# Documents about Uniform Electric Field

• 1 Pages

#### tsl197

Rhode Island, PHYS 204

Excerpt: ... Electric Dipole in Uniform Electric Field Electric dipole moment: p = q L Torque exerted by electric field: = p E Potential energy: U = -p E Z Z U () = - ()d = pE /2 sin d = -pE cos /2 Note: () and d have opposite sign. E L -q p = qL +q p tsl197 p.1/1 ...

• 2 Pages

#### Unit_02_1_Web_Lecture_Notes

Cornell, PHYS 2208

Excerpt: ... Notes from lecture: 30 January 2008. If a dielectric (of either classification) is inserted into the uniform electric field between two infinite sheets of (equal magnitude) uniform charge distribution, the charge distributions of the dielectric's molecules will be oriented in such a way as to produce an electric field which opposes the original electric field (below). Dielectrics can be further characterized by the relative strength of the induced electric field. The dielectric constant of the material, , is defined by = E 0 / E d . Since the total field magnitude E d is always less than the original field magnitude E 0 , we see that > 1. Consider a small positive test charge ( q > 0 ) in the uniform electric field produced by two uniformly charged plates (with opposite charge densities). Imagine that we were to move the small test charge very slowly in such a way as to not disturb the uniform field due to the plate. What is the work done by the electrostatic force when the charge moves from A to B (see ...

• 5 Pages

#### sg2

Uni. Worcester, PH 1120

Excerpt: ... PH 1120 STUDY GUIDE 2 Term D, 2004 -/ Objective 4 Electrical Potential (Uniform Field) \ / \ / i) Define electric potential. Calculate the potential difference \ / between two points in a uniform electric field . \ / \ / ii) Calculate the work one must do against electrical forces in moving \ / a point charge between two points in a uniform electric field . \ / \ / ii) Calculate the potential difference between two points, given the \ / value of a charge and the work involved in transporting it between \ / the two points. \ / \ / iii) Determine the motion of a charged particle accelerated through a \ / known potential difference in a uniform electric field . \ - Suggested Study Procedures Study Sec. 23-1 and 23-2 (uniform E). Please note that in defining electric potential, the lecturer will emphasize the work that you do against the electric force, while some text ...

• 21 Pages

#### Chapter 21B

Mississippi State, PH 2223

Excerpt: ... PH 2223 Vote on Date of First Exam First exam will cover Chapters 12 and 21. Vote for date of first exam: Thursday, September 4 Tuesday, September 9. 1 PH 2223 Last lecture: Electric Force Electric Todays lecture (8/28/2008): Charged Particle in Uniform Electric Field Electric Field Lines Electric Dipoles Gausss Law (Chapter 22) ReminderPH 2223 labs start next week. 2 Motion of Charged Particle in Uniform Electric Field Uniform electric field means same at all . Between parallel plates, Uniform constant acceleration use kinematic equations can 3 An electron is projected with an initial speed v0=1.60x106 m/s into the uniform field between parallel plates. Assume the field Between the plates is uniform and directed vertically downward, and that field outside the plates is zero. The electron enters the field at a point midway between the plates. If the electron just misses the upper plate as it emerges from the field, what is the 4 Exercise 21.33 Electric Field Lin ...

• 9 Pages

#### Lecture 3 jan10

University of Michigan, PHYSICS 240

Excerpt: ... ines becomes infinitesimally small then E E. In a region of uniform E field the line spacing is constant Electric Dipoles (2 equal but opposite charges +q and q separated by d) (2nd simplest charge configuration after monopole) + d +q P define P qd P points from - to (opposite of E ) Now put the dipole in a uniform electric field . -q - E How does it behave? Question #2 A dipole is completely free to move in a uniform electric field as shown. How does it move? A. It translates to the right B. It translates to the left C. It oscillates between some and - without translating D. It spins fully around clockwise without translating E. It spins fully around counter-cw without translating + P +q E -q - Torque on Electric Dipole (in uniform E) F 0 Torque r F d qE qd E P E (and F-q Fnet F +q + d F+ P PE sin ) Tends to align P with E. Calculate about any axis Potential Energy, U, of Dipole (in uniform field) U ...

