• Author / Uploaded
• KEMP

Citation preview

ENERGY AND POTENTIAL

Point Charge in an External Field • To move charge Q against the electric field, a force must be applied that counteracts the force on Q that arises from the field: E Fappl

+ Q

Fappl = - Q E

Differential Work Done on Moving a Point Charge Against an External Field • In moving point charge Q from initial position B over a differential distance dL (to final position A), the work expended is: dW = Fappl dL = QE dL = -QE dL [J] gives positive result if A (final) B (initital) charge E + + is forced F dL against the dL electric field appl

Differential Work Done on Moving a Point Charge Against an External Field • The path is along an electric field line (in the opposite direction), and over the differential path length, the field can be assumed constant.

Forcing a Charge Against the Field in an Arbitrary Direction • What matters now is the component of force in the direction of motion. A +

Force magnitude is Fappl cos(



dL

Fappl = -Q E

B

E +

Differential work in moving charge Q through distance dL will be: dW = Fappl cos( dL = QE dL

Example • An Electric field is as E = 6y2zax +12xyzay +6xy2az V/m. An incremental path is represented by ΔL = -3ax + 5ay – 2az μm. Find the work done in moving a 2 μC charge along this path if the location of the path is at: (a) PA(0,2,5); (b) PB(1,1,1); (c) PC(-2,-0,3)

Solution W  QE  L

    3 10 a

 a 

W   2 10 6 6 y 2 za x  12 xyza y  6 xy 2 a z 6

6 6  5  10 a  2  10 x y

z

W  36 10 12 y 2 z  120 10 12 xyz  24 10 12 xy 2

at PA(0,2,5) W  36  10 12 y 2 z  120  10 12 xyz  24  10 12 xy 2 W  36  10 12 (2) 2 (5)  120  10 12 (0)(2)(5)  24  10 12 (0)(2) 2 W  720  10 12 J or 720 J

Solution at PB(1,1,1) W  36  10 12 y 2 z  120  10 12 xyz  24  10 12 xy 2 W  36  10 12 (1) 2 (1)  120  10 12 (1)(1)(1)  24  10 12 (1)(1) 2 W  60  10 12 J or

 60 J

at PC(-0.7,-2,-0.3)

W  36 10 12 y 2 z  120 10 12 xyz  24 10 12 xy 2 W  36 10 12 (2) 2 (0.3)  120 10 12 (0.7)( 2)( 0.3)  24 10 12 (0.7)( 2) 2 W  60 10 12 J or

 60 J

Total Work Done over an Arbitrary Path • The integral expression for work is completely general: Any shape path may be taken, with the component of force evaluated on each differential path segment.

Total Work Done • All differential work contributions along the path are summed to give:

A (final) + 

dL

Fappl = -Q E

B (initial)

E +

Total Work Done over an Arbitrary Path

• The integral expression involving the scalar product of the field with a differential path vector is called a line integral or a contour integral.

Line Integral Evaluation • • • •

We wish to find: where and using these

Example Find the work done in moving a 5 μC charge from the origin to P(2,-1,4) thru the field E = 2xyzax + x2zay + x2yaz V/m via the path: (a) straight line segments (0,0,0) to (2,0,0) to (2,-1,0) to (2,-1,4); (b) straight line x = -2y, z = 2x; (c) curve x = -2y3, z=4y2 Answer: 80 J; 80 J; 80 J

Solution (a) A

W  Q  E  dL B

 W   5  10 W   5  10

6

 2 xyza  x za  x ya dxa  2 xyzdx  x zdy  x ydz  A

B

6

A

B

2

x

2

y

2

z

2

x

 dya y  dzaz 

Solution (a) from (0,0,0) to (2,0,0) 6





W   5  10 6









W   5  10

 2 2 xyzdx  0 x 2 zdy  0 x 2 ydz  0 0  0   x 2 2   0 0 2 2  2 yz    x yz    x yz  0 0     2  0  

 

 

