Q3: Electrons accelerated from rest through 400 V are introduced at A into a uniform electricfield E of intensity 150 V/cm as shown. If the electrons emerge at B 5x109 s later, determine(a) distance AB and (b) the angle 0.BAns: 4.98 cm; 33.6°.Q4: Two parallel plates A and B are spaced 1 cm apart. A stream of electrons is projected at avelocity corresponding to an accelerating voltage of 2 KV into the space between the platedthrough a hole in plate A at an angle of 30° to it. Find the value and the polarity of the potentialdifference which is required between A and B in order that the electrops just graze plate B.Ans: 500 V

Answered: Image /qna-images/answer/cac18c1a-a304-4b44-b762-4be7e2cb260d.jpg

Q3: Electrons accelerated from rest through 400 V are introduced at A into a uniform field E of intensity 150 V/cm as shown. If the electrons emerge at B 5x10-9 s later, determine (a) distance AB and (b) the angle e. B A.

Q3: Electrons accelerated from rest through 400 V are introduced at A into a uniform field E of intensity 150 V/cm as shown. If the electrons emerge at B 5x10-9 s later, determine (a) distance AB and (b) the angle e. B A.

University Physics Volume 2

18th Edition

ISBN:9781938168161

Author:OpenStax

Publisher:OpenStax

Chapter7: Electric Potential

Section: Chapter Questions

Problem 41P: An electron is to be accelerated in a uniform electric field having a strength of 2.00106 V/m. (a)...

See similar textbooks

Similar questions

An electron is to be accelerated in a uniform electric field having a strength of 2.00106 V/m. (a) What energy in keV is given to the electron if it is accelerated through 0.400 m? (b) Over what distance would it have to be accelerated to increase its energy by 50.0 GeV?
You are working on a research project in which you must control the direction of travel of electrons using deflection plates. You have devised the apparatus shown in Figure P22.28. The plates are of length = 0.500 m and are separated by a distance d = 3.00 cm. Electrons are fired at vi = 5.00 106 m/s into a uniform electric field from the left edge of the lower, positive plate, aimed directly at the right edge of the upper, negative plate. Therefore, if there is no electric field between the plates, the electrons will follow the broken line in the figure. With an electric field existing between the plates, the electrons will follow a curved path, bending downward. You need to determine (a) the range of angles over which the electron can leave the apparatus and (b) the electric field required to give the maximum possible deviation angle. Figure P22.28
A double charged ion is accelerated to an energy of 32.0 keV by the electric field between two parallel conducting plates separated by 2.00 cm. What is the electric field strength between the plates?
An electron is to be accelerated in a uniform electric field having a strength of 2.00106 V/m. (a) What energy in keV is given to the electron if it is accelerated through 0.400 m? (b) Over what distance would it have to be accelerated to increase its energy by 50.0 GeV?
Question 1 a) In J. J. Thomson experiment (1897), an electron moving horizontally with a constant speed vo enters in between the horizontal plates of a capacitor. The electric field strength between the plates of length L and distance d, is E. The vertical deviation of the electron at the moment of exit from the field region is measured to be Y. Derive the expression giving the electron's charge to mass ratio, i.e. e/m to be 2v,Y/CEL). (Recall that Thomson received Nobel Prize for his achievement.) b) Calculate e/m, knowing the following data. E=1.6x10* Newton/Coulomb, L=10 cm, Y=2.9 cm, v=2.19x10* km/s. (Be careful to use coherent units.)
Q6//The question in the image
350 / 746 110% 33. Four parallel plates are connected in a vacuum as in Figure 4. An electron, essentially at rest, drifts into the hole in plate X and is accelerated to the right. The vertical motion of the electron continues to be negli- gible. The electron passes through holes W and Y, then continues moving toward plate Z. Using the information given in the diagram, calculate (a) the speed of the electron at hole W (b) the distance from plate Z to the point at which the electron changes direction (c) the speed of the electron when it arrives back at plate X S a 37. I 38. f +4.0cm++4.0 cm+4.0 cm- t 3.0 x 102 V 5.0 x 10 V Figure 4 34. Two a particles, separated by an enormous distance, approach each other. Each has an initial speed of
a) Calculate the acceleration in meters per second squared of the electron if the field strength is 2.50 ✕ 104 N/C.
Oppositely charged parallel plates are used to create fairly uniform electric fields. In our case, we use a setup like this so that the field is 5000N/C (0.5% uncertainty). The positively charged plate has a hole in it. The electron is situated against the negative plate, away and is directly opposite the hole. A metre rule is used to measure the plate separate and it is found to be 1.0002m apart. Using ideas from this chapter determine the exit speed of the electron and the uncertainty on this value.
An electron is between two parallel plates. What would be the strength of the electric field in order to have an acceleration of 8.34 x 1010 m/s2. e = 1.60 x 10-19 C, melectron = 9.1 x 10-31 kg.
An electron is launched at a 45º angle and a speed of 5.0x106 m/s from the positive plate of the parallel-plate capacitor shown in the figure. The electron lands 4.0 cm away. a) What is the electric field strength inside the capacitor? b) What is the minimum spacing between the plates?
3. A particle (q = 3.0 µC, m = 10 g) has a speed of 40 m/s when it enters a region where the electric field that has a constant magnitude of 100 N/C and the direction of the velocity with the electric field is 30 degree . What is the speed of the particle 5 s after it enters this region?