PHY303L HW2

# PHY303L HW2 - thwing(jrt2329 – Homework 02 – Yao...

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: thwing (jrt2329) – Homework 02 – Yao – (59110) 1 This print-out should have 23 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points The electron gun in a television tube is used to accelerate electrons (mass of 9 . 10939 × 10- 31 kg and charge of − 1 . 60218 × 10- 19 C) from rest to 2 × 10 7 m / s within a distance of 5 . 6 cm. What electric field is required? Correct answer: 20305 . 8 N / C. Explanation: Let : m e = 9 . 10939 × 10- 31 kg , q e = 1 . 60218 × 10- 19 C , v = 2 × 10 7 m / s , and d = 5 . 6 cm . The magnitude of the force is F = q e E = m e a a = q e E m e The final velocity is v 2 f = v 2 i + 2 a d = 2 a d since v i = 0, so v 2 = 2 d q e E m e E = v 2 m e 2 d q e = parenleftBig 2 × 10 7 m / s 2 parenrightBig ( 9 . 10939 × 10- 31 kg ) 2 (5 . 6 cm) (1 . 60218 × 10- 19 C) = 20305 . 8 N / C . 002 10.0 points A 50 cm diameter loop is rotated in a uniform electric field until the position of maximum electric flux is found. The flux in this position is measured to be 6 . 58 × 10 5 N · m 2 / C. What is the electric field strength? Correct answer: 3 . 35117 × 10 6 N / C. Explanation: Let : r = 25 cm = 0 . 25 m and Φ = 6 . 58 × 10 5 N · m 2 / C . By Gauss’ law, Φ = contintegraldisplay vector E · d vector A The position of maximum electric flux will be that position in which the plane of the loop is perpendicular to the electric field; i.e. , when vector E · d vector A = E dA . Since the field is constant, Φ = E A = Eπ r 2 E = Φ π r 2 = 6 . 58 × 10 5 N · m 2 / C π (0 . 25 m) 2 = 3 . 35117 × 10 6 N / C . 003 10.0 points A (6 . 6 m by 6 . 6 m) square base pyramid with height of 4 . 06 m is placed in a vertical electric field of 26 . 1 N / C. b 6 . 6 m 4 . 06 m 26 . 1 N / C Calculate the total electric flux which goes out through the pyramid’s four slanted sur- faces. Correct answer: 1136 . 92 N m 2 / C. Explanation: Let : s = 6 . 6 m , h = 4 . 06 m , and E = 26 . 1 N / C . thwing (jrt2329) – Homework 02 – Yao – (59110) 2 By Gauss’ law, Φ = vector E · vector A Since there is no charge contained in the pyra- mid, the net flux through the pyramid must be 0 N/C. Since the field is vertical, the flux through the base of the pyramid is equal and opposite to the flux through the four sides. Thus we calculate the flux through the base of the pyramid, which is Φ = E A = E s 2 = (26 . 1 N / C) (6 . 6 m) 2 = 1136 . 92 N m 2 / C . 004 10.0 points A cubic box of side a , oriented as shown, con- tains an unknown charge. The vertically di- rected electric field has a uniform magnitude E at the top surface and 2 E at the bottom surface. a E 2 E How much charge Q is inside the box?...
View Full Document

## This note was uploaded on 11/11/2009 for the course PHY 59110 taught by Professor Yao during the Spring '09 term at University of Texas.

### Page1 / 11

PHY303L HW2 - thwing(jrt2329 – Homework 02 – Yao...

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

View Full Document
Ask a homework question - tutors are online