Idea Transcript
4.1 Energy to move a point charge through a Field Force on Q due to an electric field F E QE
Differential work done by an external source moving Q
dW QE dL Work required to move a charge a finite distance
𝟑 𝟑 𝐚𝑥 + 𝐚𝑥 𝟕 𝟕
4.2 Line Integral using vectors Work expression without W
Q
final
E L dL
in itial
EL is the component of E in the dL direction
Uniform electric field density W QE L BA
xA=1, yA=0, zA=1 xB=0.8, yB=0.6, zB=1
Same amount of work with a different path Line integrals are path independent
D4.2. 𝑓𝑖𝑛𝑎𝑙
𝑊 = −𝑄
𝑬. 𝑑𝑳 𝑖𝑛𝑡𝑖𝑎𝑙
(a) 𝑬. 𝑑𝑳 = 𝟓𝒂𝒙 . 𝒅𝒙𝒂𝒙 + 𝒅𝒚𝒂𝒙 + 𝒅𝒛𝒂𝒙 = 𝟓𝒅𝒙 𝟎
𝑾 = −𝟒
𝟎
𝟓𝒅𝒙 = −𝟐𝟎 𝟏
𝒅𝒙 = −𝟐𝟎 × −𝟏 = 𝟐𝟎 𝑱 𝟏
(b) 𝑬. 𝑑𝑳 = 𝟓𝒙𝒂𝒙 . 𝒅𝒙𝒂𝒙 + 𝒅𝒚𝒂𝒚 + 𝒅𝒛𝒂𝒛 = 𝟓𝒙𝒅𝒙 𝟎
𝑾 = −𝟒
𝟎
𝟓𝒙𝒅𝒙 = −𝟐𝟎 𝟏
𝒙𝒅𝒙 = −𝟐𝟎 × − 𝟏
𝟏 = 𝟏𝟎 𝑱 𝟐
(c) 𝑬. 𝑑𝑳 = 𝟓𝒙𝒂𝒙 + 𝟓𝒚𝒂𝒚 . 𝒅𝒙𝒂𝒙 + 𝒅𝒚𝒂𝒚 + 𝒅𝒛𝒂𝒛 = 𝟓𝒙𝒅𝒙 + 𝟓𝒚𝒅𝒚 𝟎
𝑾 = −𝟒
𝟐
𝟓𝒙𝒅𝒙 − 𝟒 𝟏
𝟎
𝟏 𝟒 𝟓𝒚𝒅𝒚 = −𝟐𝟎 × − + = −𝟑𝟎 𝑱 𝟐 𝟐
D4.3. 𝑓𝑖𝑛𝑎𝑙
𝑊 = −𝑄
𝑬. 𝑑𝑳 𝑖𝑛𝑡𝑖𝑎𝑙
𝑬. 𝑑𝑳 = 𝒚𝒂𝒙 . 𝒅𝒙𝒂𝒙 + 𝒅𝒚𝒂𝒙 + 𝒅𝒛𝒂𝒙 = 𝒚𝒅𝒙 (a) 𝑾 = −𝑸 (b) 𝑾 = −𝑸
𝒃 𝒄 𝒅 𝒚𝒅𝒙 + 𝒚𝒅𝒙 + 𝒚𝒅𝒙 𝒂 𝒃 𝒄
𝒃 𝒚𝒅𝒙 𝒂
+
𝒄 𝒚𝒅𝒙 𝒃
+
= −𝟑
𝒅 𝒚𝒅𝒙 𝒄
𝟐 𝟐 𝟐 𝒚𝒅𝒙 + 𝒚𝒅𝒙 + 𝒚𝒅𝒙 𝟏 𝟐 𝟐
= −𝟑
𝟏 𝒚𝒅𝒙 𝟏
+
𝟏 𝒚𝒅𝒙 𝟏
+
= −𝟑 𝟑 𝟐 − 𝟏 = −𝟗 𝑱
𝟐 𝒚𝒅𝒙 𝟏
= −𝟑 𝟎 𝟐 − 𝟏 = 𝟎 𝑱
4.3 Potential Difference Potential Difference
Using radial distances from the point charge
4.3 Potential Measure potential difference between a point and something which has zero potential “ground” V AB
VA VB
Example – D4.4
2
6x E( x y z) 6y 4
a) Find Vmn
2 M 6 1
3 N 3 2
M0
VMN N
0
M1
2 6x d x N
1
M2
6y d y N
4 dz 2
b) Find Vm if V=0 at Q(4,-2,-35)
4 Q 2 35
M0
VM Q
0
M1
2 6x d x Q
1
M2
6y d y Q
4 dz
VM 120
2
c) Find Vn if V=2 at P(1, 2, -4)
1 P 2 4
N0
VN P
0
N1
6x d x P 2
1
N2
6y d y P
4 dz 2 2
VN 19
VMN 139
4.4 Potential Field of a Point Charge
Let V=0 at infinity
Equipotential surface: A surface composed of all points having the same potential
Example – D4.5 9
Q 15 10
Q is located at the origin
2 P1 3 1
a) Find V1 if V=0 at (6,5,4)
6 P0 5 4
V1
12
0 8.8 510 𝑷𝟏 = −𝟐𝟐 + 𝟑𝟐 + −𝟏𝟐 = 𝟏𝟒
𝑷𝒐 =
1 1 P0 4 0 P1 Q
𝟔𝟐 + 𝟓𝟐 + 𝟒𝟐 = 𝟔𝟕
V1 20 .677
b) Find V1 if V=0 at infinity V1
Q
1
V1 36 .047
4 0 P1
c) Find V1 if V=5 at (2,0,4)
2 P5 0 4
V1
𝑷𝟓 =
1 1 5 P5 4 0 P1 Q
𝟐𝟐 + 𝟎𝟐 + 𝟒𝟐 = 𝟐𝟎 V1 10 .888
Potential field of single point charge Q1
V ( r) 4
0 r r1
A
|r - r1| Q1
Move A from infinity
Potential due to two charges V( r)
Q1 4 0 r r1
Q2 4 0 r r2
A |r - r1|
Q1
|r - r2|
Q2
Move A from infinity
Potential due to n point charges Continue adding charges
V( r)
Q1 4 0 r r1
Q2 4 0 r r2
