{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

# 14-1 - brown(twb493 Ch14-h1 chiu(56565 This print-out...

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

brown (twb493) – Ch14-h1 – chiu – (56565) 1 This print-out should have 29 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0points An electron in a region where there is an electric field experiences a force of magnitude 3 . 6 × 10 16 N. What is the magnitude of the electric filed at the location of the electron? The charge on an electron is 1 . 602 × 10 19 C Correct answer: 2247 . 19 N / C. Explanation: The field is defined to be the force per unit charge experienced by a particle (so long as the particle has a charge small enough that it does not change the background field significantly). Thus, we have: vextendsingle vextendsingle vextendsingle vector E vextendsingle vextendsingle vextendsingle = vextendsingle vextendsingle vextendsingle vector F vextendsingle vextendsingle vextendsingle q = 3 . 6 × 10 16 N 1 . 602 × 10 19 C = 2247 . 19 N / C 002(part1of8)5.0points In the region shown below there is an elec- tric field due to charged objects not shown in the diagram. A tiny glass ball with a charge of 7 × 10 9 C placed at location A experiences a force of (− 4 × 10 5 N , 4 × 10 5 N , 0 N ) , as shown in the diagram. + vector F A Which of the below arrows indicates the direction of the electric field at location A ? enter j if the field has zero magnitude. Take the z -component of the field to be zero. a b c d e f g h Correct answer: f . Explanation: The force on a positively charged particle is in the same direction as the electric field, so the correct answer is f . 003(part2of8)3.0points What is the x -component of the electric field at location A ? Correct answer: 5714 . 29 N / C. Explanation: The electric field is the force on a test parti- cle divided by the charge of that test particle, so vector E = vector F q E x = 4 × 10 5 N 7 × 10 9 C = 5714 . 29 N / C 004(part3of8)3.0points What is the y -component of the electric field at location A ? Correct answer: 5714 . 29 N / C. Explanation: The electric field is the force on a test parti- cle divided by the charge of that test particle, so

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

View Full Document
brown (twb493) – Ch14-h1 – chiu – (56565) 2 vector E = vector F q E y = 4 × 10 5 N 7 × 10 9 C = 5714 . 29 N / C 005(part4of8)3.0points What is the magnitude of this electric field? Correct answer: 8081 . 22 N / C. Explanation: The magnitude of any vector is the square root of the sum of the squares of its compo- nents, as per the Pythagorean Theorem. So, we have: vextendsingle vextendsingle vextendsingle vector E vextendsingle vextendsingle vextendsingle = |( E x , E y , E z )| = radicalBig E 2 x + E 2 y + E 2 z = radicalBig ( 5714 . 29 N / C) 2 + ( 5714 . 29 N / C) 2 + 0 2 = 8081 . 22 N / C 006(part5of8)5.0points Now, the glass ball is moved very var away, a tiny plastic ball with charge 8 × 10 9 C is placed at location A . Which arrow best indicates the direction of the electric force on this negatively charged plastic ball?
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 9

14-1 - brown(twb493 Ch14-h1 chiu(56565 This print-out...

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

View Full Document
Ask a homework question - tutors are online