solution35

# solution35 - cherubin (mcc2645) – Homework 1...

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

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.

Unformatted text preview: cherubin (mcc2645) – Homework 1 Electrostatics – Catala – (12054) 1 This print-out should have 22 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A charge of +2 . 20 × 10 − 9 C is placed at the origin, and another charge of +5 . 90 × 10 − 9 C is placed at x = 2 . 5 m. Find the point (coordinate) between these two charges where a charge of +3 . 00 × 10 − 9 C should be placed so that the net electric force on it is zero. The Coulomb constant is 8 . 98755 × 10 9 N · m 2 / C 2 . Correct answer: 0 . 947822 m. Explanation: Let : q 1 = 2 . 20 × 10 − 9 C at the origin , q 2 = 5 . 90 × 10 − 9 C , q 3 = 3 . 00 × 10 − 9 C , and x 2 = 2 . 5 m . q 3 is between q 1 and q 2 , so if r 1 , 3 = P , then r 2 , 3 = x 2 − P . F electric = k C q 1 q 2 r 2 The net electric force is zero, so F 1 , 3 = F 2 , 3 k C q 1 q 3 r 2 1 , 3 = k C q 2 q 3 r 2 2 , 3 q 1 r 2 1 , 3 = q 2 r 2 2 , 3 q 1 P 2 = q 2 ( x 2 − P ) 2 ( x 2 − P ) 2 P 2 = q 2 q 1 x 2 − P P = radicalbigg q 2 q 1 x 2 − P = P radicalbigg q 2 q 1 P parenleftbigg 1 + radicalbigg q 2 q 1 parenrightbigg = x 2 P = x 2 1 + radicalbigg q 2 q 1 = 2 . 5 m 1 + radicalbigg 5 . 9 × 10 − 9 C 2 . 2 × 10 − 9 C = . 947822 m . 002 10.0 points Find the magnitude of the electric field at a point midway between two charges +33 . 5 × 10 − 9 C and +78 . 3 × 10 − 9 C separated by a distance of 15.0 cm. The value of the Coulomb constant is 8 . 99 × 10 9 N · m 2 / C 2 . Correct answer: 71600 . 4 N / C. Explanation: Let : q 1 = 33 . 5 × 10 − 9 C , q 2 = 78 . 3 × 10 − 9 C , x = 15 . 0 cm , and k C = 8 . 99 × 10 9 N · m 2 / C 2 . r 1 = r 2 = x 2 = 15 cm 2 · parenleftbigg 1 m 100 cm parenrightbigg = 0 . 075 m The electric field is E = k C q r 2 . The magnitudes of the electric fields are E 1 = 8 . 99 × 10 9 N · m 2 / C 2 × 3 . 35 × 10 − 8 C (0 . 075 m) 2 = 53540 . 4 N / C , and E 2 = 8 . 99 × 10 9 N · m 2 / C 2 × 7 . 83 × 10 − 8 C (0 . 075 m) 2 = 1 . 25141 × 10 5 N / C . cherubin (mcc2645) – Homework 1 Electrostatics – Catala – (12054) 2 If the 33 . 5 × 10 − 9 C charge is the left one and the positive direction is to the right, E net = E 1 − E 2 = 53540 . 4 N / C − 1 . 25141 × 10 5 N / C = − 71600 . 4 N / C bardbl vector E net bardbl = 71600 . 4 N / C . The electric field at midpoint between the charges +33 . 5 × 10 − 9 C and +78 . 3 × 10 − 9 C is 71600 . 4 N / C toward the +33 . 5 × 10 − 9 C charge. 003 (part 1 of 4) 10.0 points A small cork with an excess charge of +9 . μ C is placed 0 . 24 m from another cork, which carries a charge of − 3 . 2 μ C. What is the magnitude of the electric force between the corks? The Coulomb constant is 8 . 98755 × 10 9 N · m 2 / C 2 ....
View Full Document

## This note was uploaded on 06/10/2010 for the course PHY 100 taught by Professor Pro.kiroshima during the Spring '10 term at Adelphi.

### Page1 / 8

solution35 - cherubin (mcc2645) – Homework 1...

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

View Full Document
Ask a homework question - tutors are online