n
V( r)
....
Qm
4 0 r r m m1
Qn 4 0 r r n
Potential as point charges become infinite Volume of charge
Line of charge
Surface of charge
V( r)
V( r)
V( r)
v r prime
4 0 r r prime
L r prime
4 0 r r prime
d v prime
d L prime
S r prime
4 0 r r prime
d S prime
Example Find V on the z axis for a uniform line charge L in the form of a ring
V( r)
L r prime
4 0 r r prime
d L prime
Conservative field No work is done (energy is conserved) around a closed path KVL is an application of this
4.6 Potential gradient Relationship between potential and electric field intensity V=-
EddL
Two characteristics of relationship: 1. The magnitude of the electric field intensity is given by the maximum value of the rate of change of potential with distance 2. This maximum value is obtained when the direction of E is opposite to the direction in which the potential is increasing the most rapidly
Gradient • The gradient of a scalar is a vector • The gradient shows the maximum space rate of change of a scalar quantity and the direction in which the maximum occurs • The operation on V by which -E is obtained E = - grad V = - V
Gradients in different coordinate systems The following equations are found on page 104 and inside the back cover of the text: gradV
gradV
gradV
V x V
V r
a x
a
a r
V y
a y
V z
a z
1 V V a a z z
1 V 1 V a a r r sin
Cartesian
Cylindrical
Spherical
Example 4.3 Given the potential field, V = 2x2y - 5z, and a point P(-4, 3, 6), find the following: potential V, electric field intensity E potential
VP = 2(-4)2(3) - 5(6) = 66 V
electric field intensity - use gradient operation E = -4xyax - 2x2ay + 5az EP = 48ax - 32ay + 5az
Dipole The name given to two point charges of equal magnitude and opposite sign, separated by a distance which is small compared to the distance to the point P, at which we want to know the electric and potential fields
Potential To approximate the potential of a dipole, assume R1 and R2 are parallel since the point P is very distant
V
V
Q 4 0
1
R1
Q d cos 2
4 0 r
R2 1
A Dipole in an Electric Field Although the net force on the dipole from the field is zero, and the center of mass of the dipole does not move, the forces on the charged ends do produce a net torque t on the dipole about its center of mass. The center of mass lies on the line connecting the charged ends, at some distance x from one end and a distance d -x from the other end.
