{a^2} \sin	heta$$
This elegantly matches option (b)!

Animated Solution for Physics - Electric Charges and Fields: Three charges $- q_{1}, + q_{2}$ and $- q_{3}$ are placed as shown in the figure. The x-component of the force on $- q_{1}$ is proportional to

Select Answer:

Visualized Solution

Analyzing the Setup

Force due to $+ q_{2}$

Force due to $- q_{3}$

Resolving $F_{13}$

Total x-component of Force

Final Proportionality

The Sigma Insight: Coulomb's Law

Analyzing the Setup Imagine you are standing at the origin of a coordinate system

The Pull of the Positive Charge Let's break this down using the principle of superposition

The Push of the Negative Charge Now, let's look at $- q_{3}$

Resolving the Vectors We only need the x-component of $F_{13}$

The Final Calculation Finally, we sum the x-components

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