• 2 Pages

#### PDF Equipotentials

SUNY Geneseo, PHYS 125

Excerpt: ... N W S 40V 30V 20V 10V 0V 0V 0V - 10V - 20V - 30V E 40V 30V 20V 10V 0V 0V 10V 20V 30V A - 10V - 20V B C D F - 30V - 20V - 10V 0V E 10V 20V 30V In a uniform electric field E, the indicated point X is designated 0 potential. Rank the other points from largest to smallest potential. (Note: points CDE are all the same distance from X.) X: V = 0V A B C D E (less negative) A (B&E) D C (more negative) ...

• 8 Pages

#### 1402 Potential

Lone Star College System, PHYS 1402

Excerpt: ... Electric Potential Physics 1402 Recall from Physics I Gravitational force is a conservative force. The work done by a conservative force is path independent and can be written: Wcons = -U Where U is the potential energy (PE). Near the surf ...

• 5 Pages

#### sg2

Uni. Worcester, PH 1120

Excerpt: ... PH 1120 STUDY GUIDE 2 Term D, 2009 -/ Objective 4 Electrical Potential (Uniform Field) \ / \ / i) Define electric potential. Calculate the potential difference \ / between two points in a uniform electric field . \ / \ / ii) Calculate the work one must do against electrical forces in moving \ / a point charge between two points in a uniform electric field . \ / \ / ii) Calculate the potential difference between two points, given the \ / value of a charge and the work involved in transporting it between \ / the two points. \ / \ / iii) Determine the motion of a charged particle accelerated through a \ / known potential difference in a uniform electric field . \ -Suggested Study Procedures Study Sec. 23-1 and 23-2 (uniform E). Please note that in defining electric potential, the lecturer will emphasize the work that you do against the electric force, while some textboo ...

• 2 Pages

#### DLM12_FNTs

UC Davis, PHY 7

Excerpt: ... Physics 7C DLM12 Overview DLM 12 Model: Fields and Forces Act 9.1.5 Electrical Forces and Fields (FNTs from DLM11) 45 minutes Learning Goals: Solidify understanding of connections between charges, forces, and fields. Examine analogy with gravitational fields noting similarities and differences Understand how an electric field affects a non-polar molecule Act 9.1.6 Forces on Polar Molecules 35 minutes Learning Goals: Understand forces of attraction between polar molecules (thus, understand some of the basis for hydrogen bonds) Understand motion of an electric dipole in a uniform electric field Model: Fields, Forces, Potential & Potential Energy Act 9.2.1 Electric Forces and Bond Energies 60 minutes Learning Goals: Starting making connections back to the energy language of 7A. Be able to determine forces from potential energy graphs and construct potential energy graphs from forces. Understand how two uncharged non-polar molecules or atoms can still have an attractive electrical interaction between them. ...

• 15 Pages

#### lecture-62

Excerpt: ... st Concept Quiz Which one of the diagrams represents a possible electric field pattern in a charge free region of space (which is not near any moving charges)? A. B. C. D. Uniform Electric Field e.g. Between long parallel charged plates. Characteristics: Field lines are parallel r E E is constant in magnitude and direction A charged particle, of mass m and charge r q, is placed in a uniform electric field E . (Force is shown assuming q is positive) r If E = constant r Then a = constant. r r r F = qE = ma m q r E r F Lecture Example 100 An electron is released from rest in a region where there is a uniform electric field of 5000 N/C. What is the speed of the electron after it has moved a distance of 10 cm? r E e Gausss Law Electric Flux n = unit vector to surface r Define: A = An n r E Area A Define: r r Electric Flux through A is: E = E A = EA cos r r Assumes E is constant over the small area A A1 r Total flux through a surface: E r ...

• 4 Pages

#### EM_sidescreen_4_6ppage

Toledo, PHY 132

Excerpt: ... = = (from + to ) + (from + to ) 2 o 2 o (from + to ) o Q (from + to ) o A More on Parallel Plate Capacitor Flash animation - Charged Parallel Plates: http:/www.regentsprep.org/Regents /physics/phys03/aparplate/ This is a uniform electric field . PHY132S - EM - Lecture 4 - Slide 11 PHY132S - EM - Lecture 4 - Slide 12 2 Review of Mechanical Energy Mechanical energy is conserved for system of particles interacting by conservative forces. Emech = K + U = 0 (note: Emech E !) Kinetic energy: K = Ki = mi vi2 Change in potential energy: U = Uf - Ui = - Winteraction forces (from i to f) Work done by a constant force: W = F r = F r cos (linear path) PHY132S - EM - Lecture 4 - Slide 13 Gravitational and Electric Potential Energy y yi U = Uf - Ui Ugrav = (Ugrav)f - (Ugrav)i = - Wgrav (i to f) = mgyf - mgyi and m sf Ugrav = Uo + mgy Uelec = (Uelec)f - (Uelec)i s si yf g q E PHY132S - EM - Lecture 4 - Slide 14 A Charged Particle in a Uniform ...