The net torque is:
A Dipole in an Electric Field: Potential Energy Potential energy can be associated with the orientation of an electric dipole in an electric field. The dipole has its least potential energy when it is in its equilibrium orientation, which is when its moment p is lined up with the field E. The expression for the potential energy of an electric dipole in an external electric field is simplest if we choose the potential energy to be zero when the angle (Fig.22-19) is 90°. The potential energy U of the dipole at any other value of can be found by calculating the work W done by the field on the dipole when the dipole is rotated to that value of from 90°.
Dipole Moment The dipole moment is assigned the symbol p and is equal to the product of charge and separation
p = Q*d The dipole moment expression simplifies the potential field equation
𝑯𝟐 𝑶
Example, Torque, Energy of an Electric Dipole in an Electric Field
Example: An electric dipole located at the origin in free space has a moment 𝒑 = 3𝒂𝑥 − 2𝒂𝑦 + 𝒂𝑧 𝑛𝐶. 𝑚. Find V at the points (2, 3, 4) and (2.5, 30°, 40°).
(2, 3, 4) (2.5, 30o,40o)
j r
3 10 9 p 2 10 9 9 1 10
12
0 8.85 410
𝒑 ∙ 𝑷 = 𝟑𝒙 − 𝟐𝒚 + 𝒛 ∙ 𝟐𝒙 + 𝟑𝒚 + 𝟒𝒛 = 𝟒
2 P 3 4
𝟏𝟎−𝟗 𝟒𝝅𝜺𝒐 = 𝟗
𝟐𝟐 + 𝟑𝟐 + 𝟒𝟐 = 𝟐𝟗
𝑷 = V
p 4 0 P
2.5 30 Pspherical 180 40 180
2
P
V 0.23
P
Transform this into rectangular coordinates
𝑷 =
