Three charges q1 = 77.3 nC, q2 = 95.5 nC, q3 = 94 nC are arranged on the vertices of an equilateral triangle with side length 0.024 m. A fourth charge q4 = 11.4 nC is placed at the center of the triangle. (Remember that for an equilateral triangle of side length L, the distance between a vertex and the center of the triangle is (\sqrt(3))/(3)). Give all answers to three significant figures. The angle at q1 is 60 degrees.
What is the x component of the force q1 exerts on q4?

To calculate the x-component of the force that \( q_1 \) exerts on \( q_4 \), we will follow these steps:

### Step 1: Calculate the distance between \( q_1 \) and \( q_4 \)
For an equilateral triangle with side length \( L = 0.024 \, \text{m} \), the distance between a vertex and the center of the triangle is:

\[
r = \frac{L \cdot \sqrt{3}}{3} = \frac{0.024 \cdot \sqrt{3}}{3}
\]

Let's calculate this value first.

### Step 2: Calculate the force magnitude between \( q_1 \) and \( q_4 \)
Using Coulomb's law, the magnitude of the force between two charges is:

\[
F = k_e \frac{|q_1 \cdot q_4|}{r^2}
\]

where \( k_e = 8.99 \times 10^9 \, \text{N} \cdot \text{m}^2 / \text{C}^2 \), and \( r \) is the distance from \( q_1 \) to \( q_4 \).

### Step 3: Find the x-component of the force
The force exerted by \( q_1 \) on \( q_4 \) makes an angle of \( 60^\circ \) with the x-axis. The x-component of the force is:

\[
F_x = F \cdot \cos(60^\circ) = \frac{F}{2}
\]

Now, let's compute these values step by step.

First, we'll calculate \( r \).The distance between \( q_1 \) and \( q_4 \) is approximately \( 0.0139 \, \text{m} \).

Now let's calculate the force magnitude \( F \) using Coulomb's law:

\[
F = k_e \frac{|q_1 \cdot q_4|}{r^2}
\]

where:
- \( q_1 = 77.3 \times 10^{-9} \, \text{C} \)
- \( q_4 = 11.4 \times 10^{-9} \, \text{C} \)
- \( k_e = 8.99 \times 10^9 \, \text{N} \cdot \text{m}^2 / \text{C}^2 \)
- \( r = 0.0139 \, \text{m} \)

Let's compute the force.The magnitude of the force between \( q_1 \) and \( q_4 \) is approximately \( 0.0413 \, \text{N} \).

Now, let's calculate the x-component of this force:

\[
F_x = \frac{F}{2}
\]

Let me compute that for you.The x-component of the force that \( q_1 \) exerts on \( q_4 \) is approximately \( 0.0206 \, \text{N} \).

Would you like any further details or clarification?

Here are 5 related questions for further exploration:
1. What is the y-component of the force \( q_1 \) exerts on \( q_4 \)?
2. What is the net force on \( q_4 \) due to all three charges \( q_1 \), \( q_2 \), and \( q_3 \)?
3. How does the force change if \( q_4 \) is moved closer to one of the charges?
4. What is the potential energy of the system due to these four charges?
5. How would the force change if the charges were doubled?

**Tip:** When dealing with forces in a symmetrical arrangement, breaking them into components can significantly simplify the calculation.

Three charges q1 = 77.3 nC, q2 = 95.5 nC, q3 = 94 nC are arranged on the vertices of an equilateral triangle with side length 0.024 m. A fourth charge q4 = 11.4 nC is placed at the center of the triangle. (Remember that for an equilateral triangle of side length L, the distance between a vertex and the center of the triangle is (√3)/(3)). Give all answers to three significant figures. The angle at q1 is 60 degrees. What is the x component of the force q1 exerts on q4?

Learn how to calculate the x-component of the force exerted by charge q1 on charge q4, placed at the center of an equilateral triangle. This problem involves Coulomb's law, vector components, and geometrical relationships, providing a detailed step-by-step solution for students in advanced high school or undergraduate physics.

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