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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 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
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
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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.
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.
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.
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

straight W subscript ab equals straight q left parenthesis straight V subscript straight a minus straight V subscript straight b right parenthesis

Where:

straight V subscript straight a is potential at position a, and straight V subscript straight bis potential at position b.

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