# 2 A Dielectric Sphere in a Uniform Electric Field

Save this PDF as:

Size: px
Start display at page:

## Transcription

1 Dielectric Problems and Electric Susceptability Lecture 10 1 A Dielectric Filled Parallel Plate Capacitor Suppose an infinite, parallel plate capacitor with a dielectric of dielectric constant ǫ is inserted between the plates. The field is perpendicular to the plates and to the dielectric surfaces. Thus use Gauss Law to find the field between the plates in the dielectric. For a cylindrical Gaussian surface surrounding an area of the plate surface; D d A = qfree Note the field is only between the plates with a value obtained by superposition of the field from both plates. E = σ free /ǫ The potential between the plates is therefore; V = E d l = σd/ǫ; where d is the plate separation. The capacitance is; C = Q/V = (Area)ǫ/d For a given value of V, the dielectric reduces the total field between the plates, so the capacitor stores additional charge on the plates for the same applied voltage. This is developed further in the sections below. However for the moment, you should consider how energy is conserved when it is determined using the energy density stored in the electric field. 2 A Dielectric Sphere in a Uniform Electric Field In a previous lecture the field due to a conducting sphere in a uniform electric field was found. This field caused charge to move so that there was no E component parallel to the spherical surface and no field inside the conductor. In the case of a dielectric sphere with dielectric constant ǫ = ǫ 0 ǫ r, Figure 1, charge cannot move, but charge polarization in the material occurs. Therefore there can be no free charge in, or on, the sphere. Then apply Laplace s equation with vanishing free charge density and the appropriate boundary conditions at the spherical surface. 1

2 ε o E = Eo r cos( θ ) z^ E = Eo r cos( θ ) z^ ε E = Eo r cos( θ ) z^ E = Eo r cos( θ ) z^ Figure 1: The geometry of a dielectric sphere placed in a uniform field 2 V = ρ/ǫ The above equation is solved using separation of variables with the boundary conditions; V = E 0 z = E 0 r cos(θ) as r ; Now E is finite as r = 0; and E = (ǫ /ǫ 0 ) E, E = E at r = a In spherical coordinates the solution to Laplaces s equation with azimuthal symmetry when applying separation of variables has the form; For r < a For r > a V = κ A l r l P l (x) V = κ B l r (l+1) P l (x) Then, x = cos(θ). Apply the boundary conditions to obtain the equation; V = (κ)[a 0 + A 1 r cos(θ)] for r < a V = (κ/r)[b 0 + B 1 /r] cos(θ) V 0 rcos(θ) for r > a 2

3 These solutions match the boundary conditions when r and r 0. All other coifficients, A l,b l, must vanish. Then match the potential and field when r = a. Use E = ǫ V so that; ǫ V r in = ǫ 0 V r out By definition, ǫ r = ǫ/ǫ 0 which gives; ǫ r Al l a l 1 P l = B l (l + 1) a (l+2) P l E 0 cos(θ) The requirement that tangential E is continuous is equivalent to the continuity of the potential. Al a l P l = B l a (l+1) P l E 0 P l Equate the constants for each value of l. A 0 = B 0 /a and B 0 a 2 = 0 A 1 a = B 1 a 2 E 0 a and ǫ r A 1 = 2B 1 a 3 + E 0 A 2 a 2 = B 2 a 3 and ǫ r aa 2 = 3B 2 a 4 This means that ; B 0 = 0; A 0 = 0 A 1 = B 1 /a 3 E 0 All other values of A l and B l are zero. Finally; B 1 = (ǫ r 1) (ǫ r + 2) a3 E 0 A 1 = 3 (ǫ r + 2) E 0 The potential is then; r < a V = 3 E 0 (ǫ r + 2) r cos(θ) 3

4 σ E z θ de Figure 2: The geometry used to find the field at the center of the polarized sphere r > a V = E 0 rcos(θ) + (ǫ r 1) (ǫ r + 2) E 0 (a/r) 3 r cos(θ) 3 Polarization of the Dielectric Sphere in a Uniform Electric Field. The field inside the dielectric sphere, as obtained in the last section, is; E = V with V in = 3E 0 (ǫ r + 2) r cos(θ). Thus E = 3E 0 (ǫ r + 2) ẑ The field is uniform and in the ẑ direction. The volume polarization-charge density is given by ρ = P = 0, since the field within the sphere is constant. The Polarization is given by; P = (ǫ ǫ 0 ) E Thus the volume charge density vanishes, but there is a surface charge density given by; σ = P ˆn = P cos(θ) where ˆn is the outward surface normal. The field inside the sphere is due to the surface charge which forms a dipole field, Figure 2, in addition to the applied field. Calculate this 4

5 z P y E E E E x Figure 3: The geometry used to find the field at the center of the polarized sphere field at the center of the sphere 2. The field due to a small element as shown in the figure is; de z = κ P cos(θ) r 2 cos(θ) r 2 dω Integrate over the solid angle dω; E z = P 3ǫ0 Although this field was found at the center of the sphere, it is the same for all points in the sphere, since the field inside the sphere is constant as obtained in the solution using separation of variables. Note this solution also shows the polarization field is directed opposite to the applied field, Figure Connection between the Electric Susceptibility and Atomic Polarizability An applied field induces a polarization in a dielectric material. To better understand this process consider the polarization at the center of a polarized spherical dielectric. Then combine this polarization with the applied field to obtain the field for a Class A dielectric; D = ǫ 0 E + P = ǫ E In this case the polarization is independent of surface effects. Assume the applied field is E 0, 5

