A 5.9 mC charge is accelerated by a potential difference of -7 V, which makes it go from a velocity of 4.1x103 m/s (left) to a velocity of 9.2x103 m/s (left). a) Since the particle was traveling to the left, and sped up, what was the direction of the acceleration of the charge, left or right?b) What then must have been the direction of the force on the charge?c) And then what must have been the direction of the electric field in the area?d) Given that you know the electric field has to point from high potential to low potential, is the electron traveling from a low to a high potential, or is it traveling from a high to a low potential? (check that your answer matches with the sign of the given potential difference!) 5. A 5.9 µC charge is accelerated by a potential difference of -7 V, which makes it go from a velocity of4.1x103 m/s (left) to a velocity of 9.2x10³ m/s (left).a. Since the particle was traveling to the left, and sped up, what was the direction of the acceleration of thecharge, left or right?b. What then must have been the direction of the force on the charge?c. And then what must have been the direction of the electric field in the area?d. Given that you know the electric field has to point from high potential to low potential, is the electrontraveling from a low to a high potential, or is it traveling from a high to a low potential? (check that youranswer matches with the sign of the given potential difference!

since the particle is going from region A to region B, the potential difference = VB-VA=0 -7 =-7…

Answered: 5. A 5.9 µC charge is accelerated by a…

Physics for Scientists and Engineers

10th Edition

ISBN:9781337553278

Author:Raymond A. Serway, John W. Jewett

Publisher:Raymond A. Serway, John W. Jewett

Chapter24: Electric Potential

Section: Chapter Questions

Problem 2P: (a) Find the electric potential difference Ve required to stop an electron (called a stopping...

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A 5.9 mC charge is accelerated by a potential difference of -7 V, which makes it go from a velocity of 4.1x10³ m/s (left) to a velocity of 9.2x10³ m/s (left).

a) Since the particle was traveling to the left, and sped up, what was the direction of the acceleration of the charge, left or right?

b) What then must have been the direction of the force on the charge?

c) And then what must have been the direction of the electric field in the area?

d) Given that you know the electric field has to point from high potential to low potential, is the electron traveling from a low to a high potential, or is it traveling from a high to a low potential? (check that your answer matches with the sign of the given potential difference!)

5. A 5.9 µC charge is accelerated by a potential difference of -7 V, which makes it go from a velocity of
4.1x103 m/s (left) to a velocity of 9.2x10³ m/s (left).
a. Since the particle was traveling to the left, and sped up, what was the direction of the acceleration of the
charge, left or right?
b. What then must have been the direction of the force on the charge?
c. And then what must have been the direction of the electric field in the area?
d. Given that you know the electric field has to point from high potential to low potential, is the electron
traveling from a low to a high potential, or is it traveling from a high to a low potential? (check that your
answer matches with the sign of the given potential difference!

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