Fields and Vector_calculus

7/24/2019 Fields and Vector_calculus

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Fields and vector calculus
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Vector fields and scalar fields

In many applications, we do not consider individual vectors or scalars, butfunctions that give a vector or scalar at every point. Such functions are called

vector fieldsorscalar fields.


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vector fieldsorscalar fields. For example:(a) Suppose we want to model the flow of air around an aeroplane.


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vector fieldsorscalar fields. For example:(a) Suppose we want to model the flow of air around an aeroplane. The

velocity of the air flow at any given point is a vector.
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velocity of the air flow at any given point is a vector. These vectors will bedifferent at different points, so they are functions of position (and also oftime).
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velocity of the air flow at any given point is a vector. These vectors will bedifferent at different points, so they are functions of position (and also oftime). Thus, the air velocity is a vector field.
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velocity of the air flow at any given point is a vector. These vectors will bedifferent at different points, so they are functions of position (and also oftime). Thus, the air velocity is a vector field. Similarly, the pressure andtemperature are scalar quantities that depend on position, or in other

words, they are scalar fields.
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(b) The magnetic field inside an electrical machine is a vector that depends onposition, or in other words a vector field.


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(b) The magnetic field inside an electrical machine is a vector that depends onposition, or in other words a vector field. The electric potential is a scalarfield.


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(b) The magnetic field inside an electrical machine is a vector that depends onposition, or in other words a vector field. The electric potential is a scalarfield.

Although we will mainly be concerned with scalar and vector fields in

three-dimensional space, we will sometimes use two-dimensional examplesbecause they are easier to visualise.
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The vector field u= ((1 y2)/5, 0)

y=1

y=1
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y=1

y=1

When (x, y) = (0, 0.4) we have u= ((1 0.42)/5, 0) = (0.168, 0)

2
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y=1

y=1

When (x, y) = (0, 0.4) we have u= ((1 0.42)/5, 0) = (0.168, 0)

When (x, y) = (1.2, 0.4) we again have u= (0.168, 0)

2
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y=1

y=1

When (x, y) = (0, 0.4) we have u= ((1 0.42)/5, 0) = (0.168, 0)

When (x, y) = (1.2, 0.4) we again have u= (0.168, 0), and similarly forany other value ofx.

fi (( 2)/ )
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y=1

y=1

When (x, y) = (0, 0.4) we have u= ((1 0.42)/5, 0) = (0.168, 0)


When y= 0.8 we have u= ((1 0.82)/5, 0) = (0.072, 0)

Th fi ld ((1 2)/5 0)
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y=1

y=1

When (x, y) = (0, 0.4) we have u= ((1 0.42)/5, 0) = (0.168, 0)


When y= 0.8 we have u= ((1 0.82)/5, 0) = (0.072, 0)

When y= 0.8 we haveu

= ((1 (0.8)

2

)/5, 0) = (0.072, 0)

Th fi ld ((1 2)/5 0)
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y=1

y=1

When (x, y) = (0, 0.4) we have u= ((1 0.42)/5, 0) = (0.168, 0)


When y= 0.8 we have u= ((1 0.82)/5, 0) = (0.072, 0)

When y= 0.8 we haveu

= ((1 (0.8)

2

)/5, 0) = (0.072, 0)

Th t fi ld ( /4 /4)
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The vector field u= (y/4, x/4)
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The vector field u ( y/4 x/4)


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When (x, y) = (1, 0) we have u= (0, 0.25); when (x, y) = (0, 1) we haveu= (0.25, 0);

The vector field u = ( y/4 x/4)
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When (x, y) = (1, 0) we have u= (0, 0.25); when (x, y) = (0, 1) we haveu= (0.25, 0); when (x, y) = (1, 0) we have u= (0,0.25);

The vector field u = (y/4 x/4)


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When (x, y) = (1, 0) we have u= (0, 0.25); when (x, y) = (0, 1) we haveu= (0.25, 0); when (x, y) = (1, 0) we have u= (0,0.25); when(x, y) = (0,1) we have u= (0.25, 0).