6 and E is the field due to the polarized material. The total field inside the dielectric is the superposition of these fields. Now the total field (applied plus induced) causes the polarization, so the effect is non-linear due to this self interaction. The field in the dielectric may be approximated by the applied field, vece 0 and the field due to polarization as calculated in the last section. E total = E 0 + E pol with E pol = P/3ǫ 0 The dipole moment of the sphere is obtained from the atomic polarizability, α. p = α E total = α(ǫ 0 E0 + P 3 ) The polarization is the dipole moment per unit volume, P = N p, where N is the number density of dipoles. Now solve for the polarization. P/N = α( E 0 + P/3ǫ 0 ) The electric susceptibility, χ e, is defined by; χ e = ǫ 0 (ǫ r 1) P = ǫ 0 χ e E Substitute for P in the above equation to obtain the following equation Nα = 3ǫ 0 ǫ r 1 ǫ r + 2 The above is the Clausis-Mossotti equation. The polarization is; P = ǫ 0 χ e Etotal = (ǫ ǫ 0 ) E total Now replace the total field in the material, E total, with applied field, E 0. This results in a first order polarization, P 1. P 1 = ǫ 0 χ e E 0 = (ǫ r 1)ǫ 0 E 0 However, this does not equal the above value for the final polarization, ie P P 1. Thus the polarization itself acts to create new polarization (ie a non-linear effect). Iterate the above equations, beginning with the first order polarization. This gives an electric field, E 1, due 6

7 to the polarization, which then produces the next order in the iteration of the polarization. E 1 = P 1 3ǫ = (ǫ r 1) 0 3 E 0 Thus an incremental polarization, P 2 is obtained. P 2 = (ǫ r 1)ǫ 0 E 1 = (ǫ r 1) 2 3 and thus an additional E field; E 2 = ( ǫ r 1 3 ) 2 E 0 Continuing the iterations; ǫ 0 E o P = 3 n ( ǫ r 1 3 ) n ǫ 0E0 = 3(ǫ r 1) ǫ r + 2 ǫ 0E 0 To summarize, the following solution is found for the field in the interior of a dielectric sphere in an applied uniform electric field. E total = 3E 0 ǫ r + 2 From the definition of the electric displacement, D = ǫ 0 E + P, so that P is; P = ǫ 0 (ǫ r 1) E For the dielectric sphere; P = 3 (ǫ r 1) ǫ r + 2 ǫ 0 E 0 5 Energy in a Dielectric Return to a parallel plate capacitor filled with a dielectric constant, ǫ, and plate separation, d. IN review, the capacitance is ; C = Q/V V = Ed C = Q/Ed 7

8 Use Gauss s Law to get E; D d A = Qfree E = σ/ǫ C = ǫ(area)/d Without the dielectric the capacitance is Therefore; C 0 = ǫ 0 (Area)/d C = ǫ r C 0 = ǫ 0E t + P E t C 0 In the above, E t is the total field in the capacitor. From this obtain; D = ǫ r ǫ 0 Et = (ǫ 0 Et + P) Then P points in opposition to E ǫ r = (ǫ 0 + P/E t ) = ǫ 0 (1 + χ e ) The above equation connects the permittivity (dielectric constant) to the susceptibility, χ e. The energy of a parallel plate capacitor is obtained by; W = 1/2 CV 2 = 1/2 ǫ r C 0 V 2 W = (ǫ/2) dτ E 2 When one keeps the same voltage across the capacitor, there is an increase in energy W = ǫ r W 0 in a dielectric of the capacitor. Look at this additional energy. The differential energy to polarize a dipole p in the direction of the applied field is ; dw = F d x = q E d x = Edp dw = E dp Then use the dipole density, N, to obtain the potential energy per unit volume due to the polarization, P = N p. 8

9 ε l d h Figure 4: A parallel plate capacitor dipped into a dielectric liquid dw = dp E = de ǫ0 (ǫ r 1)E W = (1/2)ǫ 0 (ǫ r 1)E 2 This is to be added to the energy density of the vacuum field (ǫ 0 /2)E 2 and finally gives the expected result (ǫ/2)e 2. Thus the missing energy in the field is stored in the polarization of the material. 6 Example of Energy in a Dielectric Suppose a charged parallel plate capacitor is dipped into a dielectric liquid. The liquid is pulled up into the capacitor. The final position of the liquid can be determined by minimizing the system energy. The geometry is shown in Figure 4. In this problem the voltage is disconnected from the capacitor so the charge remains constant, but the voltage changes as the liquid fills the volume between the plates. On the other hand, if the voltage supply remains connected to the capacitor, then the voltage remains constant, but the charge changes. In this case the battery charging the plates continues to insert energy into the system, so the total energy in the capacitor is not constant. From the figure, the system can be considered as 2 capacitors connected in parallel. Assume the width of the capacitor plates is w. The capacitance values for each capacitor are; C 1 = ǫ 0w(l h) d C 2 = ǫwh h 9

10 So that the system capacitance is C t = C 1 + C 2. C t = C 0 [1 + (h/l)(ǫ r 1)] Where C 0 = ǫ 0lw is used. Write the energy stored in the capacitor as a function of h in d terms of the stored charge, W = (1/2)Q 2 /C. W = Q 2 2C 0 [1 + (h/l)(ǫ r 1)] Since ǫ r > 1 the energy decreases as h increases. The difference in the energy goes into raising the liquid. The system energy is then; W S = W + mg(h/2) = W + g(h/2)ρ(wdh) In the above ρ is the mass density and the second term on the left represents the potential energy of the raised liquid. The minimum in the energy determining the equilibrium position is then found. W S h = 0 Q 2 l(ǫ r 1) 2C 0 [l + h(ǫ r 1)] 2 + 2ρwd(h/2) = 0 Solve for h to find the equilibrium position. This results in a cubic equation for h. h 3 + (ǫ 2 l r 1) h2 + l 2 2Q (ǫ r 1) 2 h 2 l 4ρw 2 = 0 ǫ 0 (ǫ r 1)C 0 There is one real root of the equation if q 3 + r 2 > 0 where; q = (1/3)[ l ǫ r 1 ]2 r = (1/9)( l 2Q (ǫ r 1) )3 (3/4) 2 l ρw 2 gǫ 0 (ǫ r 1) This will be the case in all physical situations. The solution is obtained as follows. a 2 = 2l ǫ r 1 s 1 = [r + (q 3 + r 2 ) 1/2 ] 1/3 s 2 = [r + (q 3 r 2 ) 1/2 ] 1/3 10