𝟎. 𝟗𝟓𝟖𝟐 + 𝟎. 𝟖𝟎𝟑𝟐 + 𝟐. 𝟏𝟔𝟓𝟐 = 𝟔. 𝟐𝟓
2.5 sin 30 cos 40 180 180 sin 40 Prectangular 2.5 sin 30 180 180 2.5 cos 30 180
0.95 8 Prectangular 0.80 3 2.16 5
𝒑 ∙ 𝑷 = 𝟑𝒙 − 𝟐𝒚 + 𝒛 ∙ 𝟎. 𝟗𝟓𝟖𝒙 + 𝟎. 𝟖𝟎𝟑𝒚 + 𝟐. 𝟏𝟔𝟓𝒛 = 𝟑. 𝟒𝟑𝟑 V
p
4 0 Prectangular
2
Prectangular Prectangular
V 1.97 3
𝒑∙𝑷 (a) 𝑽 = 𝟒𝝅𝜺 𝑷 𝒐
𝟑
𝟔𝒂𝒛 ∙ 𝟏. 𝟑𝟕𝒂𝒙 + 𝟑. 𝟕𝟔𝒂𝒛 × 𝟏𝟎−𝟗 𝟐𝟎𝟑. 𝟎𝟒 = = = 𝟑. 𝟏𝟕 𝑽 𝟏𝟎−𝟗 𝟔𝟒 𝟏𝟔 𝟑/𝟐 𝟗
𝟐𝒑𝒄𝒐𝒔𝜽 𝒑𝒔𝒊𝒏𝜽 𝟐 × 𝟔𝒄𝒐𝒔𝟐𝟎𝒐 𝟔𝒔𝒊𝒏𝟐𝟎𝒐 (b) 𝑬 = −𝛁𝑽 = 𝟒𝝅𝜺𝒐 𝑷 𝟑 𝒂𝒓 + 𝟒𝝅𝜺𝒐 𝑷 𝟑 𝒂𝜽 = 𝟒𝝅𝜺𝒐 𝑷 𝟑 𝒂𝒓 + 𝟒𝝅𝜺𝒐 𝑷 𝟑 𝒂𝜽
𝟏𝟏. 𝟐𝟖 × 𝟏𝟎−𝟗 𝟐. 𝟎𝟓 × 𝟏𝟎−𝟗 𝑬= 𝒂𝒓 + −𝟗 𝒂𝜽 −𝟗 𝟏𝟎 𝟏𝟎 𝟏𝟔 𝟑/𝟐 𝟏𝟔 𝟑/𝟐 𝟗 𝟗 𝑬 = 𝟏. 𝟓𝟖𝟒𝒂𝒓 + 𝟎. 𝟐𝟖𝟖𝒂𝜽 𝑽/𝒎
𝒄𝒐𝒔𝟐𝟎𝒐 = 𝟎. 𝟗𝟒 𝒔𝒊𝒏𝟐𝟎𝒐 = 𝟎. 𝟑𝟒
(b) Another manner: 𝑄𝑑 𝑬= 2𝑐𝑜𝑠𝜃𝒂𝑟 + 𝑠𝑖𝑛𝜃𝒂𝜃 3 4𝜋𝜖𝑜 𝑟 Recall that for V 𝑄𝑑𝑐𝑜𝑠𝜃 𝑉4𝜋𝜖𝑜 𝑟 2 𝑉= → 𝑄𝑑 = 2 4𝜋𝜖𝑜 𝑟 𝑐𝑜𝑠𝜃 Substituting for Qd in E, giving 4𝜋𝜖𝑜 𝑟 2 𝑉 𝑉 𝑬= . 2𝑐𝑜𝑠𝜃𝒂𝑟 + 𝑠𝑖𝑛𝜃𝒂𝜃 = 2𝒂𝑟 + 𝑡𝑎𝑛𝜃𝒂𝜃 3 𝑐𝑜𝑠𝜃 4𝜋𝜖𝑜 𝑟 𝑟 𝑉 𝟑. 𝟏𝟕 𝑬= 2𝒂𝑟 + 𝑡𝑎𝑛𝜃𝒂𝜃 = 2𝒂𝑟 + 𝑡𝑎𝑛20°𝒂𝜃 𝑟 𝟒 All unknown are available from (a) 𝒕𝒂𝒏𝟐𝟎𝒐 = 𝟎. 𝟑𝟔 𝑬 = 1.58𝒂𝑟 + 0.288𝒂𝜃
Potential Energy Bringing a positive charge from infinity into the field of another positive charge Requires Work. The work is done by the external source that moves the charge into position. If the source released its hold on the charge, the charge would accelerate, turning its potential energy into kinetic energy. The Potential Energy of a system is found by finding the work done by an external source in positioning the charge.
PE KE