W   5  10 6 (2) 2 yz  (0) 2 yz  x 2 (0) z  x 2 (0) z  x 2 y (0)  x 2 y (0) W 0



Solution (a) from (2,0,0) to (2,-1,0) 6





W   5  10 6









W   5  10

W   5  10 6 W 0

 2 2 xyzdx  1 x 2 zdy  0 x 2 ydz  0 0  2   x 2 2    1 0  2 yz    x 2 yz    x 2 yz  0  0     2  2    (2) 2 yz  (2) 2 yz  x 2 (1) z  x 2 (0) z  x 2 y (0)  x 2 y (0)

 

 



Solution (a) from (2,-1,0) to (2,-1,4) 6





W   5  10 6









W   5  10

 2 2 xyzdx  1 x 2 zdy  4 x 2 ydz  1 0  2   x 2 2   1 4 2 2  2 yz    x yz    x yz  1  0     2  2  

 

 

W   5  10 6 (2) 2 yz  (2) 2 yz  x 2 (1) z  x 2 (1) z  x 2 y (4)  x 2 y (0) W  20  10 6 x 2 y;

at (2,1,4)

W  20  10 6 (2) 2 (1)  80  10 6 J  80 J WT  0  0  80 J  80 J



Solution (b) x  2 y dx  2dy dx dy   2

z  2x dz  2dx

A

W  Q  E  dL B

 W   5 10 W   5 10

6

 2 xyza  x za  x ya dxa  2 xyzdx  x zdy  x ydz  A

B

6

A

B

2

x

2

y

2

z

2

x

 dya y  dzaz 

Solution (b)



W   5 10

 W   5 10 W   5 10



6

6

  x   dx  2  x  2 0  2 x  2 (2 x)dx  x (2 x)  2   x   2 2dx



2

  2 x dx  x dx  x dx   4 x dx 2

3

3

3

0

6

W   5 10 6

2

3

0

4  x  4  4 



    5 10 6 (16)  80 10 6 J  80 J  0 2





Solution (c) x  2 y 3 z  4 y2 dx  6 y 2 dy dz  8 ydy A

W  Q  E  dL B

 W   5 10 W   5 10



6

 2 xyza  x za  x ya dxa  2 xyzdx  x zdy  x ydz  A

B

6

W   5 10 6

A

2

x

2

y

2

z

x

 dya y  dzaz 

2

B



1

0

 2(2 y 3 ) y (4 y 2 )( 6 y 2 dy)  (2 y 3 ) 2 (4 y 2 )dy      (2 y 3 ) 2 y (8 ydy)   

Solution (c) W   5 10

6

W   5 10

6

 96 y dy  16 y dy  32 y dy  144 y dy 1

8

8

8

0

1

8

0

9  y W   5 10 6 144  9 

1

0

    5 10 6   144   80 10 6 J  80 J   9  

Seatwork 1. Calculate the work done in moving a 4 C charge from B(1,0,0) to A(0,2,0) along the path y = 2 – 2x, z=0 in the field E =: (a) 5ax V/m; (b) 5xax V/m; (c) 5xax + 5yay V/m. Answer: 20J; 10J; -30J

Seatwork 2. We shall see later that a time varying Efield need not be conservative. Let E = yax V/m at a certain instant of time, and calculate the work required to move a 3-C charge from (1,3,5) to (2,0,3) along the straight line segments joining: (a) (1,3,5) to (2,3,5) to (2,0,5) to (2,0,3); (b) (1,3,5) to (1,3,3) to (1,0,3) to (2,0,3). Answer: -9J; 0

Differential Path Lengths in the Three Coordinate Systems

Definition of Potential Difference • We now have the work done in moving charge Q from initial to final positions. • This is the potential energy gained by the charge as a result of this position change.