11 z ε εo x q a y Figure 5: The geometry of a problem with a point charge q placed at the center of a spherical tank of water h = (s 1 + s 2 a 2 /3 7 Point Charge placed at the center of a Spherical Tank of Water. The geometry of the problem is shown in figure 5. Use Gauss law to get the electric displacement in the water. The electric displacement (and electric field) is radial and independent of angle. Thus assume a small spherical shell centered on the charge. Because the field is radial, the electric displacement, D equals the electric displacement in the water, D. This means; D d A = Qfree = q Thus because of symmetry; D = 1 4π q r 2 ˆr and D = ǫ E. Then the polarization is, P = ǫ 0 (ǫ r 1) E = ǫ 0 (ǫ r 1) 1 4πǫ q r 2 ˆr The volume charge density is; ρ = P = 1 r 2 r [r2 P r ] = 0 11

12 Thus there is no volume charge density. The surface charge density at r = a is; σ = P ˆr = ǫ 0 (ǫ r 1) E = ǫ 0 (ǫ r 1) 1 4πǫ q a 2 Inside there is an induced charge symmetrically placed about q at a finite radius so that the total induced charge sums to zero. Outside the water tank the field is the same as the field from a point charge q in the vacuum. 8 Dipole placed at the center of a Spherical Tank of Water This problem is similar to the problem in the last section, but the point charge is replaced by a dipole aligned along the ẑ axis. The field of the dipole in vacuum is; E d = κ p r 3 [2 cos(θ)ˆr + sin(θ) ˆθ] Put this dipole inside a small spherical volume of radius, b, in the center of the tank. This keeps the solution appropriately bounded as r 0. Thus the boundary conditions at r = b are; ǫe r (water) = ǫ 0 E r (vacuum) E (water) = E (vacuum) Therefore inside the water; E(water) = κp r 3 [2(ǫ 0 /epsilon) cos(θ) ˆr + sin(θ) ˆθ] Solve for the potential in the water using separation of variables. The solution has the forms; r > a V = A l r (l+1) P l (x) b < r < a V = [B l r (l+1) + C l r l ]P l (x) Now match the boundary conditions at r = b. 12

13 ǫ[2b 1 /b 3 C 1 ]cos(θ) = 2ǫ 0 κp/b 3 cos(θ) (1/b)[B 1 /b 2 + C 1 b]sin(θ) = κp/b 3 sin(θ) All other coefficients vanish. Solve for B 1 and C 1. B 1 = κp 3ǫ r [ǫ r + 2] C 1 = 2κp 3ǫ r b 3 [ǫ r 1] This gives the potential; V = 3ǫ κp [ ǫ r + 2 r r 3 + 2(ǫ r 1) b 3 r]cos(θ) From this one gets the field; E = 3ǫ κp [ [ ǫ r + 2 r r 2 + 2(ǫ r 1)r ] cos(θ) ˆr + b 3 [ ǫ r + 2 r 3 + 2(ǫ r 1) b 3 ] sin(θ) ˆθ] The polarization is P = ǫ 0 (ǫ r 1) E. So that the volume charge density is; ρ = (ǫ 0 (ǫ r 1) E) ρ = ǫ 0(ǫ r 1) κp 3ǫ r r 2 [[ 2(ǫ r + 2) r 2 2(ǫ r 1) b 3 ] cos(θ) + [ ǫ r + 2 r 3 + 2(ǫ r 1) b 3 ]cos(θ)] ρ = ǫ 0(ǫ r 1) κp 3ǫ r r 2 [3(ǫ r + 2) + 6(ǫ r 1)(r/b) 3 ] cos(θ) The surface charge density is; r = b σ = (ǫ ǫ 0 ) 8κp 3ǫ r b 3 cos(θ) r = a σ = (ǫ ǫ 0 ) 4κp 3ǫ r a 3 [ǫ r (1 ǫ r (a/b) 3 ) + 2(a/b) 3 ] cos(θ) Matching the boundary conditions at r = a must now be carefully done. The field does not 0 as r but has a dipole form, with the potential given by; 13

14 z d ε ε 2 a x y Figure 6: The geometry used to find the capacitance of a parallel plate capacitor filled with 2 dielectrics V = 2κp(ǫ r 1) b 3 r cos(θ) This potential should be subtracted from the dipole potential. There still remains a problem in defining a dipole as a point. 9 Examples 9.1 Parallel plate capacitor filled with 2 dielectric materials A parallel plate capacitor is filled with 2 dielectric materials of dielectric constants ǫ 1, and ǫ 2 as shown in figure 6. The plates have area, A, and are separated by a distance, d. The thickness of dielectric, ǫ 2 is a. Find the capacitance. Place a potential, V 0, between the plates. By symmetry, the potential is assumed to be dependent only on z as the plate lengths are >> d. Thus the electric field is perpendicular to the plates. The solution to Laplace s equation, 2 V = 0, must have the form; V = Az + B In the above A and B are constants which are used to satisfy the boundary conditions. Thus; V 2 = A 2 z + B 2 for 0 < z < d a) V 1 = A 1 z + B 1 for d a < z < d Then when Z = 0 V 2 = B 2 = 0 14

15 and when z = d V = A 1 d + B 1 = V 0 Solving for the constants the potentials become; V 1 = [ V 0 B 1 ]z + B d 1 V 2 = A 2 z The fields are; E 1 = V 1 z = [V 0 B 1 ] d E 1 = V 2 z = A 2 The remaining constants, A 2 and B 1, are determined at the dielectric boundary, z = a, where; Also require; D 1 = ǫ 1 E 1 = D 2 = ǫ 2 E 2 A 2 = (ǫ 1 /ǫ 2 )[ V 0 B 1 ] d 0 d d l E = V 0. Then solve the 2 above equations for the constants. V B 1 = 0 a(ǫ 1 ǫ 2 ) [ǫ 1 a + ǫ 2 (d a)] A 2 = ǫ 1 V 0 [ǫ 1 a + ǫ 2 (d a)] This yields the potentials; V 1 = V 0[ǫ 2 z + (ǫ 1 ǫ 2 )a] [ǫ 1 a + ǫ 2 (d a)] V 2 = ǫ 1 V 0 [ǫ 1 a + ǫ 2 (d a)] z The fields are; E 2 = E 1 = ǫ 1 [ǫ 1 a + ǫ 2 (d a)] V 0z ǫ 2 [ǫ 1 a + ǫ 2 (d a)] V 0z 15

