solution3_pdf

solution3_pdf - maini(nm7637 hw3 Shneidman(12108 This...

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

View Full Document Right Arrow Icon
maini (nm7637) – hw3 – Shneidman – (12108) 1 This print-out should have 15 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 (part 1 oF 2) 10.0 points ±our point charges are placed at the Four cor- ners oF a square, where each side has a length a . The upper two charges have identical pos- itive charge, and the two lower charges have charges oF the same magnitude as the frst two but opposite sign. That is, q 1 = q 2 = q and q 3 = q 4 = q where q > 0. b b b b b q 4 q 2 q 3 q 1 P j i Determine the direction oF the electric feld at the center (point P ). 1. i 2. j correct 3. j 4. 1 2 ( i j ) 5. 1 2 ( i + j ) 6. i 7. 1 2 ( i + j ) 8. 1 2 ( i j ) Explanation: The direction is already clear: all the x - components cancel, and the lower charges at- tract and the top ones repel, so the answer is j . 002 (part 2 oF 2) 10.0 points Determine the magnitude oF the electric feld at point P . 1. 8 k e q a 2 2. 4 k e q a 2 3. k e q a 2 4. 2 k e q a 2 5. 4 2 k e q a 2 correct 6. 2 2 k e q a 2 Explanation: The electric feld due to a point charge is E = k q r 2 . The feld at P must be Found through super- position oF felds due to the Four point par- ticles. Recall that the direction oF a feld at a point P is the direction in which a positive test charge at that point would Feel a Force. The top two charges ( q 1 and q 2 ) give rise to repulsive Forces on a positive test charge at P . The felds E 1 and E 2 are directed as in the fgure P E 2 E 1 We see that the x -components cancel. The distance From the center to one oF the particles is a 2 . ThereFore, the magnitudes are E 1 y = E 2 y = 1 2 k q p a 2 P 2 = 2 k q a 2 The bottom two charges are negative, so the Forces on a positive test charge at the center are attractive:
Background image of page 1

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

View Full Document Right Arrow Icon
maini (nm7637) – hw3 – Shneidman – (12108) 2 P E 4 E 3 Similarly, E 3 y = E 4 y = 2 k q a 2 so the total magnitude at P (since the x - components cancel) is E = 4 E 1 y = 4 2 k q a 2 . 003 (part 1 of 3) 10.0 points Coulomb constant is 8 . 98755 × 10 9 N m 2 / C 2 . The 1 . 68 μ C charge is at the origin and a 8 . 54 μ C charge is 10 cm to the right, as shown in the Fgure. x O I II III y 1 . 68 μ C 8 . 54 μ C 10 cm Identify the direction of v E in the region II (0 < x < 10 cm, along the x -axis). 1. all possibilities: right, left, or zero 2. down 3. None of these 4. left 5. right correct 6. up Explanation: Let : q 1 = 1 . 68 μ C , q 2 = 8 . 54 μ C , and a = 10 cm . The direction of the electric Feld at a point P is the direction that a positive charge would move if placed at P . A positive charge placed in region II would be attracted to q 2 and repelled by q 1 . Thus the direction is to the right.
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

Page1 / 7

solution3_pdf - maini(nm7637 hw3 Shneidman(12108 This...

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