Definition of Potential Difference • The potential difference is defined as the work done (or potential energy gained) per unit charge. • We express this quantity in units of Joules/Coulomb, or volts:

Definition of Potential Difference • Finally:

Example Let E = (-6y/x2)ax + (6/x)ay + 5az V/m and calculate: (a) VPQ given P(-7,2,1) and Q(4,1,2); (b) Vp if V=0 at Q; (c) Vp if V = 0 at (2,0,1)

Solution (a) V   A E  dL AB B P

VPQ    E  dL Q

VPQ VPQ

6   6y      2 ax  ay  5az   dxax  dyay  dzaz 4 x  x  7   6 y 6      2 dx  dy  5dz  4 x  x  7

VPQ  

7

4

6  6y  dx  dy  5 dz  2  x x 

Solution Equation of the line x  xQ y  yQ z  zQ   4  (7) 1  2 2 1 x  4 y 1 z  2   11 1 1 x  4 y 1 x4 z2   11 1 11 1  x  4  11 y  11 x  4 11z  22 15  x 18  x y z 11 11 dx dx dy   dz  11 11

Solution VPQ VPQ VPQ VPQ

7

7 6y 6 7  dx   dy  5 dz 2 4 x 4 x 4  7 6  15  x   7 dx 6  7 dx  2 dx     5  4 x 4 11 x 4 11  11  90  7  2 6  7 dx 6  7 dx 5  7   x dx       dx 11 4 11 4 x 11 4 x 11 4 90  7  2 5 7   x dx   dx 11 4 11 4 1  7

VPQ

90 x 5  7 90   x4   5  8.21V 11  1 4 11 28

Solution (b) V AB  V A  VB

V PQ  V P  VQ 8.21  VP  0 V P  8.21V

Solution (c) Equation of the line

x  xV y  yV z  zV   2  (7) 0  2 11 x  2 y  0 z 1   9 2 2 y x2 z 1 x  2   2 9 2 9 4  2x  2x  5 y z 9 9 2 2 dy   dx dz  dx 9 9

Solution VPQ VPQ VPQ VPQ

7

7 6y 6 7  dx   dy  5 dz 2 2 x 2 x 2 7 6  4  2 x  7  2 6 7 2  2 dx    dx  5 dx  2 x 2 x 2 9 9  9  24  7  2 12  7 dx 12  7 dx 10  7  x dx       dx  9 2 9 2 x 9 2 x 9 2 24  7  2 10  7  x dx   dx  9 2 9 2 1  7

VPQ

24 x 10  7 24   x2   10  8.29V 9 1 2 9 14

Solution V PQ  V P  VQ  8.29  V P  0 V P  8.29V

Seatwork An electric field is expressed in cartesian coordinates by E = 6x2ax + 6yay + 4az V/m. Find: (a) VMN if points M and N are specified by M(2,6,-1) and N(-3,-3,2); (b) VM if V = 0 at Q(4,-2,-35); (c) VN if V = 2 at P(1,2,-4). Answer: -139 V; -120 V; 19V

Seatwork A 15 nC point charge is at the origin in free space. Calculate V1 if point P1 is located at P1(-2,3,-1) and: (a) V = 0 at (6,5,4); (b) V = 0 at infinity; (c) V = 5V at (2,0,4) Answer: 20.7 V; 36 V; 10.89 V

Potential Field of a Point Charge Off-Origin • The setup here is the same in the electric field of an off-origin point charge. 1

P

1 1

1

1

1

Potential Field of a Point Charge OffOrigin • Introduce a second point charge, and the two scalar potentials simply add:

• For n charges, the process continues:

Potential Associated with Continuous Charge Distributions • If each point charge is now represented as a small element of a continuous volume charge distribution ρvΔv, then

• As we allow the number of elements to become infinite, we obtain the integral expression:

Potential Functions Associated with Line, Surface, and Volume Charge Distributions • Line Charge: • Surface Charge:

• Volume Charge:

Potential Functions Associated with Line, Surface, and Volume Charge Distributions • Compare to our earlier expression for electric field --- generally a more difficult integral to evaluate:

Example Assume a zero reference at infinity and find the potential at P(0,0,10) that is caused by the charge configuration in free space; (a) 20 nC at the origin; (b) 10 nC/m along the line x = 0, z = 0, -1 < y < 1; (c) 10 nC/m along the line x = 0, y = 0, -1 < z < 1.

