{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

inclasstest 03

# inclasstest 03 - Version 124/ABDDA inclasstest 03...

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

Version 124/ABDDA – inclasstest 03 – Turner – (59070) 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 102 cm. The total chargeon the arc is - 38 . 8 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. 976432.0 2. 371869.0 3. 628695.0 4. 2321950.0 5. 519869.0 6. 395715.0 7. 213380.0 8. 713156.0 9. 582775.0 10. 112174.0 Correct answer: 2 . 1338 × 10 5 N / C. Explanation: Let : q = - 38 . 8 μ C , r = 102 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 ]}

### Page1 / 2

inclasstest 03 - Version 124/ABDDA inclasstest 03...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online