16 These solutions should be checked for the limiting cases when a = 0, a = d, and ǫ 1 = ǫ 2 Now the charges on the plates are obtained from Gauss law, D d A = Qfree. Bottom Plate; σ F = ǫ 2 E 2 = ǫ 1 ǫ 2 ǫ 1 a + ǫ 2 (d a) V 0 Top Plate; σ F = ǫ 1 E 1 = ǫ 1 ǫ 2 ǫ 1 a + ǫ 2 (d a) V 0 So the free charge on the top and bottom plates are equal as they should be. Then using the free charge, the capacitance per unit area is; C/A = σ F /V 0 = ǫ 1 E 1 = ǫ 1 ǫ 2 ǫ 1 a + ǫ 2 (d a) V The field of a polarized sphere Suppose a sphere of radius, R, composed of polarized material. The polarization has only and angular dependence given by; P = P 0 cos(θ) ˆr There is a surface charge density; σ = P ˆr = P 0 cos(θ) The volume charge density is; ρ = P = 2P 0 cos(θ) r Then find the potential outside the sphere, Figure 7. The volume contribution to the potential has the form;. V = κ dω r 2 ρ r r Substitute for the volume charge density and use the addition theorem to expand 1 r r 16

17 r P o cos ( q ) r Figure 7: Geometry to obtain the potential of a polarized sphere 1 r r = l,m 4π 2l + 1 (r l /r (l+1) ) Yl m (θ, φ ) Yl m (θ, φ). Note, cos(θ) = 4π/3Yl 0. In the following use the orthogonality of the spherical harmonics. V = 8πκ(P 0 /r) 1 2l + 1 V = 8πκ( P 0R 3 3r 3 ) cos(θ) [ dω cos(θ, φ ) Y m l The integral for the surface proceeds in a similar way. V = κ R 2 dω P 0 cos(θ) r r V = 4πκ P 0R 3 cos(θ) 3r 3 (θ, φ )][ dr (r l+2 /r l )] Y m l The potential is the sum of the potentials from the volume and surface charge densities. 17

### Chapter 4. Electrostatic Fields in Matter

Chapter 4. Electrostatic Fields in Matter 4.1. Polarization A neutral atom, placed in an external electric field, will experience no net force. However, even though the atom as a whole is neutral, the

### Chapter 22: Electric Flux and Gauss s Law

22.1 ntroduction We have seen in chapter 21 that determining the electric field of a continuous charge distribution can become very complicated for some charge distributions. t would be desirable if we

### Chapter 7: Polarization

Chapter 7: Polarization Joaquín Bernal Méndez Group 4 1 Index Introduction Polarization Vector The Electric Displacement Vector Constitutive Laws: Linear Dielectrics Energy in Dielectric Systems Forces

### HW6 Solutions Notice numbers may change randomly in your assignments and you may have to recalculate solutions for your specific case.

HW6 Solutions Notice numbers may change randomly in your assignments and you may have to recalculate solutions for your specific case. Tipler 22.P.053 The figure below shows a portion of an infinitely

### Exercises on Voltage, Capacitance and Circuits. A d = (8.85 10 12 ) π(0.05)2 = 6.95 10 11 F

Exercises on Voltage, Capacitance and Circuits Exercise 1.1 Instead of buying a capacitor, you decide to make one. Your capacitor consists of two circular metal plates, each with a radius of 5 cm. The

### Mechanics lecture 7 Moment of a force, torque, equilibrium of a body

G.1 EE1.el3 (EEE1023): Electronics III Mechanics lecture 7 Moment of a force, torque, equilibrium of a body Dr Philip Jackson http://www.ee.surrey.ac.uk/teaching/courses/ee1.el3/ G.2 Moments, torque and

### State of Stress at Point

State of Stress at Point Einstein Notation The basic idea of Einstein notation is that a covector and a vector can form a scalar: This is typically written as an explicit sum: According to this convention,

### 11. Rotation Translational Motion: Rotational Motion:

11. Rotation Translational Motion: Motion of the center of mass of an object from one position to another. All the motion discussed so far belongs to this category, except uniform circular motion. Rotational

### Problem 1 (25 points)

MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Physics 8.02 Spring 2012 Exam Three Solutions Problem 1 (25 points) Question 1 (5 points) Consider two circular rings of radius R, each perpendicular

### AP Physics C: Electricity and Magnetism 2011 Scoring Guidelines

AP Physics C: Electricity and Magnetism 11 Scoring Guidelines The College Board The College Board is a not-for-profit membership association whose mission is to connect students to college success and

### Section 11.1: Vectors in the Plane. Suggested Problems: 1, 5, 9, 17, 23, 25-37, 40, 42, 44, 45, 47, 50

Section 11.1: Vectors in the Plane Page 779 Suggested Problems: 1, 5, 9, 17, 3, 5-37, 40, 4, 44, 45, 47, 50 Determine whether the following vectors a and b are perpendicular. 5) a = 6, 0, b = 0, 7 Recall

### Section 9.5: Equations of Lines and Planes

Lines in 3D Space Section 9.5: Equations of Lines and Planes Practice HW from Stewart Textbook (not to hand in) p. 673 # 3-5 odd, 2-37 odd, 4, 47 Consider the line L through the point P = ( x, y, ) that

### Midterm Solutions. mvr = ω f (I wheel + I bullet ) = ω f 2 MR2 + mr 2 ) ω f = v R. 1 + M 2m

Midterm Solutions I) A bullet of mass m moving at horizontal velocity v strikes and sticks to the rim of a wheel a solid disc) of mass M, radius R, anchored at its center but free to rotate i) Which of

