A) An electron is to be accelerated from a velocity of 5.00×106 m/s to a velocity...

A) An electron is to be accelerated from a velocity of 5.00×106 m/s to a velocity of 7.00×106 m/s . Through what potential difference must the electron pass to accomplish this?

B) Through what potential difference must the electron pass if it is to be slowed from 7.00×106 m/s to a halt?

Solutions

Expert Solution

genius_generous answered 2 hours ago

Next > < Previous

Related Solutions

When an energetic electron with the velocity of 1.00 x 106 m s-1 was collided with...

When an energetic electron with the velocity of 1.00 x 106 m s-1 was collided with Rb atom (assume stationary), Rb- anion moves at 313 m s-1 and released 382.4 nm of visible light. Calculate the energy change during the formation of Rb- anion from Rb atom in kJ mol-1

A proton is traveling horizontally to the right at 5.00×106 m/s . Find (a)the magnitude and...

A proton is traveling horizontally to the right at 5.00×106 m/s . Find (a)the magnitude and (b) direction of the weakest electric field that can bring the proton uniformly to rest over a distance of 3.40 cm . a. E =___N/C b.∘ counterclockwise from the left direction c. How much time does it take the proton to stop after entering the field? d. What minimum field ((a)magnitude and (b)direction) would be needed to stop an electron under the conditions of...

An electron is to be accelerated in a uniform electric field having a strength of 4.58×106...

An electron is to be accelerated in a uniform electric field having a strength of 4.58×106 V/m. (a) What energy in keV is given to the electron if it is accelerated through 0.562 m? (b) Over what distance would it have to be accelerated to increase its energy by 58.0 GeV? Draw a diagram and show your parameters and all your work.

A proton is traveling horizontally to the right at 5.00×106 m/s Part A, B: Find (a)the...

A proton is traveling horizontally to the right at 5.00×106 m/s Part A, B: Find (a)the magnitude and (b) direction of the weakest electric field that can bring the proton uniformly to rest over a distance of 3.40 cm . Part C: How much time does it take the proton to stop after entering the field? Part D, E: What minimum field ((a)magnitude and (b)direction) would be needed to stop an electron under the conditions of part (a)?

a 5.00 kg ball, moving to the right at a velocity of 2.00 m/s on the...

a 5.00 kg ball, moving to the right at a velocity of 2.00 m/s on the frictionless table, collide head on with a stationary 7.50 kg ball. find the final velocities of the balls if the collision is elastic, completely inelastic the balls stick together

A proton has an initial velocity of 7.6 × 106 m/s in the horizontal direction. It...

A proton has an initial velocity of 7.6 × 106 m/s in the horizontal direction. It enters a uniform electric field of 8800 N/C directed vertically. a) Ignoring gravitational effects, find the time it takes the proton to travel 0.127 m horizontally. The mass of the proton is 1.6726 × 10−27 kg and the fundamental charge is 1.602 × 10−19 C . Answer in units of ns. b)What is the vertical displacement of the proton after the electric field acts...

A proton with an initial velocity of 3.9 106 m/s enters a parallel-plate capacitor at an...

A proton with an initial velocity of 3.9 106 m/s enters a parallel-plate capacitor at an angle of 46° through a small hole in the bottom plate as shown in the figure below. If the plates are separated by a distance d = 2.1 cm, what is the magnitude of the minimum electric field to prevent the proton from hitting the top plate? ___ N/C Compare your answer with the results obtained if the proton was replaced with an electron...

An electron in a TV camera tube is moving at 7.20×106 m/s in a magnetic field...

An electron in a TV camera tube is moving at 7.20×106 m/s in a magnetic field of strength 59 mT. Without knowing the direction of the field, what can you say about the greatest and least magnitude of the force acting on the electron due to the field? Maximum force? Minimum force? At one point the acceleration of the electron is 6.327×1016 m/s2. What is the angle between the electron velocity and the magnetic field? (deg)

An electron in a TV camera tube is moving at 6.20×106 m/s in a magnetic field...

An electron in a TV camera tube is moving at 6.20×106 m/s in a magnetic field of strength 75 mT. Without knowing the direction of the field, what can you say about the greatest and least magnitude of the force acting on the electron due to the field? Maximum force? Minimum force? At one point the acceleration of the electron is 7.767×1016 m/s2. What is the angle between the electron velocity and the magnetic field? (deg)

A point charge q = 40 nC has a velocity 3* 106 m/s in the direction...

A point charge q = 40 nC has a velocity 3* 106 m/s in the direction av = 0.2 ax + 0.75 ay – 0.3 az. Calculate; a) The magnitude of the force on the charge due to the field B = 3ax + - 4ay + 6az mT. b) The magnitude of the force on the charge due to the field E = -2ax + 4ay + 5az kV/m c) The Lorentz force due to the two fields in...

Subjects