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The vector field u= ( y/4, x/4)


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The vector field u= ( y/4, x/4)


When (x, y) = (1.2, 0) we have u= (0, 0.3);

The vector field u = (y/4, x/4)
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The vector field u ( y/4, x/4)


When (x, y) = (1.2, 0) we have u= (0, 0.3); when (x, y) = (0, 1.2) wehave u= (0.3, 0);

The vector field u = (y/4, x/4)


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The vector field u ( y/4, x/4)


When (x, y) = (1.2, 0) we have u= (0, 0.3); when (x, y) = (0, 1.2) wehave u= (0.3, 0); when (x, y) = (1.2, 0) we have u= (0,0.3);



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( y/ , / )


When (x, y) = (1.2, 0) we have u= (0, 0.3); when (x, y) = (0, 1.2) wehave u= (0.3, 0); when (x, y) = (1.2, 0) we have u= (0,0.3); when

(x, y) = (0,1.2) we have u= (0.3, 0).

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( y/ , / )



(x, y) = (0,1.2) we have u= (0.3, 0).

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( y/ , / )



(x, y) = (0,1.2) we have u= (0.3, 0).

The vector field u= (x/4,y/4)
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( / / )

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When (x, y) = (0.87, 0.50) we have u= (0.22,0.12);

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When (x, y) = (0.87, 0.50) we have u= (0.22,0.12);

when (x, y) = (0.50, 0.87) we have u= (0.12,0.22);

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When (x, y) = (0.87, 0.50) we have u= (0.22,0.12);

when (x, y) = (0.50, 0.87) we have u= (0.12,0.22);

when (x, y) = (0.00, 1.00) we have u= (0.00,0.25);

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When (x, y) = (0.87, 0.50) we have u= (0.22,0.12);

when (x, y) = (0.50, 0.87) we have u= (0.12,0.22);

when (x, y) = (0.00, 1.00) we have u= (0.00,0.25);

when (x, y) = (0.50, 0.87) we have u= (0.12,0.22).

http://find/


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When (x, y) = (0.87, 0.50) we have u= (0.22,0.12);

when (x, y) = (0.50, 0.87) we have u= (0.12,0.22);

when (x, y) = (0.00, 1.00) we have u= (0.00,0.25);

when (x, y) = (0.50, 0.87) we have u= (0.12,0.22).

http://find/


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When (x, y) = (0.87, 0.50) we have u= (0.22,0.12);

when (x, y) = (0.50, 0.87) we have u= (0.12,0.22);

when (x, y) = (0.00, 1.00) we have u= (0.00,0.25);

when (x, y) = (0.50, 0.87) we have u= (0.12,0.22).

http://find/


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When (x, y) = (0.87, 0.50) we have u= (0.22,0.12);

when (x, y) = (0.50, 0.87) we have u= (0.12,0.22);

when (x, y) = (0.00, 1.00) we have u= (0.00,0.25);

when (x, y) = (0.50, 0.87) we have u= (0.12,0.22).

http://find/


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When (x, y) = (0.87, 0.50) we have u= (0.22,0.12);

when (x, y) = (0.50, 0.87) we have u= (0.12,0.22);

when (x, y) = (0.00, 1.00) we have u= (0.00,0.25);

when (x, y) = (0.50, 0.87) we have u= (0.12,0.22).

http://find/


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When (x, y) = (0.87, 0.50) we have u= (0.22,0.12);

when (x, y) = (0.50, 0.87) we have u= (0.12,0.22);

when (x, y) = (0.00, 1.00) we have u= (0.00,0.25);

when (x, y) = (0.50, 0.87) we have u= (0.12,0.22).

http://find/


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When (x, y) = (0.87, 0.50) we have u= (0.22,0.12);

when (x, y) = (0.50, 0.87) we have u= (0.12,0.22);

when (x, y) = (0.00, 1.00) we have u= (0.00,0.25);

when (x, y) = (0.50, 0.87) we have u= (0.12,0.22).

http://find/


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When (x, y) = (0.87, 0.50) we have u= (0.22,0.12);

when (x, y) = (0.50, 0.87) we have u= (0.12,0.22);

when (x, y) = (0.00, 1.00) we have u= (0.00,0.25);

when (x, y) = (0.50, 0.87) we have u= (0.12,0.22).