### A vector is a directed line segment used to represent a vector quantity.

Chapters and 6 Introduction to Vectors A vector quantity has direction and magnitude. There are many examples of vector quantities in the natural world, such as force, velocity, and acceleration. A vector

### MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Physics. 8.02 Spring 2013 Conflict Exam Two Solutions

MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Physics 802 Spring 2013 Conflict Exam Two Solutions Problem 1 (25 points): answers without work shown will not be given any credit A uniformly charged

### Solutions for Review Problems

olutions for Review Problems 1. Let be the triangle with vertices A (,, ), B (4,, 1) and C (,, 1). (a) Find the cosine of the angle BAC at vertex A. (b) Find the area of the triangle ABC. (c) Find a vector

### ... ... . (2,4,5).. ...

12 Three Dimensions ½¾º½ Ì ÓÓÖ Ò Ø ËÝ Ø Ñ So far wehave been investigatingfunctions ofthe form y = f(x), withone independent and one dependent variable Such functions can be represented in two dimensions,

### A wave lab inside a coaxial cable

INSTITUTE OF PHYSICS PUBLISHING Eur. J. Phys. 25 (2004) 581 591 EUROPEAN JOURNAL OF PHYSICS PII: S0143-0807(04)76273-X A wave lab inside a coaxial cable JoãoMSerra,MiguelCBrito,JMaiaAlves and A M Vallera

### LINES AND PLANES IN R 3

LINES AND PLANES IN R 3 In this handout we will summarize the properties of the dot product and cross product and use them to present arious descriptions of lines and planes in three dimensional space.

### Dot product and vector projections (Sect. 12.3) There are two main ways to introduce the dot product

Dot product and vector projections (Sect. 12.3) Two definitions for the dot product. Geometric definition of dot product. Orthogonal vectors. Dot product and orthogonal projections. Properties of the dot

### Physics 53. Kinematics 2. Our nature consists in movement; absolute rest is death. Pascal

Phsics 53 Kinematics 2 Our nature consists in movement; absolute rest is death. Pascal Velocit and Acceleration in 3-D We have defined the velocit and acceleration of a particle as the first and second

### Edmund Li. Where is defined as the mutual inductance between and and has the SI units of Henries (H).

INDUCTANCE MUTUAL INDUCTANCE If we consider two neighbouring closed loops and with bounding surfaces respectively then a current through will create a magnetic field which will link with as the flux passes

### Review of Vector Analysis in Cartesian Coordinates

R. evicky, CBE 6333 Review of Vector Analysis in Cartesian Coordinates Scalar: A quantity that has magnitude, but no direction. Examples are mass, temperature, pressure, time, distance, and real numbers.

### Eðlisfræði 2, vor 2007

[ Assignment View ] [ Pri Eðlisfræði 2, vor 2007 28. Sources of Magnetic Field Assignment is due at 2:00am on Wednesday, March 7, 2007 Credit for problems submitted late will decrease to 0% after the deadline

### Chapter 23 Electric Potential. Copyright 2009 Pearson Education, Inc.

Chapter 23 Electric Potential 23-1 Electrostatic Potential Energy and Potential Difference The electrostatic force is conservative potential energy can be defined. Change in electric potential energy is

### PHY114 S11 Term Exam 3

PHY4 S Term Exam S. G. Rajeev Mar 2 20 2:0 pm to :45 pm PLEASE write your workshop number and your workshop leader s name at the top of your book, so that you can collect your graded exams at the workshop.

### Math 241, Exam 1 Information.

Math 241, Exam 1 Information. 9/24/12, LC 310, 11:15-12:05. Exam 1 will be based on: Sections 12.1-12.5, 14.1-14.3. The corresponding assigned homework problems (see http://www.math.sc.edu/ boylan/sccourses/241fa12/241.html)

### COURSE: PHYSICS DEGREE: COMPUTER ENGINEERING year: 1st SEMESTER: 1st

COURSE: PHYSICS DEGREE: COMPUTER ENGINEERING year: 1st SEMESTER: 1st WEEKLY PROGRAMMING WEE K SESSI ON DESCRIPTION GROUPS GROUPS Special room for LECTU PRAC session RES TICAL (computer classroom, audiovisual

### Capacitors in Circuits

apacitors in ircuits apacitors store energy in the electric field E field created by the stored charge In circuit apacitor may be absorbing energy Thus causes circuit current to be reduced Effectively

### THREE DIMENSIONAL GEOMETRY

Chapter 8 THREE DIMENSIONAL GEOMETRY 8.1 Introduction In this chapter we present a vector algebra approach to three dimensional geometry. The aim is to present standard properties of lines and planes,

### Columbia University Department of Physics QUALIFYING EXAMINATION

Columbia University Department of Physics QUALIFYING EXAMINATION Monday, January 13, 2014 1:00PM to 3:00PM Classical Physics Section 1. Classical Mechanics Two hours are permitted for the completion of

### DEGREE: Bachelor's Degree in Industrial Electronics and Automation COURSE: 1º TERM: 2º WEEKLY PLANNING

SESSION WEEK COURSE: Physics II DEGREE: Bachelor's Degree in Industrial Electronics and Automation COURSE: 1º TERM: 2º WEEKLY PLANNING DESCRIPTION GROUPS (mark ) Indicate YES/NO If the session needs 2

### 1.3. DOT PRODUCT 19. 6. If θ is the angle (between 0 and π) between two non-zero vectors u and v,

1.3. DOT PRODUCT 19 1.3 Dot Product 1.3.1 Definitions and Properties The dot product is the first way to multiply two vectors. The definition we will give below may appear arbitrary. But it is not. It

### EE301 Lesson 14 Reading: 10.1-10.4, 10.11-10.12, 11.1-11.4 and 11.11-11.13

CAPACITORS AND INDUCTORS Learning Objectives EE301 Lesson 14 a. Define capacitance and state its symbol and unit of measurement. b. Predict the capacitance of a parallel plate capacitor. c. Analyze how

### Concepts in Calculus III

Concepts in Calculus III Beta Version UNIVERSITY PRESS OF FLORIDA Florida A&M University, Tallahassee Florida Atlantic University, Boca Raton Florida Gulf Coast University, Ft. Myers Florida International

