{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

# 6 - roofner(bar784 – Homework 6 – Weathers –(22202 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: roofner (bar784) – Homework 6 – Weathers – (22202) 1 This print-out should have 16 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points At distance r from a point charge q , the elec- tric potential is 431 V and the magnitude of the electric field is 310 N / C. Determine the value of q . Correct answer: 6 . 66732 × 10 − 8 C. Explanation: Let : k e = 8 . 98755 × 10 9 N · m 2 / C 2 , V = 431 V , and e = 310 N / C . E = k e q r 2 and V = k e q r , so that V E = r . The potential is V = k e q r = k e q V E = k e q E V q = V 2 k e E = (431 V) 2 (8 . 98755 × 10 9 N · m 2 / C 2 ) (310 N / C) = 6 . 66732 × 10 − 8 C . 002 10.0 points Four identical particles each have charge . 5 μ C and mass 0 . 03 kg. They are released from rest at the vertices of a square of side . 34 m. How fast is each charge moving when their distances from the center of the square dou- bles? Correct answer: 0 . 546045 m / s. Explanation: Let : m = 0 . 03 kg , q = 0 . 5 μ C , and L = 0 . 34 m . Each charge moves off on its diagonal line, and by symmetry each has the same speed. The initial potential energy of the system of charges is U i = k e q 2 parenleftbigg 4 L + 2 L √ 2 parenrightbigg . The final potential energy is obtained by dou- bling the distances. By conservation of energy ( K + U ) i = ( K + U ) f 4 k e q 2 L + 2 k e q 2 L √ 2 = 4 2 mv 2 + 4 k e q 2 2 L + 2 k e q 2 2 L √ 2 k e q 2 L parenleftbigg 2 + 1 √ 2 parenrightbigg = 2 mv 2 v = radicalBigg k e q 2 mL parenleftbigg 1 + 1 2 √ 2 parenrightbigg Since k e q 2 mL = (8 . 98755 × 10 9 N · m 2 / C 2 ) (0 . 03 kg)(0 . 34 m) × (5 × 10 − 7 C) 2 = 0 . 220283 m 2 / s 2 , v = radicalBigg (0 . 220283 m 2 / s 2 ) parenleftbigg 1 + 1 2 √ 2 parenrightbigg = . 546045 m / s . 003 (part 1 of 4) 10.0 points The electric potential over a certain region of space is given by V = ax 2 y- bxz- cy 2 , where a = 3 V / m 3 , b = 8 V / m 2 , and c = 1 V / m 2 . Find the electric potential at the point ( x,y,z ) = (1 m , 4 m , 8 m). Correct answer:- 68 V. Explanation: roofner (bar784) – Homework 6 – Weathers – (22202) 2 Let : a = 3 V / m 3 , b = 8 V / m 2 , c = 1 V / m 2 , and ( x,y,z ) = (1 m , 4 m , 8 m) . The electric potential is V = ax 2 y- bxz- cy 2 = ( 3 V / m 3 ) (1 m) 2 (4 m)- ( 8 V / m 2 ) (1 m) (8 m)- ( 1 V / m 2 ) (4 m) 2 =- 68 V . 004 (part 2 of 4) 10.0 points Find the x-component of the electric field at the same point. Correct answer: 40 V / m. Explanation: The electric field is the gradient of the po- tential, so E x =- ∂ V ∂x =- 2 axy + bz =- 2 ( 3 V / m 3 ) (1 m) (4 m) + ( 8 V / m 2 ) (8 m) = 40 V / m . 005 (part 3 of 4) 10.0 points Find the y-component of the electric field at the same point....
View Full Document

{[ snackBarMessage ]}

### Page1 / 8

6 - roofner(bar784 – Homework 6 – Weathers –(22202 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