Fluid flow around a cylinder
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The formula for this vector field is u= 1

5

1

x2 y2

(x2 +y2)2, 2xy

(x2 +y2)2

.



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5

1

x2 y2

(x2 +y2)2, 2xy

(x2 +y2)2

.

Far from the origin where x2 +y2 is large we have u (1/5, 0).

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5

1

x2 y2

(x2 +y2)2, 2xy

(x2 +y2)2

.

Far from the origin where x2 +y2 is large we have u (1/5, 0).

On the unit circle where x2 +y2 = 1 we haveu= (1 x2 +y2,2xy)/5 = 2y

5(y, x) (tangent to the circle).

The gradient of a scalar field


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If f is a scalar field, then we define (f) = (fx, fy, fz) =

f

x,f

y,f

z

.

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f

x,f

y,f

z

.

(A vector field, the gradientoff, sometimes written grad(f) rather than (f).)



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f

x,f

y,f

z

.


(a) For the function f =x3 +y4 +z5, we have (f) = (3x2, 4y3, 5z4).

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f

x,f

y,f

z

.


(a) For the function f =x3 +y4 +z5, we have (f) = (3x2, 4y3, 5z4).(b) For the function f= sin(x) sin(y)sin(z) we have

(f) = (cos(x)sin(y) sin(z), sin(x)cos(y)sin(z), sin(x) sin(y)cos(z)).

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f

x,f

y,f

z

.




(c) For the function r= (x2 +y2 +z2)1

2 we have

rx = 1

2(x2 +y2 +z2)

12 .2x= x

(x2 +y2 +z2)1

2

=x/r

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f

x,f

y,f

z

.





2 we have

rx = 1

2(x2 +y2 +z2) 12 .2x= x

(x2 +y2 +z2)1

2

=x/r,

and similarly ry =y/r and rz=z/r.


f f f
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f

x,f

y,f

z

.





2 we have

rx = 1

2(x2 +y2 +z2) 12 .2x= x

(x2 +y2 +z2)1

2

=x/r,

and similarly ry =y/r and rz=z/r. This means that

(r) = (x/r, y/r, z/r).


f f f
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f

x,f

y,f

z

.





2 we have

rx = 1

2(x2 +y2 +z2) 12 .2x= x

(x2 +y2 +z2)1

2

=x/r,


(r) = (x/r, y/r, z/r).

More generally, for any n we have

(rn)x =nrn1

rx =nrn1

x/r =nrn2x.


f f f
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f

x,f

y,f

z

.


(a) For the function f =x3

+y4

+z5

, we have (f) = (3x2

, 4y3

, 5z4

).(b) For the function f= sin(x) sin(y)sin(z) we have



2 we have

rx = 1

2(x2 +y2 +z2) 12 .2x= x

(x2 +y2 +z2)1

2

=x/r,


(r) = (x/r, y/r, z/r).

More generally, for any n we have

(rn)x =nrn1

rx =nrn1

x/r =nrn2x.

The other two derivatives work in the same way, so

(rn) =nrn2 (x, y, z).

Geometry of the gradient

( )
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Fact: The vector (f) points in the direction of maximum increase of f.


F Th (f ) i i h di i f i i f f I i
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Fact: The vector (f) points in the direction of maximum increase of f. It isperpendicular to the surfaces where f is constant.