• 22 Pages

#### EM_sidescreen_4_1ppage

Toledo, PHY 132

Excerpt: ... (from + to ) + (from + to ) = 2 o 2 o = (from + to ) o Q (from + to ) = o A This is a uniform electric field . PHY132S - EM - Lecture 4 - Slide 11 More on Parallel Plate Capacitor Flash animation - Charged Parallel Plates: http:/www.regentsprep.org/Regents /physics/phys03/aparplate/ PHY132S - EM - Lecture 4 - Slide 12 Review of Mechanical Energy Mechanical energy is conserved for system of particles interacting by conservative forces. Emech = K + U = 0 (note: Emech E !) Kinetic energy: K = Ki = mi vi2 Change in potential energy: U = Uf - Ui = - Winteraction forces (from i to f) Work done by a constant force: W = F r = F r cos (linear path) PHY132S - EM - Lecture 4 - Slide 13 Gravitational and Electric Potential Energy y yi U = Uf - Ui s si Ugrav = (Ugrav)f - (Ugrav)i = - Wgrav (i to f) = mgyf - mgyi and m sf Ugrav = Uo + mgy Uelec = (Uelec)f - (Uelec)i yf g q E PHY132S - EM - Lecture 4 - Slide 14 A Charged Particle in a U ...

• 3 Pages

#### Ch 16 hw 1

N.C. State, PY 208M

Excerpt: ... 1. 1/1 points | 1/4 submissions Last Response | Show Details All Responses Notes What is the kinetic energy of a proton that is traveling at a speed of 2400 m/s? K = 4.8096e-21 [4.90e-21] J 2. 1/1 points | 1/4 submissions Last Response | Show Details All Responses Notes If the kinetic energy of an electron is 5.2e-18 J, what is the speed of the electron? (You can use the approximate (nonrelativistic) formula here.) v = 3399346.342 [3.40e+06] m/s 3. 5.5/5.5 points | 3/4 submissions Last Response | Show Details All Responses Notes Locations A, B, and C are in a region of uniform electric field , as shown in the diagram above. Location A is at < -0.5, 0, 0>m. Location B is at < 0.4, 0, 0>m. In the region the electric field = < 850, 0, 0> N/C. For a path starting at B and ending at A, calculate: (a) The displacement vector = < -0.9 [-0.9], 0 [0], 0 [0] > m (a) the change in electric potential: V = 765 [765] volts (b) the potential energy change for the system when a proton moves from B to A: U = 1.224e ...

• 1 Pages

#### LQ17

Cornell, P 213

Excerpt: ... Physics 213 Lecture questions October 29, 2002 A. A positively charged particle travels from left to right through a uniform magnetic field (pointing upward) and a uniform electric field (direction unknown). The sum of electric and magnetic forces is zero. What must be the direction of the electric field? r B +q r v 1. 2. 3. 4. down up out of the page (toward you) into the page (away from you) B. The work done by a magnetic force on a moving charge is: r 1. positive if B points in the direction of motion (for a positive charge). r 2. positive if B points opposite the direction of motion (for a positive charge). 3. always zero. ...

• 8 Pages

#### 2426 Potential

Lone Star College System, PHYS 2426

Excerpt: ... Electric Potential Recall from Physics I Gravitational force is a conservative force. The work done by a conservative force is path cons = U W independent and can be written: Where U is the potential energy (PE). Near the surface of a planet, the change U = PE mgh in gravitationalmgh2 is: 1 Consider moving a ball of mass m from point A to B in a uniform gravitational field. The change in potential energy is: U = mghB mghA The work done by gravity in moving a particle from point A to point B is WAB = U = ( path independent.mghB mghA ) Recall that mgh is meaningless, only the change in PE has meaning. Consider moving a positive charge q from point A to B in a uniform electric field . (ignore gravity) The force that the field exerts on this charge is: F = qE B B Consider a Uniform Electric Field h B From the hdefinition of work we see WAB = F dr = ( qE ) ( dy ) j = qE dy = qE (hB ...

