Textbook: University Physics with Modern Physics, 11th Edition

Excerpt: ... 22.61: a) For a sphere NOT at the coordinate origin: Qencl 4r r r b 4r 2 E 0 0 3 ^ in the r - direction. ( r b ) E . 30 3 E r , 30 b) The electric field inside a hole in a charged insulating sphere is: r (r b ) b Ehole Esphere E(a) . 30 30 30 Note that E is uniform. ...

Pittsburgh, PHYS 0175

Excerpt: ... Reminder: Homework Assignments: PHYS 0175 LON-CAPA HW #01 Due on Sunday(@10pm), January 18th . PHYS 0175 LON-CAPA HW #02 Now Open: Due on Sunday(@10pm), January 25 Lecture 6 (Jan. 16, 2009): Chapter 22 (conclusion): The electric field above the center of a rod with uniformly distributed net charge The electric field above one end of a rod with uniformly distributed net charge General rules for calculating E field The electric field inside a conductor Torque experienced by a dipole in an electric field Example 6.1: A positive net charge Q is distributed uniformly over the length L of a thin non-conducting rod. What is the electric field E at point P located a perpendicular distance R above the rods center? Will E be zero? Will Ex be zero? Will Ey be zero? Example 6.2: A positive net charge Q is distributed uniformly over the length L of a thin non-conducting rod. What is the electric field E at point P located a perpendicular distance R above one end of the rod? y P R + ...

University of Illinois, Urbana Champaign, ECE 329

Excerpt: ... ECE 329 Lecture 21 Conductors ECE 329 Oct 17, 2005 Conductors in an electric field - E0 + + + + + + + + ECE 329: Introduction to Electromagnetic Fields 2 Conductors in an electric field - E0 + + + + + + + + - + + + + + + + + ECE 329: Introduction to Electromagnetic Fields 3 Conductors in an electric field The field outside of the conductor is the superposition of the applied field and the secondary field (in this case, and most simple cases, the secondary fields cancel out outside of conductor) - E0 + + + + + + + + - + + + + + + + + The total + charge and - charge on the surfaces of the conductor must be equal to the total charge of the conducting slab. The surface charge density may be different, depending on geometries. ECE 329: Introduction to Electromagnetic Fields 4 Spherical Conducting Shell What is the electric field everywhere, including inside of the neutral conducting shell (Ri < R < Ro)? +Q We know that the electric field inside of the conducting shell (R ...

UVA, PHYS 632

Excerpt: ... q n r dA = ndA dA 2 1 -12 C 4k = where 0 = 8.85 x10 0 Nm 2 net qenc = e0 Gauss' Law r 15 Gauss' Law = 0 This result can be extended to any shape surface with any number of point charges inside and outside the surface as long as we evaluate the net flux through it. 16 Approximate Flux = r r E A r r = E rdA r dA = ndA Circle means you integrate over a closed surface. Exact Flux r 17 You should know how to prove these things 1 2 1 2 5 6 7 8 9 The electric field inside a conductor is 0. The total net charge inside a conductor is 0. It resides on the surface. Find electric field just outside the surface of a conductor. Find electric field around two parallel flat conducting planes. Find electric field of a large non-conducting sheet of charge. Find electric field of an infinitely long uniformly line of charge. Find E inside and outside of a long non-conducting solid cylinder of uniform charge density. Find E for a thin cylindrical shell of surface charge density . Find E inside and ...

Sewanee, PHYSICS 104

Excerpt: ... , and its radius is 0.05 m. The current through coil 1 is changing with time. The distance between the centers of the coils is 0.14 m. At t=0 s, the current through coil 1 is 17 A. At t=0.4 s, the current through coil 1 is 6 A. Inside coil 2, what is the direction of d /dt during this interval? -Select- 6 1 of 2 05/31/2005 14:36 Preview http:/www.webassign.net/v4cgirwchabay@ncsu/assignments/preview.tp. What is the direction of the electric field inside the wire of coil 2, at a location on the top of coil 2? -Select- 6 At time t=0, what is the magnetic flux through one turn of coil 2? At t=0 1 turn = T m2 At t=0.4 s, what is the magnetic flux through one turn of coil 2? At t=0.4 s 1 turn = T m2 What is the average emf in one turn of coil 2 during this time interval? |emf1 turn| = V The voltmeter is connected across all turns of coil 2. What is the reading on the voltmeter during this time interval? voltmeter reading is V During this interval, what is the magnitude of the non-Coulomb electric ...