F t Th (f ) i i h di i f i i f f I i


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The picture below illustrates the two-dimensional version of this fact in thecase where f =

x2/9 +y2/4.


F t Th t (f ) i t i th di ti f i i f f It i


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x2/9 +y2/4.

The four red ovals are given by f = 1, f = 2, f= 3 and f = 4.


Fact: The vector (f ) points in the direction of maximum increase of f It is


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x2/9 +y2/4.

The four red ovals are given by f = 1, f = 2, f= 3 and f= 4. The blue arrowsshow the vector field (f)
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To see why the above fact is true, remember that if we make small changes x,y and z to x, y and z, then the resulting change in f is approximately givenby

f =fxx+fyy+fzz.


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f =fxx+fyy+fzz.

If we write r for the vector (x, y, z), this becomes

f = (f).r


Fact: The vector (f ) points in the direction of maximum increase of f . It is
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f =fxx+fyy+fzz.


f = (f).r= |(f)||r| cos(),

where is the angle between r and (f).


Fact: The vector (f) points in the direction of maximum increase of f. It is
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( ) pperpendicular to the surfaces where f is constant.


f =fxx+fyy+fzz.


f = (f).r= |(f)||r| cos(),

where is the angle between r and (f). If we move along a surface where fis constant, then f will be zero




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f =fxx+fyy+fzz.


f = (f).r= |(f)||r| cos(),

where is the angle between r and (f). If we move along a surface where fis constant, then fwill be zero so we must have cos() = 0


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f =fxx+fyy+fzz.


f = (f).r= |(f)||r| cos(),

where is the angle between r and (f). If we move along a surface where fis constant, then fwill be zero so we must have cos() = 0, so = /2


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perpendicular to the surfaces where f is constant.


f =fxx+fyy+fzz.


f = (f).r= |(f)||r| cos(),

where is the angle between r and (f). If we move along a surface where fis constant, then fwill be zero so we must have cos() = 0, so = /2, sor is perpendicular to (f).




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f =fxx+fyy+fzz.


f = (f).r= |(f)||r| cos(),

where is the angle between r and (f). If we move along a surface where fis constant, then fwill be zero so we must have cos() = 0, so = /2, sor is perpendicular to (f). This means that (f) is perpendicular to thesurfaces of constant f, as we stated before.


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f =fxx+fyy+fzz.


f = (f).r= |(f)||r| cos(),

where is the angle between r and (f). If we move along a surface where fis constant, then fwill be zero so we must have cos() = 0, so = /2, sor is perpendicular to (f). This means that (f) is perpendicular to thesurfaces of constant f, as we stated before. On the other hand, to make f as

large as possible (for a fixed step size |r|) we need to maximise cos()


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f =fxx+fyy+fzz.


f = (f).r= |(f)||r| cos(),


large as possible (for a fixed step size |r|) we need to maximise cos(), whichmeans taking = 0


Fact: The vector (f) points in the direction of maximum increase of f. It isdi l h f h f i
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f =fxx+fyy+fzz.


f = (f).r= |(f)||r| cos(),


large as possible (for a fixed step size |r|) we need to maximise cos(), whichmeans taking = 0, so that r is in the same direction as (f).


Fact: The vector (f) points in the direction of maximum increase of f. It isdi l t th f h f i t t
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f =fxx+fyy+fzz.


f = (f).r= |(f)||r| cos(),


large as possible (for a fixed step size |r|) we need to maximise cos(), whichmeans taking = 0, so that r is in the same direction as (f). In otherwords,(f) points in the direction of maximum increase of f.

Directional derivatives

Suppose we have a function f(x, y, z), a point a and a unit vector n.
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Th di i l d i i f f i di i i h f h f f


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The directional derivative of f at ain direction n is the rate of change of fwhen we move away from a in direction n at speed one.

It can be calculated as n.(f) (where (f) must be evaluated at a).



