Multiple-Concept Example 4 deals with the concepts that are important in this problem. As illustrated in the figure, a negatively charged particle is released from rest at point B and accelerates until it reaches point A. The mass and charge of the particle are 1.23 x 104 kg and -1.04 x 105 C, respectively. Only the gravitational force and the electrostatic force act on the particle, which moves on a horizontal straight line without rotating. The electric potential at A is 36.0 V greater than that at B; in other words, VA - VB = 36.0 V. What is the translational speed of the particle at point A?

College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
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Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
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Multiple-Concept Example 4 deals with the concepts that are important in this problem. As illustrated in the figure, a negatively
charged particle is released from rest at point B and accelerates until it reaches point A. The mass and charge of the particle are 1.23 x
104 kg and -1.04 x 105 C, respectively. Only the gravitational force and the electrostatic force act on the particle, which moves on a
horizontal straight line without rotating. The electric potential at A is 36.0 V greater than that at B; in other words, VA - VB = 36.0 V.
What is the translational speed of the particle at point A?
Number
Units
✪
B
VB=0 m/s
Transcribed Image Text:Multiple-Concept Example 4 deals with the concepts that are important in this problem. As illustrated in the figure, a negatively charged particle is released from rest at point B and accelerates until it reaches point A. The mass and charge of the particle are 1.23 x 104 kg and -1.04 x 105 C, respectively. Only the gravitational force and the electrostatic force act on the particle, which moves on a horizontal straight line without rotating. The electric potential at A is 36.0 V greater than that at B; in other words, VA - VB = 36.0 V. What is the translational speed of the particle at point A? Number Units ✪ B VB=0 m/s
Expert Solution
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Given

m=1.23×10-4Kg (mass of particle)q=1.04×10-5 C (charge of particle)vB=0 (velocity at point B)VA-VB=36 V (potential difference between A and B)vA= velocity at A

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