Question 1 of 6 Question 2 of 6 Question 3 of 6 Part 1 of 2 - Electric Field and

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Question 3 of 6

Part 1 of 2 - Electric Field and Potential A constant voltage of V is maintained between two metallic parallel plates.
An electron is sent through a small hole in the parallel plates and is accelerated by the electric field between them, cf. left figure. Answer the following questions.

Question 1 of 6

5.0 Points By going through the electric field, the electron gained a kinetic energy of 10.0 eV (electron Volts).
What was the potential difference between the two metallic plates? A. 1.60×10-19 V B. 5.0 V C. 10.0 V D. 1.60×10-18 V E. 20.0 V Reset Selection

Question 2 of 6

5.0 Points What is the speed of the electron after going through the apparatus? Assume that the electron started from rest. A. 9.37×105 m/s B. 2.65×106 m/s C. 1.87×106 m/s D. 1.33×106 m/s E. 2.30×106 m/s Reset Selection

Question 3 of 6

5.0 Points A proton is sent through the apparatus in the opposite direction as shown in the right figure. What is the speed of the proton after acceleration? Assume the proton started from rest. A. 3.10×104 m/s B. 4.38×104 m/s C. 2.19×104 m/s D. 5.36×104 m/s E. 6.19×104 m/s

Explanation / Answer

part 1 )

KE = PE

10 eV = 1.6 x 10^-18 J

1.6 x 10^-18 J = qV

q = 1.6 x 10^-19 J

V = 10 V

part b )

KE = 1/2mv^2

1.6 x 10^-18 J = mv^2/2

m = mass of electron = 9.31 x 10^-31 kg

v = sqrt(2*1.6*10^-18/9.31*10^-31)

v = 1.87 x 10^6 m/s

part 3 )

v = sqrt(2*1.6*10^-18/m)

m = mass of proton = 1.67 x 10^-27 kg

v = 4.38 x 10^4 m/s

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