Th di i l d i i f f i di i i h f h f f
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It can be calculated as n.(f) (where (f) must be evaluated at a).In other words, it is the component of(f) in the direction n.



Th di ti l d i ti f f t i di ti i th t f h f f
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Example: take f(x, y, z) =xy2 +z3 and a= (1, 2, 3) and n= (0.6, 0, 0.8).



The directional derivative of f at a in direction n is the rate of change of f


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Example: take f(x, y, z) =xy2 +z3 and a= (1, 2, 3) and n= (0.6, 0, 0.8).Then

(f) =fx,

f

y,

f

z



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(f) =fx,

f

y,

f

z

= (y

2

, 2xy, 3z

2

)



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(f) =fx,

f

y,

f

z

= (y

2

, 2xy, 3z

2

)

= (22, 2 1 2, 3 32) at (1, 2, 3)





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(f) =fx,

f

y,

f

z

= (y

2

, 2xy, 3z

2

)

= (22, 2 1 2, 3 32) = (4, 4, 27) at (1, 2, 3)



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(f) =fx,

f

y,

f

z

= (y

2

, 2xy, 3z

2

)

= (22, 2 1 2, 3 32) = (4, 4, 27) at (1, 2, 3)

n.(f) = (0.6, 0, 0.8).(4, 4, 27)



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(f) =fx,

f

y,

f

z

= (y

2

, 2xy, 3z

2

)

= (22, 2 1 2, 3 32) = (4, 4, 27) at (1, 2, 3)

n.(f) = (0.6, 0, 0.8).(4, 4, 27) = 0.6 4 + 0.8 27



The directional derivative of f at ain direction n is the rate of change of f
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gwhen we move away from a in direction n at speed one.



(f) =fx,

f

y,

f

z

= (y

2

, 2xy, 3z

2

)

= (22, 2 1 2, 3 32) = (4, 4, 27) at (1, 2, 3)

n.(f) = (0.6, 0, 0.8).(4, 4, 27) = 0.6 4 + 0.8 27 = 24.

Applications of the gradient

(a) We write E for the electric field (which is a vector field) and for theelectric potential (which is a scalar field).
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(a) We write E for the electric field (which is a vector field) and for theelectric potential (which is a scalar field). These are related by the

( )
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equation E= ().



i E () (All hi i lid l h h i ifi
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equation E= (). (All this is valid only when there are no significant

time-varying magnetic fields.)



ti E () (All thi i lid l h th i ifi t
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time-varying magnetic fields.)(b) Similarly, there is a gravitational potential function



ti E () (All thi i lid l h th i ifi t
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time-varying magnetic fields.)(b) Similarly, there is a gravitational potential function , and the

gravitational force field is proportional to ().


(a) We write E for the electric field (which is a vector field) and for theelectric potential (which is a scalar field). These are related by theequation E () (All this is valid only when there are no significant
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equation E= (). (All this is valid only when there are no significanttime-varying magnetic fields.)

(b) Similarly, there is a gravitational potential function , and thegravitational force field is proportional to ().

(c) The net force on a particle of air involves (p), where p is the pressure.

Electric field of a point charge

If we have a single charge at the origin, then the resulting electric potential

function is = Ar1 for some constant A, where r= (x2 +y2 +z2)1

2 as usual.
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2 as usual.N h
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Note that

(r1

)x = 1

2 (x2

+y2

+z2

)

3

2

.2x




2 as usual.N t th t
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Note that

(r1

)x = 1

2 (x2

+y2

+z2

)

3

2

.2x= x/r3




2 as usual.N t th t


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Note that

(r1

)x = 1

2 (x2

+y2

+z2

)

3

2

.2x= x/r3

and similarly (r1)y = y/r3 and (r1)z= z/r

3.




