Homework 4-solutions - soto(rrs946 – Homework 4 –...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: soto (rrs946) – Homework 4 – Hoffmann – (56605) 1 This print-out should have 20 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 3) 10.0 points A proton is released from rest in a uniform electric field of magnitude 40000 V / m di- rected along the positive x axis. The proton undergoes a displacement of 0 . 2 m in the di- rection of the electric field as shown in the figure. The mass of a proton is 1 . 672623 × 10 − 27 kg. + + + + + + + + + + + + +------------- v A = 0 + . 2 m v 40000 V / m A B Find the change in the electric potential if the proton moves from the point A to B . Correct answer:- 8000 V. Explanation: Let : q p = 1 . 60218 × 10 − 19 C , m p = 1 . 67262 × 10 − 27 kg , E = 40000 V / m , and x = 0 . 2 m . From the equation Δ V =- vector E · integraldisplay B A dvectors or in a non-calculus way for the straight path AB , Δ V =- vector E · vector AB we obtain Δ V =- E x =- (40000 V / m) (0 . 2 m) =- 8000 V . The negative result tells us that the electric potential of the proton decreases as it moves from A to B . 002 (part 2 of 3) 10.0 points Find the change in potential energy of the proton for this displacement. Correct answer:- 1 . 28174 × 10 − 15 J. Explanation: The change in the potential energy of the proton is Δ U = q p Δ V = (1 . 60218 × 10 − 19 C) (- 8000 V) =- 1 . 28174 × 10 − 15 J . The negative sign here means that the poten- tial energy of the proton decreases as it moves in the direction of the electric field. This makes sense since, as the proton accelerates in the direction of the field, it gains kinetic energy and at the same time (the field) loses potential energy (the energy is conserved). 003 (part 3 of 3) 10.0 points Apply the principle of energy conservation to find the speed of the proton after it has moved . 2 m, starting from rest. Correct answer: 1 . 23799 × 10 6 m / s. Explanation: Conservation of energy in this case is Δ K + Δ U = 0 . K f- K i + Δ U = 0 1 2 m p v 2 f + Δ U = 0 v f = radicalBigg- 2 Δ U m p = radicalBigg- 2 (- 1 . 28174 × 10 − 15 J) 1 . 67262 × 10 − 27 kg = 1 . 23799 × 10 6 m / s . soto (rrs946) – Homework 4 – Hoffmann – (56605) 2 004 (part 1 of 2) 10.0 points An electron moving parallel to the x axis has an initial speed of 5 × 10 6 m / s at the origin. Its speed is reduced to 1 × 10 5 m / s at the point x P , 5 cm away from the origin. The mass of the electron is 9 . 10939 × 10 − 31 kg and the charge of the electron is- 1 . 60218 × 10 − 19 C. Calculate the absolute value of the poten- tial difference between this point and the ori- gin. Correct answer: 71 . 042 V....
View Full Document

Page1 / 9

Homework 4-solutions - soto(rrs946 – Homework 4 –...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online