### Level Set Framework, Signed Distance Function, and Various Tools

Level Set Framework Geometry and Calculus Tools Level Set Framework,, and Various Tools Spencer Department of Mathematics Brigham Young University Image Processing Seminar (Week 3), 2010 Level Set Framework

### Persuasion by Cheap Talk - Online Appendix

Persuasion by Cheap Talk - Online Appendix By ARCHISHMAN CHAKRABORTY AND RICK HARBAUGH Online appendix to Persuasion by Cheap Talk, American Economic Review Our results in the main text concern the case

### Eðlisfræði 2, vor 2007

[ Assignment View ] [ Print ] Eðlisfræði 2, vor 2007 30. Inductance Assignment is due at 2:00am on Wednesday, March 14, 2007 Credit for problems submitted late will decrease to 0% after the deadline has

### The Technical Archer. Austin Wargo

The Technical Archer Austin Wargo May 14, 2010 Abstract A mathematical model of the interactions between a long bow and an arrow. The model uses the Euler-Lagrange formula, and is based off conservation

### 2.1 Three Dimensional Curves and Surfaces

. Three Dimensional Curves and Surfaces.. Parametric Equation of a Line An line in two- or three-dimensional space can be uniquel specified b a point on the line and a vector parallel to the line. The

### Last time : energy storage elements capacitor.

Last time : energy storage elements capacitor. Charge on plates Energy stored in the form of electric field Passive sign convention Vlt Voltage drop across real capacitor can not change abruptly because

### Q24.1 The two conductors a and b are insulated from each other, forming a capacitor. You increase the charge on a to +2Q and increase the charge on b

Q24.1 The two conductors a and b are insulated from each other, forming a capacitor. You increase the charge on a to +2Q and increase the charge on b to 2Q, while keeping the conductors in the same positions.

### Worked Examples from Introductory Physics Vol. I: Basic Mechanics. David Murdock Tenn. Tech. Univ.

Worked Examples from Introductory Physics Vol. I: Basic Mechanics David Murdock Tenn. Tech. Univ. February 24, 2005 2 Contents To the Student. Yeah, You. i 1 Units and Vectors: Tools for Physics 1 1.1

### Heterogeneous Catalysis and Catalytic Processes Prof. K. K. Pant Department of Chemical Engineering Indian Institute of Technology, Delhi

Heterogeneous Catalysis and Catalytic Processes Prof. K. K. Pant Department of Chemical Engineering Indian Institute of Technology, Delhi Module - 03 Lecture 10 Good morning. In my last lecture, I was

### Physics 2A, Sec B00: Mechanics -- Winter 2011 Instructor: B. Grinstein Final Exam

Physics 2A, Sec B00: Mechanics -- Winter 2011 Instructor: B. Grinstein Final Exam INSTRUCTIONS: Use a pencil #2 to fill your scantron. Write your code number and bubble it in under "EXAM NUMBER;" an entry

### MATH 275: Calculus III. Lecture Notes by Angel V. Kumchev

MATH 275: Calculus III Lecture Notes by Angel V. Kumchev Contents Preface.............................................. iii Lecture 1. Three-Dimensional Coordinate Systems..................... 1 Lecture

### C;Ri.N. (o3s8: Université Louis Pasteur de Strasbourg. Institut National de Physique Nucléaire et de Physique des Particules t...

(o3s8: C;Ri.N CWMB 7»-fi CO i\ ANGULAR ASSÏMBTRIBS IN A SHIELDED -^ PAIR LINE CO CJ H.-FANMarri, C.A. GMCIA CANAL «nd H. VUCETICH o \. CHRKE DE «CH.1RCHES KKXEAIRES UHIVERSITB LOUIS PASTEUR STRASBOURG

### Learning Objectives for AP Physics

Learning Objectives for AP Physics These course objectives are intended to elaborate on the content outline for Physics B and Physics C found in the AP Physics Course Description. In addition to the five

### A) F = k x B) F = k C) F = x k D) F = x + k E) None of these.

CT16-1 Which of the following is necessary to make an object oscillate? i. a stable equilibrium ii. little or no friction iii. a disturbance A: i only B: ii only C: iii only D: i and iii E: All three Answer:

### Chapter 30 Inductance

Chapter 30 Inductance - Mutual Inductance - Self-Inductance and Inductors - Magnetic-Field Energy - The R- Circuit - The -C Circuit - The -R-C Series Circuit . Mutual Inductance - A changing current in

### Numerical Analysis Lecture Notes

Numerical Analysis Lecture Notes Peter J. Olver 5. Inner Products and Norms The norm of a vector is a measure of its size. Besides the familiar Euclidean norm based on the dot product, there are a number

### Lecture 6. Weight. Tension. Normal Force. Static Friction. Cutnell+Johnson: 4.8-4.12, second half of section 4.7

Lecture 6 Weight Tension Normal Force Static Friction Cutnell+Johnson: 4.8-4.12, second half of section 4.7 In this lecture, I m going to discuss four different kinds of forces: weight, tension, the normal

### INVESTIGATION OF ELECTRIC FIELD INTENSITY AND DEGREE OF UNIFORMITY BETWEEN ELECTRODES UNDER HIGH VOLTAGE BY CHARGE SIMULATIO METHOD

INVESTIGATION OF ELECTRIC FIELD INTENSITY AND DEGREE OF UNIFORMITY BETWEEN ELECTRODES UNDER HIGH VOLTAGE BY CHARGE SIMULATIO METHOD Md. Ahsan Habib, Muhammad Abdul Goffar Khan, Md. Khaled Hossain, Shafaet

### Lecture 3: Linear methods for classification

Lecture 3: Linear methods for classification Rafael A. Irizarry and Hector Corrada Bravo February, 2010 Today we describe four specific algorithms useful for classification problems: linear regression,

### VELOCITY, ACCELERATION, FORCE

VELOCITY, ACCELERATION, FORCE velocity Velocity v is a vector, with units of meters per second ( m s ). Velocity indicates the rate of change of the object s position ( r ); i.e., velocity tells you how

### Slide 1 / 26. Inductance. 2011 by Bryan Pflueger

Slide 1 / 26 Inductance 2011 by Bryan Pflueger Slide 2 / 26 Mutual Inductance If two coils of wire are placed near each other and have a current passing through them, they will each induce an emf on one

### Linear algebra and the geometry of quadratic equations. Similarity transformations and orthogonal matrices

MATH 30 Differential Equations Spring 006 Linear algebra and the geometry of quadratic equations Similarity transformations and orthogonal matrices First, some things to recall from linear algebra Two

### 3 Concepts of Stress Analysis

3 Concepts of Stress Analysis 3.1 Introduction Here the concepts of stress analysis will be stated in a finite element context. That means that the primary unknown will be the (generalized) displacements.

