HW #2-solutions

# HW #2-solutions - zapata (jz883) HW #2 Antoniewicz (56495)...

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

zapata (jz883) – HW #2 – Antoniewicz – (56495) 1 This print-out should have 23 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 (part 1 oF 4) 3.0 points A charge 1 . 1 × 10 5 C is fxed on the x -axis at 9 m, and a charge 8 × 10 6 C is fxed on the y -axis at 5 m, as shown on the diagram. 9 m 1 . 1 × 10 5 C 5 m 8 × 10 6 C Calculate the magnitude oF the resultant electric feld v E at the origin. Correct answer: 3124 . 29 N / C. Explanation: Let : k = 8 . 98755 × 10 9 N · m 2 / C 2 , Q x = 1 . 1 × 10 5 C , Q y = 8 × 10 6 C , Q = 3 × 10 6 C , x = 9 m , and y = 5 m . ±or charge Q x , | E x | = k | Q x | x 2 = (8 . 98755 × 10 9 N · m 2 / C 2 ) × v v 1 . 1 × 10 5 C v v (9 m) 2 = 1220 . 53 N / C . ±or charge Q y , | E y | = k | Q y | y 2 = (8 . 98755 × 10 9 N · m 2 / C 2 ) × 8 × 10 6 C (5 m) 2 = 2876 . 02 N / C . The resultant electric feld at the origin is shown on the diagram. 1220 . 53 N / C 2876 . 02 N / b v E Thus, b v E b = r E 2 x + E 2 y = r (1220 . 53 N / C) 2 + (2876 . 02 N / C) 2 = 3124 . 29 N / C . 002 (part 2 oF 4) 3.0 points Calculate the electric potential at the origin. Correct answer: 3395 . 3 V. Explanation: The electric potential oF a point charge is V = k q r , so the electric potential at the origin (due to charges Q x and Q y ) is thus V = k p Q x x + Q y y P = (8 . 98755 × 10 9 N · m 2 / C 2 ) × p 1 . 1 × 10 5 C 9 m + 8 × 10 6 C 5 m P = 3395 . 3 V . 003 (part 3 oF 4) 2.0 points A 3 × 10 6 C charge is brought From a very distant point by an external Force and placed at the origin. Calculate the magnitude oF the electric Force on this charge.

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

View Full Document
zapata (jz883) – HW #2 – Antoniewicz – (56495) 2 Correct answer: 0 . 00937286 N. Explanation: We know that v F = q v E . The electric force on Q is shown on the diagram. x y b v F b v F b = | Q | b v E b = v v 3 × 10 6 C v v (3124 . 29 N / C) = 0 . 00937286 N . 004 (part 4 of 4) 2.0 points Calculate the work that had to be done by an external force to bring Q to the origin from the distant point. Correct answer: 0 . 0101859 J. Explanation: The external force must do work equal in magnitude to that done by the electric Feld; i.e. , equal to the change in electric potential energy: W = Q Δ V = ( 3 × 10 6 C) (3395 . 3 V 0) = 0 . 0101859 J . 005 10.0 points Two charges, 2 Q and + Q , are located on the x -axis, as shown below. Point P , at a distance of 3 D from the origin O , is one of two points on the positive x -axis at which the electric potential is zero. x 0 2 Q + Q P D D D How far d from the origin O is the other point?
This is the end of the preview. Sign up to access the rest of the document.

## This note was uploaded on 10/01/2010 for the course PHYSICS PHY 303 L taught by Professor Antoniewicz during the Spring '10 term at Texas Pan American.

### Page1 / 10

HW #2-solutions - zapata (jz883) HW #2 Antoniewicz (56495)...

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

View Full Document
Ask a homework question - tutors are online