Solution (a)

Q V 4 0 | r  r ' | 9

20  10 V  17.975V 12 4 (8.854  10 )(10  0)

Solution (b)

 L 1 dL V  4 0 1 r 9

1 10  10 V dy 12  4 (8.854  10 )(10) 1

10  10  9 1 V y 1  17.97V 12 4 (8.854  10 )(10)

Solution (c)

 L 1 dL V 4 0 1 r 1 10  10  9 V dz 12  4 (8.854  10 )(10) 1 9

10  10 1 V z 1  17.97V 12 4 (8.854  10 )(10)

Seatwork If we take the zero reference for potential at infinity, find the potential at (0,0,2) caused by this charge configuration in free space; (a) 12 nC/m on the line ρ = 2.5 m, z = 0; (b) point charge of 18 nC at (1,2,-1); (c) 12 nC/m on the line y = 2.5, z = 0. Answer: 529 V; 43.2 V; 67.4 V

Change in Voltage over an Incremental Distance • The change in potential occurring over distance L depends on the angle between this vector and the electric field; i.e., the projection of the field along the path: or

Change in Voltage over an Incremental Distance from which: whose maximum value is: when the path vector lies along the electric field direction.

Relation Between Potential and Electric Field • The maximum rate of increase in potential should occur in a direction exactly opposite the electric field:

Relation Between Potential and Electric Field • The maximum rate of increase in potential should occur in a direction exactly opposite the electric field: unit vector normal to an equipotential surface and in the direction of increasing potential

Change in Voltage over an Incremental Distance • The change in potential occurring over distance L depends on the angle between this vector and the electric field; i.e., the projection of the field along the path: or from which:

Change in Voltage over an Incremental Distance

whose maximum value is: when the path vector lies along the electric field direction.

Relation Between Potential and Electric Field Equipotential surfaces

aN

E

• E points in the direction of maximum rate of decrease in potential -- in the direction of the negative gradient of V.

Electric Field in Terms of V in Rectangular Coordinates • The differential voltage change can be written as the sum of changes of V in the three coordinate directions:

• We also know that:

Electric Field in Terms of V in Rectangular Coordinates

We therefore identify:

So that:

Electric Field as the Negative Gradient of the Potential Field • We now have the relation between E and V

• This is obtained by using the del operator on V

Electric Field as the Negative Gradient of the Potential Field • A more compact relation therefore emerges, which is applicable to static electric fields: E is equal to the negative gradient of V • The direction of the gradient is that of the maximum rate of increase in the scalar field, or normal to all equipotential surfaces.

Direction of the Gradient Vector • The gradient of V is a directional derivative that represents spatial rate of change. • This is a vector which we would assume must be in some fixed direction at a given point.

Direction of the Gradient Vector • The projection of the gradient along a direction tangent to an equipotential surface must give a result of zero, as the potential by definition is constant along that surface: In other words,

Direction of the Gradient Vector • Therefore, must be perpendicular to t, or normal to an equipotential surface, and in the direction of maximum increase in V.

Gradient of V in the Three Coordinate Systems

Electric Dipole • The objective is to find the potential due to both charges at point P, and then from the potential function, determine the electric field.

Electric Dipole • The potential will be just the sum of the two potential functions associated with each point charge:

Far-Field Approximation • Under the condition r >>d, the three position vectors are approximately parallel. • This means that we may use the approximations: and to get finally:

Far-Field Approximation

Far-Field Approximation • Having found the potential: • Electric field is found by taking the negative gradient: or.. from which finally:

Electric Dipole Field and Equipotentials Equipotential surface Electric field streamline

Example A dipole of moment p = -4ax + 5ay + 3az nC-m is located at D(1,2,-1) in free space. Find V at: (a) PA(0,0,0) (b) PB(1,2,0) (c) PC(1,2,-2) (d) PD(2,6,1)

Solution (a)

V 

1 r  r'  P  | r  r '| 4 0 | r  r ' | 2

find V

at PA (0,0,0)

r  r '  (0  1)ax  (0  2)ay  (0  1)az r  r '  ax  2ay  az | r  r ' | 12  2 2  12  6 | r  r '|  2