UAB, PH 222

Excerpt: ... PH 222-3A Spring 2007 222 3A Gauss Law Lectures 3-4 Chapter 23 (Halliday/Resnick/Walker, Fundamentals of Physics 8th edition) 1 Chapter 23 Gauss Law G L In this chapter we will introduce the following new concepts: The flux (symbol ) of the electric field Symmetry Gauss law Ga ss la We will then apply Gauss law and determine the electric field generated by: An infinite, uniformly charged insulating plane An infinite, uniformly charged insulating rod A uniformly charged spherical shell A uniform spherical charge distribution We will also apply Gauss law to determine the electric field inside and 2 outside charged conductors. Flux of a Vector. Consider an airstream of velocity v that is aimed at a loop of area A . The velocity vector v is at angle with respect to the loop normal n. The product = vA cos is known as the flux. In this example the flux is equal to the volume flow rate through the loop (thus the name flux). Note 1 : depends on . It is maximum and equal ...

Sewanee, PHYSICS 104

Excerpt: ... The magnitude of the charge on each disk is 53 C. Consider points A, B, C, and D inside the capacitor, as shown in the diagram. The MI02 16.Q.02 [414889] distance s1 = 1.5 mm, and the distance s2 = 0.5 mm. = 8.85e-12 C2/(Nm2) Figure 16.55 (a) What is the electric field inside the capacitor? =< , , > N/C 1 of 2 2005-05-15 11:36 Preview http:/www.webassign.net/v4cgirwchabay@ncsu/assignments/preview.tp. (b) First, calculate the potential difference VB - VA. What is m along this path? =< , , > What is VB - VA? VB - VA = volts (c) Next, calculate the potential difference VC - VB What is m along this path? =< , , > What is VC - VB? VC - VB = volts (d) Finally, calculate the potential difference VA - VC What is m along this path? =< , , > What is VA - VC? VA - VC = volts Submit for Testing 2 of 2 2005-05-15 11:36 ...

N.C. State, PY 208M

Excerpt: ... t all on the wire near location G. [x] [x] There is a large gradient of surface charge on the wire between locations C and E. [x] [x] The magnitude of the electric field at location D is larger than the magnitude of the electric field at location G http:/www.webassign.net/v4cgiwcwebb@ncsu/student.pl?l=20060505025936wcwebb@ncsu651652069 PDF created with pdfFactory Pro trial version www.pdffactory.com 5/4/2006 Ch 18 HW 3 F2006 1 2 pts 1/1 1/1 2/2 Page 3 of View: All Responses Notes pts 1/1 1/1 1/1 3/3 3. A Nichrome wire 44 cm long and 0.25 mm in diameter is connected to a 1.6 volt flashlight battery. What is the electric field inside the wire? 3.636363 [3.64] V/m The wire is replaced by a different Nichrome wire with the same length, but diameter 0.45 mm. Now what is the electric field inside the wire? 3.636363 [3.64] V/m 1 2 3 View: All Responses Notes 4. A Nichrome wire 96 cm long and 0.25 mm in diameter is connected to a 1.3 volt flashlight battery. What is the electric field in ...

N.C. State, PY 208M

Excerpt: ... dge of the plastic slab, location 3 is at the right edge of the slab, and location 4 is at the right plate of the capacitor. All of these locations are near the center of the capacitor. Calculate the following potential differences. V1 - V2 = 207.5 [208] V V2 - V3 = 106.4 V3 - V4 = 207.5 [106] V [208] V V1 - V4 = 521.4 [521] V What assumptions or approximations did you make in this calculation? [x] [x] Assume the radius of the plates was much larger than the distance between the plates [x] [x] Assume that the field due to the polarized plastic slab was negligible in the air gaps [_] [_] Assume the electric field inside the plastic slab is zero [_] [_] Assume the positive plate of the capacitor does not contribute to the electric field in the gap between the plastic and the negative plate [_] [_] Assume the plastic slab did not polarize Solution or Explanation: Without the plastic, the electric field inside the capacitor is V/s, assuming that the field is uniform, which is true if the radius of the plate ...