### Chapter 2. Derivation of the Equations of Open Channel Flow. 2.1 General Considerations

Chapter 2. Derivation of the Equations of Open Channel Flow 2.1 General Considerations Of interest is water flowing in a channel with a free surface, which is usually referred to as open channel flow.

### Objectives. Capacitors 262 CHAPTER 5 ENERGY

Objectives Describe a capacitor. Explain how a capacitor stores energy. Define capacitance. Calculate the electrical energy stored in a capacitor. Describe an inductor. Explain how an inductor stores energy.

### Solutions to old Exam 1 problems

Solutions to old Exam 1 problems Hi students! I am putting this old version of my review for the first midterm review, place and time to be announced. Check for updates on the web site as to which sections

### PHY121 #8 Midterm I 3.06.2013

PHY11 #8 Midterm I 3.06.013 AP Physics- Newton s Laws AP Exam Multiple Choice Questions #1 #4 1. When the frictionless system shown above is accelerated by an applied force of magnitude F, the tension

### (Most of the material presented in this chapter is taken from Thornton and Marion, Chap. 7)

Chapter 4. Lagrangian Dynamics (Most of the material presented in this chapter is taken from Thornton and Marion, Chap. 7 4.1 Important Notes on Notation In this chapter, unless otherwise stated, the following

### Inductance and Magnetic Energy

Chapter 11 Inductance and Magnetic Energy 11.1 Mutual Inductance... 11-3 Example 11.1 Mutual Inductance of Two Concentric Coplanar Loops... 11-5 11. Self-Inductance... 11-5 Example 11. Self-Inductance

### Section 4: The Basics of Satellite Orbits

Section 4: The Basics of Satellite Orbits MOTION IN SPACE VS. MOTION IN THE ATMOSPHERE The motion of objects in the atmosphere differs in three important ways from the motion of objects in space. First,

### THEORETICAL MECHANICS

PROF. DR. ING. VASILE SZOLGA THEORETICAL MECHANICS LECTURE NOTES AND SAMPLE PROBLEMS PART ONE STATICS OF THE PARTICLE, OF THE RIGID BODY AND OF THE SYSTEMS OF BODIES KINEMATICS OF THE PARTICLE 2010 0 Contents

### 9 Multiplication of Vectors: The Scalar or Dot Product

Arkansas Tech University MATH 934: Calculus III Dr. Marcel B Finan 9 Multiplication of Vectors: The Scalar or Dot Product Up to this point we have defined what vectors are and discussed basic notation

### XI / PHYSICS FLUIDS IN MOTION 11/PA

Viscosity It is the property of a liquid due to which it flows in the form of layers and each layer opposes the motion of its adjacent layer. Cause of viscosity Consider two neighboring liquid layers A

### LINEAR ALGEBRA W W L CHEN

LINEAR ALGEBRA W W L CHEN c W W L Chen, 1982, 2008. This chapter originates from material used by author at Imperial College, University of London, between 1981 and 1990. It is available free to all individuals,

### Dispersion diagrams of a water-loaded cylindrical shell obtained from the structural and acoustic responses of the sensor array along the shell

Dispersion diagrams of a water-loaded cylindrical shell obtained from the structural and acoustic responses of the sensor array along the shell B.K. Jung ; J. Ryue ; C.S. Hong 3 ; W.B. Jeong ; K.K. Shin

### Equations Involving Lines and Planes Standard equations for lines in space

Equations Involving Lines and Planes In this section we will collect various important formulas regarding equations of lines and planes in three dimensional space Reminder regarding notation: any quantity

### 1. Introduction. O. MALI, A. MUZALEVSKIY, and D. PAULY

Russ. J. Numer. Anal. Math. Modelling, Vol. 28, No. 6, pp. 577 596 (2013) DOI 10.1515/ rnam-2013-0032 c de Gruyter 2013 Conforming and non-conforming functional a posteriori error estimates for elliptic

### Solving Simultaneous Equations and Matrices

Solving Simultaneous Equations and Matrices The following represents a systematic investigation for the steps used to solve two simultaneous linear equations in two unknowns. The motivation for considering

### The performance improvement by ferrite loading means - increasing, - increasing of ratio, implicitly related to the input impedance.

3.2.3. Ferrite Loading Magnetic ferrite loading can enhance a transmitting signal as high as 2 to 10 db for MHz [Devore and Bohley, 1977]. There is an optimum frequency range where ferrite loading is beneficial.

