Part C: A point charge q3 = -5.00 μC moves from point a to point b. How much work is done on q3 by the electric forces exerted by q1 and q2? Express your answer with the appropriate units. Wab=__
Part C: A point charge q3 = -5.00 μC moves from point a to point b. How much work is done on q3 by the electric forces exerted by q1 and q2? Express your answer with the appropriate units. Wab=__
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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Part C: A point charge q3 = -5.00 μC moves from point a to point b. How much work is done on q3 by the electric forces exerted by q1 and q2?
Express your answer with the appropriate units.
Wab=__
![## Part A
**Question:**
What is the electric potential at point \( a \) due to \( q_1 \) and \( q_2 \)?
**Answer:**
\[ V_a = 0 \, \text{V} \]
### Explanation:
**Correct**
**IDENTIFY:** The potential at any point is the scalar sum of the potentials due to individual charges.
**SET UP:**
\[ V = kq/r \]
\[ W_{ab} = q(V_a - V_b) \]
**EXECUTE:**
\[ r_{a1} = r_{a2} = \frac{1}{2} \sqrt{(0.0100 \, \text{m})^2 + (0.0100 \, \text{m})^2} = 0.0071 \, \text{m} \]
\[ V_a = k\left(\frac{q_1}{r_{a1}} + \frac{q_2}{r_{a2}}\right) = 0 \, \text{V} \]
---
## Part B
**Question:**
What is the electric potential at point \( b \)?
**Answer:**
\[ V_b = -5.27 \times 10^5 \, \text{V} \]
### Explanation:
**Correct**
Given:
\[ r_{b1} = 0.0141 \, \text{m}, \, r_{b2} = 0.0100 \, \text{m} \]
Calculate:
\[ V_b = k\left(\frac{q_1}{r_{b1}} + \frac{q_2}{r_{b2}}\right) \]
Using:
\[ k = 8.99 \times 10^9 \, \text{N} \cdot \text{m}^2/\text{C}^2 \]
\[ V_b = (8.99 \times 10^9 \, \text{N} \cdot \text{m}^2/\text{C}^2) \left(\frac{+2.00 \times 10^{-6} \, \text{C}}{0.0141 \, \text{m}} + \frac{-2.00 \times 10^{-6} \, \text{C}}{0.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb2f5b0ad-df1d-48c6-a9a3-5338b8167143%2Ffbb35a4e-7be0-4dc1-8cbb-324dcd800f0d%2F9obuob_processed.png&w=3840&q=75)
Transcribed Image Text:## Part A
**Question:**
What is the electric potential at point \( a \) due to \( q_1 \) and \( q_2 \)?
**Answer:**
\[ V_a = 0 \, \text{V} \]
### Explanation:
**Correct**
**IDENTIFY:** The potential at any point is the scalar sum of the potentials due to individual charges.
**SET UP:**
\[ V = kq/r \]
\[ W_{ab} = q(V_a - V_b) \]
**EXECUTE:**
\[ r_{a1} = r_{a2} = \frac{1}{2} \sqrt{(0.0100 \, \text{m})^2 + (0.0100 \, \text{m})^2} = 0.0071 \, \text{m} \]
\[ V_a = k\left(\frac{q_1}{r_{a1}} + \frac{q_2}{r_{a2}}\right) = 0 \, \text{V} \]
---
## Part B
**Question:**
What is the electric potential at point \( b \)?
**Answer:**
\[ V_b = -5.27 \times 10^5 \, \text{V} \]
### Explanation:
**Correct**
Given:
\[ r_{b1} = 0.0141 \, \text{m}, \, r_{b2} = 0.0100 \, \text{m} \]
Calculate:
\[ V_b = k\left(\frac{q_1}{r_{b1}} + \frac{q_2}{r_{b2}}\right) \]
Using:
\[ k = 8.99 \times 10^9 \, \text{N} \cdot \text{m}^2/\text{C}^2 \]
\[ V_b = (8.99 \times 10^9 \, \text{N} \cdot \text{m}^2/\text{C}^2) \left(\frac{+2.00 \times 10^{-6} \, \text{C}}{0.0141 \, \text{m}} + \frac{-2.00 \times 10^{-6} \, \text{C}}{0.

Transcribed Image Text:Point charges \( q_1 = +2.00 \, \mu \text{C} \) and \( q_2 = -2.00 \, \mu \text{C} \) are placed at adjacent corners of a square for which the length of each side is 1.00 cm. Point \( a \) is at the center of the square, and point \( b \) is at the empty corner closest to \( q_2 \). Take the electric potential to be zero at a distance far from both charges.
Expert Solution

Step 1: Work is done on a charge
We know the work is done on a charge q to move from position a to b is
Where:
is potential at position a, and
is potential at position b.
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