This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Reilly (mar3978) – homework 08 – Turner – (56725) 1 This printout should have 12 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 2) 10.0 points A proton is accelerated through a potential difference of 3 . 5 × 10 6 V. a) How much kinetic energy has the proton acquired? Correct answer: 5 . 6 × 10 − 13 J. Explanation: Let : Δ V = 3 . 5 × 10 6 V and q = 1 . 60 × 10 − 19 C . Δ K = Δ U = q Δ V = (1 . 60 × 10 − 19 C) (3 . 5 × 10 6 V) = 5 . 6 × 10 − 13 J . 002 (part 2 of 2) 10.0 points b) If the proton started at rest, how fast is it moving? Correct answer: 2 . 58738 × 10 7 m / s. Explanation: Let : m = 1 . 673 × 10 − 27 kg . Since K i = 0 J , Δ K = K f = 1 2 mv 2 f v f = radicalbigg 2 K f m = radicalBigg 2 (5 . 6 × 10 − 13 J) 1 . 673 × 10 − 27 kg = 2 . 58738 × 10 7 m / s . 003 10.0 points Points A (3 m, 3 m) and B (6 m, 8 m) are in a region where the electric field is uniform and given by vector E = E x ˆ ı + E y ˆ , where E x = 5 N / C and E y = 4 N / C. What is the potential difference V A V B ? Correct answer: 35 V. Explanation: Let : E x = 5 N / C , E y = 4 N / C , ( x A ,y A ) = (3 m , 3 m) , and ( x B ,y B ) = (6 m , 8 m) . We know V ( A ) V ( B ) = integraldisplay A B vector E · dvectors = integraldisplay B A vector E · dvectors For a uniform electric field vector E = E x ˆ ı + E y ˆ . Now consider the term E x ˆ ı · dvectors in the inte grand. E x is just a constant and ˆ ı · dvectors may be interpreted as the projection of dvectors onto x , so that E x ˆ ı · dvectors = E x dx. Likewise E y ˆ · dvectors = E y dy . Or more simply, dvectors = dx ˆ ı + dy ˆ dotting it with E x ˆ ı + E y ˆ gives the same result as above. Therefore V A V B = E x integraldisplay x B x A dx + E y integraldisplay y B y A dy = (5 N / C) (6 m 3 m) + (4 N / C) (8 m 3 m) = 35 V . Note that the potential difference is inde pendent of the path taken from A to B. 004 10.0 points Reilly (mar3978) – homework 08 – Turner – (56725) 2 Consider a circular arc of constant linear charge density λ as shown below. x y 3 4 π + + + + + + + + + + + + + + + + + + + O r What is the potential V O at the origin O due to this arc? 1. V O = 3 20 λ ǫ 2. V O = 0 3. V O = 5 32 λ ǫ 4. V O = 1 7 λ ǫ 5. V O = 5 24 λ ǫ 6. V O = 5 36 λ ǫ 7. V O = 1 5 λ ǫ 8. V O = 3 28 λ ǫ 9. V O = 3 16 λ ǫ correct 10. V O = 5 28 λ ǫ Explanation: The potential at a point due to a continuous charge distribution can be found using V = k e integraldisplay dq r . In this case, with linear charge density λ, dq = λds = λr dθ , so V = k e integraldisplay 3 4 π λdθ = 1 4 π ǫ integraldisplay 3 4 π λdθ = λ 4 π ǫ θ vextendsingle vextendsingle vextendsingle vextendsingle 3 4 π = λ 4 π ǫ parenleftbigg 3 4 π parenrightbigg = 3 16 λ ǫ ....
View
Full
Document
This note was uploaded on 11/30/2010 for the course PHY 60230 taught by Professor Turner,j during the Spring '10 term at University of Texas at Austin.
 Spring '10
 TURNER,J

Click to edit the document details