{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

midterm 01-solutions

# midterm 01-solutions - Version 068/ABABA midterm 01...

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

Version 068/ABABA – midterm 01 – turner – (56705) 1 This print-out should have 19 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Three identical point charges, each of mass 120 g and charge + q , hang from three strings, as in the figure. 9 . 8 m / s 2 10 . 4 cm 120 g + q 10 . 4 cm 120 g + q 37 120 g + q If the lengths of the left and right strings are each 10 . 4 cm, and each forms an an- gle of 37 with the vertical, determine the value of q . The acceleration of gravity is 9 . 8 m / s 2 , and the Coulomb constant is 8 . 98755 × 10 9 N · m 2 / C 2 . 1. 0.618551 2. 0.567833 3. 0.488958 4. 0.651312 5. 0.775448 6. 1.014 7. 0.555881 8. 0.934425 9. 0.711318 10. 0.517167 Correct answer: 0 . 555881 μ C. Explanation: Let : θ = 37 , m = 120 g = 0 . 12 kg , L = 10 . 4 cm = 0 . 104 m , g = 9 . 8 m / s 2 , and k e = 8 . 98755 × 10 9 N · m 2 / C 2 . Consider the forces acting on the charge on the right. There must be an electrostatic force F acting on this charge, keeping it balanced against the force of gravity m g . The electro- static force is due to the other two charges and is therefore horizontal. In the x -direction F - T sin θ = 0 . In the y -direction T cos θ - m g = 0 . F sin θ F cos θ = F e m g F = m g tan θ (1) = (0 . 12 kg) (9 . 8 m / s 2 ) tan 37 = 0 . 88618 N . The distance between the right charge and the middle charge is L sin θ , and the distance to the left one is twice that. Since all charges are of the same sign, both forces on the right charge are repulsive (pointing to the right). We can add the magnitudes F = k e q q ( L sin θ ) 2 + k e q q (2 L sin θ ) 2 F = 5 k e q 2 4 L 2 sin 2 θ (2) q 2 = 4 F L 2 sin 2 θ 5 k e (3) q = 2 L sin θ F 5 k e = 2 (0 . 104 m) sin 37 × 0 . 88618 N 5 (8 . 98755 × 10 9 N · m 2 / C 2 ) = 5 . 55881 × 10 - 7 C = 0 . 555881 μ C . 002 10.0 points Three small spheres carry equal amounts of electric charge. They are equally spaced and lie along the same line. + - - What is the direction of the net electric force on each charge due to the other charge?

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

View Full Document
Version 068/ABABA – midterm 01 – turner – (56705) 2 1. + - - 2. + - - 3. + - - 4. + - - 5. + - - correct 6. + - - 7. + - - 8. + - - 9. + - - 10. + - - Explanation: Since like charges repel and unlike charge attract, + - - 003 10.0 points A charge of 8 . 4 μ C is 20 cm above the center of a square of side length 40 cm. The permittivity of free space is 8 . 85 × 10 - 12 C 2 / N · m 2 . Find the flux through the square. 1. 154426.0 2. 73446.3 3. 116761.0 4. 158192.0 5. 139360.0 6. 82862.5 7. 90395.5 8. 165725.0 9. 124294.0 10. 94162.0 Correct answer: 1 . 58192 × 10 5 N · m 2 / C. Explanation: Let : q = 8 . 4 μ C = 8 . 4 × 10 - 6 C , = 20 cm = 0 . 2 m , s = 40 cm = 0 . 4 m , and 0 = 8 . 85 × 10 - 12 C 2 / N · m 2 . The point charge is at the center of the cube. Applying Gauss’ law, the flux through one face of the square is Φ square = 1 6 Φ total = 1 6 q inside 0 = 1 6 8 . 4 × 10 - 6 C 8 . 85 × 10 - 12 C 2 / N · m 2 = 1 . 58192 × 10 5 N · m 2 / C . keywords: 004 10.0 points A charge moves a distance of 2 . 3 cm in the direction of a uniform electric field having a magnitude of 239 N / C. The electrical potential energy of the charge decreases by 72 . 3544 × 10 - 19 J as it moves.
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 13

midterm 01-solutions - Version 068/ABABA midterm 01...

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

View Full Document
Ask a homework question - tutors are online