• 4 Pages

#### 40-43. Electric Flux- Application

Rhode Island, PHYS 204

Excerpt: ... Electric Flux: Application (1) Consider a rectangular sheet oriented perpendicular to the yz plane as shown and positioned in a uniform electric field E = (2^ j)N/C. z 2m 3m 4m x (a) Find the area A of the sheet. (b) Find the angle between A and E. (c) Find the electric flux through the sheet. A E y tsl40 p.1/4 Electric Flux: Application (2) Consider a plane sheet of paper whose orientation in space is described by the area ^ vector A = (3^ + 4k)m2 positioned in a region of uniform electric field j ^ E = (1^ + 5^ - 2k)N/C. i j z A E y x (a) Find the area of the sheet. (b) Find the magnitude of the electric field. (c) Find the electric flux through the sheet. (d) Find the angle between A and E. tsl41 p.2/4 Electric Flux: Application (3) ^ The room shown below is positioned in an electric field E = (3^ + 2^ + 5k)N/C. i j z 1m 2m x (a) What is the electric flux E through the closed door? (b) What is the electric flux E through the door opened at = 90 ? (c) At what angl ...

• 29 Pages

#### Lecture 2-3 Ch22

UAB, PH 222

Excerpt: ... flat, charged sheet 23 The Electric Field of the Two Charged Sheets 24 ED=(1.1x105N/C)[-1-1+1] = -1.1x105 N/C 25 Example 3 : Electric field lines generated by an electric dipole (a positive and a negative point charge of the same size but of opposite sign) Example 4 : Electric field lines generated by two equal positive point charges 26 F+ Forces and Torques Exerted on Electric Dipoles by a Uniform Electric Field Consider the electric dipole shown in the figure in the presence of a uniform (constant magnitude and direction) electric field E along the x-axis. axis The electric field exerts a force F+ = qE on the Fx-axis i p positive charge and a force F = qE on the g q negative charge. The net force on the dipole is Fnet = qE qE = 0. The net torque generated by F+ and F about the dipole center is = + + = F+ d d sin F sin = qEd sin = pE sin q p 2 2 In vector form: = p E p The electric dipole in a uniform electric field does not mo ...

• 4 Pages

#### electric potential 1

Penn State, PHYS 212L

Excerpt: ... Alessandro Volta (1745-1827) Last Time E=0 in a conductor, and E is to conducting surfaces Gauss' Law can be used to determine E when E is constant and perpendicular to a Gaussian surface (imaginary). The Electric Potential Today This lecture: HRW 24.1-24.8 For next time: HRW 24.9-24.12 1 The Electric Potential Equipotential surfaces The electric potential of point charges 2 Potential energy, U(r), describes a conservative force. The reference point, where U=0, is chosen for convenience. Change in potential energy f i 1 Potential Energy (review) r r dW = F ds Electric Potential Energy Interactive Lecture Question #7.1 What is the change in potential energy of a 3 C charge in a uniform electric field of 10 N/C when it is moved along the path shown? 5m y x initial final U Work by the field U = Uf -Ui = -W 2 Work is path independent. In a round trip, U is unchanged (conserved). kinetic energy total energy U Fx = - f r r dU ; U = - F ds dx i For con ...

• 12 Pages

#### lecture-63

Excerpt: ... Electric Potential Energy U = -Wq0 xB In 1-dim: A q0 B x B r ds U = -Wq0 = - Fx dx = - q0 E x dx A xA B More generally: r r U = -q0 E d s B A r E Path Integral q0 A Because the electric force is a conservative force, this result is independent of the path. i.e. U depends only on the initial and final positions. So we can write U = UB UA Where UA = PE of q0 at point A. Electric Potential Energy e.g. Uniform Electric Field Particle with charge q0 moves from A to B in the direction of the electric field. r E Force drawn x r q0 assuming q0 A B F x positive r r F = q0 E = constant Fx = q0 E x U = -Wq0 = - Fx x = - q0 E x x U B - U A = -q0 E x x Remember: We can define the zero of potential energy wherever it is convenient. (Just as for gravitation PE) e.g. If we let UB = 0 then U A = q0 E x x Note: UA depends on q0 (Just as gravitational PE depends on the mass) Electric Potential Define: UA Electric Potential VA = q0 The electric potential energy per unit charge at point A Electric Potent ...