St. Mary MD, PHYS 2520

Excerpt: ... s. Those lines should 1) point in the direction of the electric field vector E at every point; 2) start at positive charges or at infinity; 3) end at negative charges or at infinity; 4) be most dense where E has a greater magnitude. Note that electric field lines are not the closed lines. They are always starting at the charged particle and ending at the charged particle. This means, that electrostatic field is a potential field of conservative force, in same way as gravitational field. The fact that electrostatic forces are conservative forces means that the work done by electrostatic force on the moving charge around the closed path is zero and this work does not depend on the form of trajectory but only on the final and original position of the charge. This knowledge about electric field helps to resolve some problems. For instance if we consider an electric field inside of the charged conductor in electrostatic equilibrium, we can prove that this field is zero. Indeed if we have several charges of the sam ...

Stevens, PEP 112

Excerpt: ... PEP112 Spring 2008 Prof. Svetlana Malinovskaya 28 January 2008 The Parallel-Plate Capacitor. Motion in an Electric Field. A Disk of Charge zQi Eiz = 2 4 0 (ri + z 2 )3 / 2 1 Edisk z Edisk z z Qi = 2 i =1 4 0 (ri + z 2 )3 / 2 z N z = 1 - 2 2 2 0 z +R 1 Q = 4 0 z 2 R Edisk z A Plane of Charge Many electronic devices use charged, flat surfaces disks, squares, rectangles, etc. These charged surfaces are called electrodes. We can often model an electrode as an infinite plane of charge. Edisk z z = 1 - 2 2 0 z + R2 = 2 0 2 0 = - 2 0 R E plane z>0 z<0 E plane The Parallel-Plate Capacitor Two electrodes with charge +Q and Q placed face-to-face a distance d apart are called a parallel-plate capacitor. Our goal is to find the electric field inside and outside the capacitor. The net charge of the capacitor is zero. Capacitors are charged by transferring electrons from one plate to another. The plate that gains electrons has charge Q=N(-e). ...

Texas, PHY 303L

Excerpt: ... Topics to be covered in class: Electric fields and conductors Electrostatic energy and electrostatic potential How do we calculate electrostatic potential from the electrostatic field? Example: electrostatic potential due to several point charges Example: electrostatic potential due to continuous charge distribution Independent reading: Potential in conductors (Ch.25.3) No electric field inside a conductor no charges inside the conductor If an excess charge is placed on an isolated conductor, that amount of charge will move to the surface of the conductor! Outer surface, that is! Electric field outside of conductor is always normal (perpendicular) to the conductor's surface Fields inside conductors with cavities +q Fields inside conductors with cavities +q Why introduce electric potential? Work done by the electric field on a charge: from A to B Electric field does the work potential energy changes Note: the field is generated by all charges in the worl ...

ECCD, DOE 315

Excerpt: ... 97.315 Course Notes Page 17 Winter 2001 At the conductor's surface, we can relate the charge density to the electric field using Gauss' Law again. Construct a Gaussian surface as shown below: E - + + + E + + + + Gaussian Surface Surface Charge Density s The area of conductor surface inside the Gaussian box is A so that the enclosed charge is Q = s A . The electric field inside the conductor must be zero, as discussed above, and the flux must be the same regardless of how closely the box surface follows the conductor, so the electric field must be perpendicular to the conductor surface. s A Gauss' law relates the electric field to the charge by E A = - , or 0 s E = -0 (eq. 1.10) For a hollow conductive shell, the same thing will happen, since the electrons will simply flow around the hole in the conductor. We can also show that the field inside the hollow is zero using the conservative nature of the electrostatic field. ...