### Fundamentals of Electromagnetic Fields and Waves: I

Fundamentals of Electromagnetic Fields and Waves: I Fall 2007, EE 30348, Electrical Engineering, University of Notre Dame Mid Term II: Solutions Please show your steps clearly and sketch figures wherever

### THE NUMBER OF GRAPHS AND A RANDOM GRAPH WITH A GIVEN DEGREE SEQUENCE. Alexander Barvinok

THE NUMBER OF GRAPHS AND A RANDOM GRAPH WITH A GIVEN DEGREE SEQUENCE Alexer Barvinok Papers are available at http://www.math.lsa.umich.edu/ barvinok/papers.html This is a joint work with J.A. Hartigan

### Notes on Elastic and Inelastic Collisions

Notes on Elastic and Inelastic Collisions In any collision of 2 bodies, their net momentus conserved. That is, the net momentum vector of the bodies just after the collision is the same as it was just

### Lecture notes on linear algebra

Lecture notes on linear algebra David Lerner Department of Mathematics University of Kansas These are notes of a course given in Fall, 2007 and 2008 to the Honors sections of our elementary linear algebra

### Induced voltages and Inductance Faraday s Law

Induced voltages and Inductance Faraday s Law concept #1, 4, 5, 8, 13 Problem # 1, 3, 4, 5, 6, 9, 10, 13, 15, 24, 23, 25, 31, 32a, 34, 37, 41, 43, 51, 61 Last chapter we saw that a current produces a magnetic

### PHYSICS 111 HOMEWORK SOLUTION #10. April 8, 2013

PHYSICS HOMEWORK SOLUTION #0 April 8, 203 0. Find the net torque on the wheel in the figure below about the axle through O, taking a = 6.0 cm and b = 30.0 cm. A torque that s produced by a force can be

### FINAL EXAM SOLUTIONS Math 21a, Spring 03

INAL EXAM SOLUIONS Math 21a, Spring 3 Name: Start by printing your name in the above box and check your section in the box to the left. MW1 Ken Chung MW1 Weiyang Qiu MW11 Oliver Knill h1 Mark Lucianovic

### Two vectors are equal if they have the same length and direction. They do not

Vectors define vectors Some physical quantities, such as temperature, length, and mass, can be specified by a single number called a scalar. Other physical quantities, such as force and velocity, must

### Some Comments on the Derivative of a Vector with applications to angular momentum and curvature. E. L. Lady (October 18, 2000)

Some Comments on the Derivative of a Vector with applications to angular momentum and curvature E. L. Lady (October 18, 2000) Finding the formula in polar coordinates for the angular momentum of a moving

### The Role of Electric Polarization in Nonlinear optics

The Role of Electric Polarization in Nonlinear optics Sumith Doluweera Department of Physics University of Cincinnati Cincinnati, Ohio 45221 Abstract Nonlinear optics became a very active field of research

### Introduction Assignment

PRE-CALCULUS 11 Introduction Assignment Welcome to PREC 11! This assignment will help you review some topics from a previous math course and introduce you to some of the topics that you ll be studying

### PRIMARY CONTENT MODULE Algebra I -Linear Equations & Inequalities T-71. Applications. F = mc + b.

PRIMARY CONTENT MODULE Algebra I -Linear Equations & Inequalities T-71 Applications The formula y = mx + b sometimes appears with different symbols. For example, instead of x, we could use the letter C.

### SCHWEITZER ENGINEERING LABORATORIES, COMERCIAL LTDA.

Pocket book of Electrical Engineering Formulas Content 1. Elementary Algebra and Geometry 1. Fundamental Properties (real numbers) 1 2. Exponents 2 3. Fractional Exponents 2 4. Irrational Exponents 2 5.

### Electromagnetic waves

Chapter 8 lectromagnetic waves David Morin, morin@physics.harvard.edu The waves we ve dealt with so far in this book have been fairly easy to visualize. Waves involving springs/masses, strings, and air

### Basic Electrical Technology Dr. L. Umanand Department of Electrical Engineering Indian Institute of Science, Bangalore. Lecture - 33 3 phase System 4

Basic Electrical Technology Dr. L. Umanand Department of Electrical Engineering Indian Institute of Science, Bangalore Lecture - 33 3 phase System 4 Hello everybody. So, in the last class we have been

### A QUICK GUIDE TO THE FORMULAS OF MULTIVARIABLE CALCULUS

A QUIK GUIDE TO THE FOMULAS OF MULTIVAIABLE ALULUS ontents 1. Analytic Geometry 2 1.1. Definition of a Vector 2 1.2. Scalar Product 2 1.3. Properties of the Scalar Product 2 1.4. Length and Unit Vectors

### Mathematics Notes for Class 12 chapter 10. Vector Algebra

1 P a g e Mathematics Notes for Class 12 chapter 10. Vector Algebra A vector has direction and magnitude both but scalar has only magnitude. Magnitude of a vector a is denoted by a or a. It is non-negative

### Solutions to Homework 5

Solutions to Homework 5 1. Let z = f(x, y) be a twice continously differentiable function of x and y. Let x = r cos θ and y = r sin θ be the equations which transform polar coordinates into rectangular

### INJECTION MOLDING COOLING TIME REDUCTION AND THERMAL STRESS ANALYSIS

INJECTION MOLDING COOLING TIME REDUCTION AND THERMAL STRESS ANALYSIS Tom Kimerling University of Massachusetts, Amherst MIE 605 Finite Element Analysis Spring 2002 ABSTRACT A FEA transient thermal structural

### Lecture 6 - Boundary Conditions. Applied Computational Fluid Dynamics

Lecture 6 - Boundary Conditions Applied Computational Fluid Dynamics Instructor: André Bakker http://www.bakker.org André Bakker (2002-2006) Fluent Inc. (2002) 1 Outline Overview. Inlet and outlet boundaries.

### Practice Final Math 122 Spring 12 Instructor: Jeff Lang

Practice Final Math Spring Instructor: Jeff Lang. Find the limit of the sequence a n = ln (n 5) ln (3n + 8). A) ln ( ) 3 B) ln C) ln ( ) 3 D) does not exist. Find the limit of the sequence a n = (ln n)6

### A SURVEY ON CONTINUOUS ELLIPTICAL VECTOR DISTRIBUTIONS

A SURVEY ON CONTINUOUS ELLIPTICAL VECTOR DISTRIBUTIONS Eusebio GÓMEZ, Miguel A. GÓMEZ-VILLEGAS and J. Miguel MARÍN Abstract In this paper it is taken up a revision and characterization of the class of

### 5 VECTOR GEOMETRY. 5.0 Introduction. Objectives. Activity 1

5 VECTOR GEOMETRY Chapter 5 Vector Geometry Objectives After studying this chapter you should be able to find and use the vector equation of a straight line; be able to find the equation of a plane in