Empty Universe Positioning the first charge, Q1, requires no work (no field present) Positioning more charges does take work Total positioning work = potential energy of field = WE = Q2V2,1 + Q3V3,1 + Q3V3,2 + Q4V4,1 + Q4V4,2 + Q4V4,3 + ... Manipulate this expression to get WE = 0.5(Q1V 1 + Q2V2 + Q3V3 + …)
Empty Universe
n-2
n
n-1
: : : : : : 4
1 2 3
𝛁 ∙ 𝑫 = 𝝆𝒗 𝑾𝑬 =
𝟏 𝟐
𝒗𝒐𝒍
Maxwell’s Equation 𝟏 𝝆𝒗 𝑽𝒅𝒗 = 𝛁 ∙ 𝑫 𝑽𝒅𝒗 𝟐 𝒗𝒐𝒍
𝛁 ∙ 𝑽𝑫 ≡ 𝑽 𝛁 ∙ 𝑫 + 𝑫 ∙ 𝛁𝑽
Vector identity
𝛁 ∙ 𝑽𝑫 − 𝑫 ∙ 𝛁𝑽 ≡ 𝑽 𝛁 ∙ 𝑫 𝑾𝑬 =
𝑾𝑬 =
𝟏 𝟐
𝟏 𝟐
𝟏 𝑾𝑬 = 𝟐
𝛁 ∙ 𝑽𝑫 − 𝑫 ∙ 𝛁𝑽 𝒅𝒗 𝒗𝒐𝒍
𝛁 ∙ 𝑽𝑫 𝒅𝒗 − 𝒗𝒐𝒍
𝟏 𝑽𝑫 ∙ 𝒅𝑺 − 𝟐
𝟏 𝟐
𝑫 ∙ 𝛁𝑽 𝒅𝒗
dS ∝ 𝒓𝟐
𝒗𝒐𝒍
𝑫 ∙ 𝛁𝑽 𝒅𝒗
𝟏 𝒓 𝟏 𝑫∝ 𝟐 𝒓 𝑽∝
Divergence theorem
𝒗𝒐𝒍
𝑬 = −𝛁𝑽 When r approach to infinity the surface integral equal to zero, so 𝟏 ; 𝑫 = 𝜺𝒐 𝑬 𝑾𝑬 = 𝑫 ∙ 𝑬𝒅𝒗 𝟐 𝒗𝒐𝒍
𝒂𝝆𝒔 𝒂𝝆𝒔 𝑫𝝆 = →𝑬= 𝒂𝝆 𝝆 𝝆𝜺𝒐 𝟏 𝑾𝑬 = 𝟐
𝑳 𝟐𝝅 𝒃
𝜺𝒐 𝟎 𝟎 𝒂
𝑽𝒂 = −
𝒂
𝒂𝝆𝒔 𝝆𝜺𝒐 𝒃
𝑬 ∙ 𝒅𝝆 = − 𝒃
𝒃
𝟐
𝝅𝑳𝒂𝟐 𝝆𝒔 𝟐 𝒃 𝝆𝒅𝝆𝒅𝝋𝒅𝒛 = 𝒍𝒏 𝜺𝒐 𝒂
𝒂𝝆𝒔 𝒂𝝆𝒔 𝒃 𝒅𝝆 = 𝒍𝒏 𝝆𝜺𝒐 𝜺𝒐 𝒂
𝑸 = 𝟐𝛑𝒂𝑳𝝆𝒔
𝒂𝝆𝒔 𝒃 𝑽𝒂 = 𝒍𝒏 𝜺𝒐 𝒂
𝟏 𝑾𝑬 = 𝑸𝑽𝒂 𝑱 𝟐
𝝏𝑽 −𝟐𝟎𝟎 𝟐𝟎𝟎 𝒂 𝑬 = −𝛁𝑽 = − =− 𝟐 = 𝟐 𝝏𝒓 𝒓 𝒓 𝟏 𝑾𝑬 = 𝜺𝒐 𝟐
𝑬
𝟐 𝒅𝒗
𝟏 = 𝜺𝒐 𝟐
𝝅𝝅 𝟐 𝟐 𝟎.𝟎𝟎𝟑
𝟎 𝟎 𝟎.𝟎𝟎𝟐
𝟒𝟎𝟎𝟎𝟎 𝟐 𝒓 𝒔𝒊𝒏𝜽𝒅𝜽𝒅𝒓𝒅𝝋 = 𝟏. 𝟑𝟗𝟏 𝒑𝑱 𝒓𝟒
𝝏𝑽 𝟏 𝝏𝑽 𝟔𝟎𝟎𝒄𝒐𝒔𝜽 𝟑𝟎𝟎𝒔𝒊𝒏𝜽 𝒃 𝑬 = −𝛁𝑽 = − + = 𝒂𝒓 + 𝒂𝜽 𝝏𝒓 𝒓 𝝏𝜽 𝒓𝟑 𝒓𝟑 𝟏 𝑾𝑬 = 𝜺𝒐 𝟐 𝟏 𝑾𝑬 = 𝜺𝒐 𝟐 𝟏 + 𝜺𝒐 𝟐
𝟏 𝑬 𝟐 𝒅𝒗 = 𝜺𝒐 𝟐 𝝅/𝟐 𝝅/𝟐 𝟎.𝟎𝟎𝟑
𝟎 𝟎 𝟎.𝟎𝟎𝟐 𝝅/𝟐 𝝅/𝟐 𝟎.𝟎𝟎𝟑
𝟎
= 𝟑𝟔. 𝟕 𝑱
𝟎
𝟎.𝟎𝟎𝟐
𝝅/𝟐 𝝅/𝟐 𝟎.𝟎𝟎𝟑
𝟎
𝟎
𝟎.𝟎𝟎𝟐
𝟔𝟎𝟎𝒄𝒐𝒔𝜽 𝒓𝟑
𝟐
𝟑𝟎𝟎𝒔𝒊𝒏𝜽 𝒓𝟑
𝟐
𝟔𝟎𝟎𝒄𝒐𝒔𝜽 𝒓𝟑
𝟐
𝒓𝟐 𝒔𝒊𝒏𝜽𝒅𝜽𝒅𝒓𝒅𝝋
𝒓𝟐 𝒔𝒊𝒏𝜽𝒅𝜽𝒅𝒓𝒅𝝋
𝟑𝟎𝟎𝒔𝒊𝒏𝜽 + 𝒓𝟑
𝟐
𝒓𝟐 𝒔𝒊𝒏𝜽𝒅𝜽𝒅𝒓𝒅𝝋