**Physics Problem on Electric Potential and Work Done****Description:**This problem involves calculating electric potential and work related to charges positioned in a coordinate system.**Problem Statement:**1. **Charge $ q_1 = 2.5 \, \mu C$ is located at the coordinate system origin.** - Charge $ q_2 = 1 \, \mu C$ is placed at $(a, 0)$ where $ a = 1.85 \, \text{m} $. The point $ P $ is located at $(0, a)$. - A third charge $ q_3 = 3.5 \, \mu C$ will be placed later. **Diagram Explanation:** - There is a coordinate system with $ q_1 $ at the origin $(0, 0)$. - $ q_2 $ is on the x-axis at position $(a, 0)$. - Point $ P $ is on the y-axis at position $(0, a)$.**Tasks:**(a) **Find the electric potential $ V_p $ at point $ P $ in volts.** Assume the potential is zero at infinity. - Equation: $ V_p = $(b) **Determine how much work $ W_i $, in joules,** would you need to do to bring the third charge $ q_3 $ from very far away to the point $ P $. - Equation: $ W_i = $(c) **Calculate the total potential energy $ U $ (in joules)** of the final configuration of the three charges. - Equation: $ U = $**Instructions for Solving:**- Use the formula for electric potential, $ V = \frac{k \cdot q}{r} $, where $ k = 8.99 \times 10^9 \, \text{N m}^2/\text{C}^2 $.- Calculate the potential contributions from each charge at point $ P $ and sum them.- Use the work-energy principle to find the work done in bringing charge $ q_3 $ to point $ P $.- Determine the total potential energy by considering pairwise interactions.

The charges are q1=-2.5 nC=-2.5×10-9 Cq2=0.11 nC=0.11×10-9 C c=a2+b2c=20.85 m2c=1.2 m

Answered: Charge 9=-2,5,c is charge q =, llnct is…

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Charge 9=-2,5,c is charge q =, llnct is a oordinares. la,b)igcher be later, placed

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Chapter1: Units, Trigonometry. And Vectors

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Question

**Physics Problem on Electric Potential and Work Done**

**Description:**
This problem involves calculating electric potential and work related to charges positioned in a coordinate system.

**Problem Statement:**

1. **Charge $ q_1 = 2.5 \, \mu C$ is located at the coordinate system origin.**
- Charge $ q_2 = 1 \, \mu C$ is placed at $(a, 0)$ where $ a = 1.85 \, \text{m} $. The point $ P $ is located at $(0, a)$.
- A third charge $ q_3 = 3.5 \, \mu C$ will be placed later.

**Diagram Explanation:**
- There is a coordinate system with $ q_1 $ at the origin $(0, 0)$.
- $ q_2 $ is on the x-axis at position $(a, 0)$.
- Point $ P $ is on the y-axis at position $(0, a)$.

**Tasks:**

(a) **Find the electric potential $ V_p $ at point $ P $ in volts.** Assume the potential is zero at infinity.
- Equation: $ V_p = $

(b) **Determine how much work $ W_i $, in joules,** would you need to do to bring the third charge $ q_3 $ from very far away to the point $ P $.
- Equation: $ W_i = $

(c) **Calculate the total potential energy $ U $ (in joules)** of the final configuration of the three charges.
- Equation: $ U = $

**Instructions for Solving:**
- Use the formula for electric potential, $ V = \frac{k \cdot q}{r} $, where $ k = 8.99 \times 10^9 \, \text{N m}^2/\text{C}^2 $.
- Calculate the potential contributions from each charge at point $ P $ and sum them.
- Use the work-energy principle to find the work done in bringing charge $ q_3 $ to point $ P $.
- Determine the total potential energy by considering pairwise interactions.

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