 6

2

6

1  ax  2ay  az V   (4ax  5ay  3az)  12 4 (8.854  10 (6) 6 V  1.835V

Solution (b) find V at PB (1,2,0) r  r '  (1  1)ax  (2  2)ay  (0  1)az r  r '  az | r  r ' | 12  1 | r  r ' | 2  (1) 2  1 1 az V  (4ax  5ay  3az)  12 1 4 (8.854  10 (1) V  27V

Solution (c) find V at PC (1,2,2) r  r '  (1  1)ax  (2  2)ay  (2  1)az r  r '  az | r  r ' | 12  1 | r  r ' |  (1)  1 1  az V  (4ax  5ay  3az)  12 1 4 (8.854  10 (1) V  27V 2

2

Solution (d)

find V

at PD (2,6,1)

r  r '  (2  1)ax  (6  2)ay  (1  1)az r  r '  ax  4ay  2az | r  r ' | 12  4 2  2 2  21 | r  r '|  2

 21

2

 21

1 ax  4ay  2az V   (4ax  5ay  3az)  12 4 (8.854  10 (21) 21 V  2.05V

Example Point charges of +3 μC and -3 μC are located at (0,0,1mm) and (0,0,-1mm) respectively in free space. (a) Find p. (b) Find E in spherical components at P(r =2, θ = 40O, Φ = 50O). (c) Find E in spherical component at (1,2,1.5)

Solution (a) P  Qd 6

3

P  3  10 [1  (1)](10 )az 6

3

P  3  10 (2  10 )az 9

P  6  10 az  6nC  m

Solution (b)

Qd E (2 cosar  sin a ) 3 4 0 r (6  10 )( 2 cos 40 ar  sin a ) E 12 3 4 (8.854  10 )( 2) E  10.33a r  4.33a V / m 9

Solution (c) r  12  2 2  1.5 2  2.692 1.5   cos  56.145 0 2.692 Qd E (2 cosar  sin a ) 3 4 0 r 1

(6  10 9 )(2 cos56.145 0 ar  sin 56.145 0 a ) E 4 (8.854  10 12 )(2.692 ) 3 E  3.08a r  2.29a V / m

Seatwork An electric dipole located at the origin in free space has a moment p = 3ax – 2ay + az nC-m. (a) Find V at PA(2,3,4). (b) Find V at r = 2.5, θ = 30O, Φ = 40O. Answer: 0.230 V; 1.973 V

Seatwork A dipole of moment p = 6az nC-m is located at the origin in free space. (a) Find V at P(r = 4, θ = 20O, Φ = 0O). (b) Find E at P. Answer: 3.17 V; 1.584 ar + 0.288aθ V/m

Electric Dipole Moment • The dipole moment vector is directed from the negative charge to the positive charge, and is defined as: • In the charge configuration we have used, the direction of p is az , and therefore:

Electric Dipole Moment • so we may write: which would account for any orientation of p. • or in general, for a dipole at any orientation, positioned off-origin:

Potential Energy in a System of Two Point Charges Q1 +

Q1 has zero energy if isolated

R2,1

Q2 +

Charge Q2 is brought into position from infinity.

The work done in bringing Q2 into position is:

Potential Energy in a System of Two Point Charges • This is the stored energy in the “system”, consisting of the two assembled charges.

Potential Energy in a System of Three Point Charges Q1 +

Q2 +

R2,1

R3,2

R3,1 +

Charge Q3 is brought into position from infinity, with Q1 and Q2 already situated.

Q3

The system energy is now the previous 2charge energy plus the work done in bringing Q3 into position:

Potential Energy in a System of Three Point Charges Where

and

Extension to an n-Charge Ensemble • Extending the previous result, we can write the energy expression for n charges:

• where the local potential (at the position of charge m) is:

Extension to an n-Charge Ensemble • Note that this is the potential due to all charges except charge m, evaluated at the location of charge m.

Stored Energy in a Continuous Charge Distribution • If we have a continuous charge, characterized by a charge density function, we use implicitly the expression

Stored Energy in a Continuous Charge Distribution • but the charge Q is replaced by the quantity dq = v dv, and the summation becomes an integral over the charge volume:

• where V is the position-dependent potential function within the charge volume.

