In the figure particle 1 of charge +4e is above a floor by distance d, = 2.60 mm and particle 2 of charge +8e is on the floor, at distance d2 = 7.30 mm horizontally from particle 1. What is the x component of the electrostatic force on particle 2 due to particle 1?

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### Electrostatic Force: Calculating X Component

#### Problem Statement:
In the figure, particle 1 of charge \(+4e\) is above a floor by a distance \(d_1 = 2.60 \, \text{mm}\) and particle 2 of charge \(+8e\) is on the floor, at a distance \(d_2 = 7.30 \, \text{mm}\) horizontally from particle 1. What is the \(x\)-component of the electrostatic force on particle 2 due to particle 1?

#### Diagram Explanation:
The diagram involves two particles and a coordinate system:
- Particle 1 is positioned vertically above the floor at a distance \(d_1 = 2.60 \, \text{mm}\).
- Particle 2 is placed on the floor, horizontally from particle 1 at a distance \(d_2 = 7.30 \, \text{mm}\).

The text also includes a standard coordinate system with the \(x\)-axis along the horizontal plane and an origin presumably at the floor directly below particle 1.

#### Calculation Steps:
1. **Determine Horizontal Distance (x-component)**:
   - The electrostatic force between two charges is given by Coulomb's Law:
     \[
     F = k_e \frac{|q_1 q_2|}{r^2}
     \]
     where:
     - \(k_e\) is Coulomb's constant \((8.99 \times 10^9 \, \text{Nm}^2/\text{C}^2)\),
     - \(q_1 = +4e\),
     - \(q_2 = +8e\),
     - \(r\) is the distance between the charges.

2. **Calculate \(r\)**:
   \[
   r = \sqrt{d_1^2 + d_2^2}
   \]
   \[
   r = \sqrt{(2.60 \times 10^{-3})^2 + (7.30 \times 10^{-3})^2} \, \text{m}
   \]
   \[
   r \approx 7.74 \times 10^{-3} \, \text{m}
   \]

3. **Calculate the Force**:
   Using the values, compute the force magnitude \(F\):
   \
Transcribed Image Text:### Electrostatic Force: Calculating X Component #### Problem Statement: In the figure, particle 1 of charge \(+4e\) is above a floor by a distance \(d_1 = 2.60 \, \text{mm}\) and particle 2 of charge \(+8e\) is on the floor, at a distance \(d_2 = 7.30 \, \text{mm}\) horizontally from particle 1. What is the \(x\)-component of the electrostatic force on particle 2 due to particle 1? #### Diagram Explanation: The diagram involves two particles and a coordinate system: - Particle 1 is positioned vertically above the floor at a distance \(d_1 = 2.60 \, \text{mm}\). - Particle 2 is placed on the floor, horizontally from particle 1 at a distance \(d_2 = 7.30 \, \text{mm}\). The text also includes a standard coordinate system with the \(x\)-axis along the horizontal plane and an origin presumably at the floor directly below particle 1. #### Calculation Steps: 1. **Determine Horizontal Distance (x-component)**: - The electrostatic force between two charges is given by Coulomb's Law: \[ F = k_e \frac{|q_1 q_2|}{r^2} \] where: - \(k_e\) is Coulomb's constant \((8.99 \times 10^9 \, \text{Nm}^2/\text{C}^2)\), - \(q_1 = +4e\), - \(q_2 = +8e\), - \(r\) is the distance between the charges. 2. **Calculate \(r\)**: \[ r = \sqrt{d_1^2 + d_2^2} \] \[ r = \sqrt{(2.60 \times 10^{-3})^2 + (7.30 \times 10^{-3})^2} \, \text{m} \] \[ r \approx 7.74 \times 10^{-3} \, \text{m} \] 3. **Calculate the Force**: Using the values, compute the force magnitude \(F\): \
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