• 9 Pages

#### handout15

Cornell, ECE 4070

Excerpt: ... Handout 15 Dynamics of Electrons in Energy Bands In this lecture you will learn: The behavior of electrons in energy bands subjected to uniform electric field s The dynamical equation for the crystal momentum The effective mass tensor and inertia of electrons in energy bands Examples Magnetic fields ECE 407 Spring 2009 Farhan Rana Cornell University Electron Dynamics in Energy Bands 1) The quantum states of an electron in a crystal are given by Bloch functions that obey the Schrodinger equation: r r r r r H n ,k (r ) = E n k n ,k (r ) r where the wavevector k is confined to the FBZ and n is the band index r r n,k (r + R ) = e i k . R n,k (r ) () 2) Under a lattice translation, Bloch functions obey the relation: r r r r r Now we ask the following question: if an external potential is added to the crystal Hamiltonian, r H + U (r , t ) then what happens? How do the electrons behave? How do we find the new energies and eigenstates? The extern ...

• 1 Pages

#### lec0221-011-07S

St. Marys CA, PHYS 011

Excerpt: ... The two spheres are made to touch. What charge is now on qA ? qB ? (d) Do the spheres repel the less, the same, or more than before? 3. A point-like object with a charge of +2C sits, fixed, on the y - axis at point R; at the same time, a uniform electric field exists in the region (made by some parallel plates not shown) with magnitude of 500 N/C directed in the +y direction. (a) Calculate the electric field (both magnitude and direction) made by the point charge and the uniform field at point S. (b) An object charged with qC = +4C is placed at S. Determine the electrostatic force on this object. (c) Without doing a calculation, explain what happens to your answers to (a) and (b) if qC = -4C. 4. Estimate the net force between the CO group and the HN group shown below. The C and O have charges 0.40e and the H and N have charges 0.20e. [Hint: do not include the internal forces between C and O, or between H and N.] ...

• 28 Pages

#### Lecture_03_2097

Pittsburgh, PHYS 01750

Excerpt: ... the charge. The relevant electric field is that from other charges (not q), also know as the `external electric field'. Avoid confusion: q is not affected by its own electric field! 11 An electron is placed in the electric field at point P as shown. The direction of the force on the charge is: a) right b) left c) up d) down e) none of the above P 12 Uniform Electric Field Charge two parallel conducting plates by connecting them up to a battery. Away from the edges, the electric field between the plates is uniform, i.e., it has the same magnitude and direction throughout the interior region. 13 Halliday, Resnick and Walker: Problem 22-45 Beams of high-speed protons can be produced in "guns" using electric fields to accelerate the protons. (a) What acceleration would a proton experience if the gun's electric field were 2.00 104 N/C? (b) What speed would the proton attain if the field accelerated the proton through a distance of 1.00 cm? 14 Halliday, Resnick and Walker: Problem 22-48 At some i ...

• 4 Pages

#### lecture02A

Penn State, PHYS 212L

Excerpt: ... applied electric field. In the demo, electric field is created by a 5000 V power supply connected across two conductors "Don't try this at home!" The dipoles tend to align with the field. Why? We'll see in a moment. 11 Electric Dipole in a Uniform Electric Field +Q Uniform E Field -Q (Charge separation = d) FCalculate torque about dipole center F+ r+ r- E 5000V + r r r = r F = 2 ( d/2 ) ( QE ) sin j^ Net force on dipole = 0; ^ center of mass stays where = ( Qd ) Esin j k^ r it is. = p Esin j^ j^ r Net torque is into page. r = p E i^ r r r Dipole rotates to align with = p E E! 12 3 Potential Energy of an Electric Dipole F A dipole in an electric field rotates to align with E, due to the torque, . Forces, torques always act to reduce potential energy the energy of a dipole aligned with a field must be less than the unaligned energy. (things roll downhill) + E F An electric dipole is placed in a non- uniform electric field as Non-uniform +Q 2 E F ...