mat-sln-asn-hwk9-spr02

# mat-sln-asn-hwk9-spr02 - Solution for Homework 9 Computing...

This preview shows pages 1–4. Sign up to view the full content.

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

View Full Document

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: Solution for Homework 9 Computing Potential Difference Solution to Homework Problem 9.1(Gauss’ Law for Parallel Planes) Problem: Two infinite parallel planes of fixed charge 2 σ and- 3 σ are held a distance d apart as shown below. A neutral conducting slab of thickness d/ 4 is inserted between the plates. Compute the magnitude of the potential difference between the plates. Select One of the Following: (a) | Δ V | = σd ε (b) | Δ V | = 3 d 4 3 σ 2 ε (c) | Δ V | = σd 2 ε (d-Answer) | Δ V | = 3 d 4 5 σ 2 ε (e) | Δ V | = 5 σd 2 ε d/4 I II III IV V conductor-3 σ +2 σ Solution (a) Compute the Fields: Apply Gauss’ Law to a Gaussian cylinder with end area A with ends in region i and iii . E V A- E I A = Q enc ε = (2 σ- 3 σ ) A ε E V- E I =- σ ε By symmetry E I =- E V vector E V =- σ 2 ε ˆ x vector E I = σ 2 ε ˆ x 1 Now apply Gauss’ Law to a surface enclosing only the left plane. E I A- E I A = Q enc ε = 2 σA ε E II = E I + 2 σ ε = σ 2 ε + 2 r ε = 5 σ 2 ε vector E II = 5 σ 2 ε ˆ x (b) The potential difference in a uniform field is given by Δ V = Ed . The potential difference across the conductor is zero, and since E IV = E II , | Δ V | = 3 d 4 E II = 3 d 4 5 σ 2 ε Total Points for Problem: 8 Points Solution to Homework Problem 9.2(Velocity of Electron Escaping Hydrogen) Problem: An electron and a proton are on average 5 × 10 − 11 m apart in a hydrogen atom. Suppose (disasterously) the electron’s charge changed sign and the atom blew apart. The charge of an electron is 1 . 6 × 10 − 19 C and the mass of an electron is 9 . 11 × 10 − 31 kg . How fast would the electron be going after it was very far from the proton. You may assume the proton is stationary. Neglect any relativistic effects. Select One of the Following: (a) 3 × 10 8 m s (b-Answer) 3 × 10 6 m s (c) 9 × 10 12 m s (d) 4 × 10 − 18 m s (e) 4 × 10 3 m s Solution The potential energy of the electron/proton system is the work to assemble it, U = q Δ V = kq 2 r = k (1 . 6 × 10 − 19 C) 2 5 × 10 − 11 m = 4 . 61 × 10 − 18 J All this potential energy is converted into kinetic energy once the electron is far from the proton. U = 1 2 mv 2 or solving for the velocity v = radicalbigg 2 U m = radicalBigg 2(3 × 10 6 m s ) 9 . 11 × 10 − 31 kg = 3 × 10 6 m s Total Points for Problem: 3 Points 2 Solution to Homework Problem 9.3(Work Along Various Paths) Problem: Refer to the figure below. Which path, 1 , 2 or 3 , from point...
View Full Document

## This note was uploaded on 03/29/2010 for the course PHYS 2469 taught by Professor Stewat during the Spring '10 term at University of Arkansas Community College at Batesville.

### Page1 / 10

mat-sln-asn-hwk9-spr02 - Solution for Homework 9 Computing...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online