2 as usual.Note that


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Note that

(r1

)x = 1

2 (x2

+y2

+z2

)

3

2

.2x= x/r3


3. This gives the electricfield:

E= ()




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Note that

(r1

)x = 1

2 (x2

+y2

+z2

)

3

2

.2x= x/r3



E= () = Ar3(x, y, z)




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Note that

(r1

)x = 1

2 (x2

+y2

+z2

)

3

2

.2x= x/r3



E= () = Ar3(x, y, z) = Ar/r3.

Electric field of a line charge

Suppose instead that we have a whole line of charges distributed along thez-axis.
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Suppose instead that we have a whole line of charges distributed along thez-axis. It works out that the corresponding electric potential function is= 1

2A ln(x2 + y2) for some constant A.
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2



2A ln(x2 + y2) for some constant A. This is independent ofz, soz= 0.
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On the other hand, we have

x = 1

2A

x2 +y2.2x




O h h h d h
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x = 1

2A

x2 +y2.2x=

Ax

x2 +y2.




O h h h d h
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x = 1

2A

x2 +y2.2x=

Ax

x2 +y2.

By a similar calculation we have y = Ay/(x2 +y2)




O th th h d h
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x = 1

2A

x2 +y2.2x=

Ax

x2 +y2.

By a similar calculation we have y = Ay/(x2 +y2), so

E= () = Axx2 +y2 , Ayx2 +y2 , 0

.

Uniform electric field

Suppose we have an electric potential of the form = ax+ by+ cz, where a, band c are constant.
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Suppose we have an electric potential of the form = ax+ by+ cz, where a, band care constant. The corresponding electric field is

E = () = (a, b, c).
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E () (a, b, c).



E= () = (a, b, c).
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() ( , , )

In other words, we have a uniform electric field everywhere.



E= () = (a, b, c).
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() ( , , )

In other words, we have a uniform electric field everywhere. If we putu= (a, b, c) we can write the above in vector notation as = u.r



E= () = (a, b, c).
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() ( , , )

In other words, we have a uniform electric field everywhere. If we putu= (a, b, c) we can write the above in vector notation as = u.r and() = (u.r) = u.

grad()

Consider the function

(x, y, z) = angle between thex-axis and (x, y, 0) = arctan(y/x)
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(as used in polar coordinates).

grad()


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(as used in polar coordinates).It is a standard fact that arctan(t) = 1/(1 +t2).

grad()


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(as used in polar coordinates).It is a standard fact that arctan(t) = 1/(1 +t2). Using this, we get

x= arctan

y

x

x

y

x

grad()


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x= arctan

y

x

x

y

x

=

1

1 + (y/x)2y

x2

grad()


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x= arctan

y

x

x

y

x

=

1

1 + (y/x)2y

x2 =

y

x2 +y2

grad()


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116/120


x= arctan

y

x

x

y

x

=

1

1 + (y/x)2y

x2 =

y

x2 +y2

y= arctan

y

x

y

y

x

grad()


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117/120


x= arctan

y

x

x

y

x

=

1

1 + (y/x)2y

x2 =

y

x2 +y2

y= arctan

y

x

y

y

x =

1

1 + (y/x)2

1

x

grad()


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x= arctan

y

x

x

y

x

=

1

1 + (y/x)2y

x2 =

y

x2 +y2

y= arctan

y

x

y

y

x =

1

1 + (y/x)2

1

x

= x

x2

+y2

grad()


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x= arctan

y

x

x

y

x

=

1

1 + (y/x)2y

x2 =

y

x2 +y2

y= arctan

y

x

y

y

x =

1

1 + (y/x)2

1

x

= x

x2

+y2

z = 0

grad()


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120/120


x= arctan

y

x

x

y

x

=

1

1 + (y/x)2y

x2 =

y

x2 +y2

y= arctan

y

x

y

y

x =

1

1 + (y/x)2

1

x

= x

x2

+y2

z = 0,

so

() =

y

x2 +y2,

x

x2 +y2, 0

.
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Fields and Vector_calculus

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Transcript of Fields and Vector_calculus