Stored Energy in the Electric Field • We use Maxwell’s first equation to express volume charge density in terms of D:

Stored Energy in the Electric Field

(continued) • where the vector identity, , has been used. • Next, the divergence theorem is used on the first term, replacing the volume integral by an integral over the surface that surrounds the volume:

Stored Energy in the Electric Field (continued) • We now have: in which the region of integration now includes all space, or wherever the field and potential exist. • We are no longer constrained to the volume taken up by the charge. This means that the surface of integration in general lies at infinity, or at an infinite radius from the otherwise compact charge.

Stored Energy in the Electric Field (continued) • At the infinite distance, the potential and D fields begin to resemble those of a point charge: and therefore • This means that the surface integral will : vanish, because the inverse cube dependence in the integrand falls off at a more rapid rate with r than the surface area increases (surface area increases only as the square of the radius).

Stored Energy in the Electric Field (continued) • This means that the surface integral will vanish, because the inverse cube dependence in the integrand falls off at a more rapid rate with r than the surface area increases (surface area increases only as the square of the radius).

Electric Field Energy and Energy Density • The field energy expression now reads: • but we know that: • which leads us to the final result: • where the energy density in the electric field is defined as:

Example Find the energy stored in free space for the region, 0 < ρ < a, 0 < Φ < π, 0 < z < 2, given the potential field V =: (a) Vo ρ/a; (b) Vo (ρ/a)cos2Φ.

Solution (a) E  V   V a  1 V a  V az       z   V0 E  /a  V0 E a 2 V0 2 E  2 a

Solution 1 (a) WE   0 Edv  2 vol 2  0V0 a  2 WE  dddz 2 0 0 0 2a WE 

V  2 0 0 2

2a

2 a

2

WE  1.571 V

0

2 0 0



0 z0

2

Solution  V 1 V V  (b) E  V    a    a  z az     V0 (  / a) cos2  1 V0 (  / a) cos2    E         V0 V0 2 E   cos   (2) cos ( sin  ) a a V0 V0 E  2 sin  cos  cos2  a a 2 2 2 4 V 4 V V 2 2 2 3 4 0 0 0 E  2 sin  cos   2 cos  sin   2 cos  a a a

Solution (b) 1 WE    0 E 2 dv 2 vol 2 2  0V02 a  2 4V02 2 4 V V 2 3 4  0 0 WE  sin  cos   2 cos  sin   2 cos   dv; 2 0 0 0  2 2a a a  a  but

4 sin 2  cos2   sin 2 2 ; dv  dddz

2 V02 2 4V02 V 3 4  0 WE  sin 2  2 cos  sin   2 cos   dddz 2 0 0 0  2 2a a a a   0V02 a  2 2 4 0V02 a  2 3 WE  sin 2dddz  cos  sin dddz 2 0 0 0 2 0 0 0 2a 2a  0V02 a  2 4  cos dddz 2 0 0 0 2a

 0V02

a



2

Solution  0V02

(b)For : 

V  2a



0

0



2a 2

2 0 0 2

a

2 a

1 0 sin 2dddz ; sin 2  2 1  cos4  2

2

2



1 2  0  sin 4 z 0 2 0 4 0 

 0.7854 0V02 4 0V02 For :  2a 2 2 a

4 V   2a 2 2 0 0 2

0

a



0

0



2

3 cos   sin dddz 0



cos  2 z0 4 0 0 4

Solution (b) For : 

 0V02 2a 2

V  2 0 0 2

2a

a



0

0



2 a

1 2 cos  d  d  dz ; cos   ( 1  cos 2  ) 0 4 2

4

4







1 1 1 1 2  0 2  sin 2   sin 4 z 0 2 0 2 0 2 0 2 4 0 

 0.5890 0V02 WE  0.7854 0V02  0  0.5890 0V02  1.374 0V02

Seatwork Find the energy stored in free space for the region 2mm < r < 3mm, 0 < θ < 90O, 0 < Φ < 90O, given the potential field V =: (a) 200/r V; (b) (300cosθ)/r2 V. Answer: 1.391 pJ; 36.7 J