{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

solut56ion4_pdf

# solut56ion4_pdf - Version 097/ABCAB inclasstest 03...

This preview shows pages 1–2. Sign up to view the full content.

Version 097/ABCAB – inclasstest 03 – Turner – (56725) 1 This print-out should have 2 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Consider the setup shown in the figure be- low, where the arc is a semicircle with radius of 123 cm. The total chargeon the arc is - 37 . 7 μ C, and distributed uniformly on the semicircle. The Coloumb constant is 8 . 98755 × 10 9 N m 2 / C 2 . x y - - - - - - - - - - - - - - - - - - Δ θ θ r x y I II III IV B A O Determine the magnitude of the electric field at O . 1. 223145.0 2. 1550570.0 3. 113107.0 4. 395715.0 5. 125362.0 6. 371869.0 7. 780225.0 8. 1125790.0 9. 876921.0 10. 142578.0 Correct answer: 1 . 42578 × 10 5 N / C. Explanation: Let : q = - 37 . 7 μ C , r = 123 cm , and k = 8 . 98755 × 10 9 N m 2 / C 2 . x y - - - - - - - - - - - - - - - - - - Δ θ θ r x y I II III IV B A O E By symmetry of the semicircle, the y - component of the electric field at the center is E y = 0 . There we need consider only the x -component of the electric Field. Using ds = r dθ , we have Δ q = λ ds = λ r dθ = q π r r dθ = q π dθ , so Δ E x = k | Δ q | cos θ r 2 = k | q | π r 2 cos θ Δ θ .

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

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern