Chapter(2(ii)(((

Gradient(of(a(scalar(field(

1(

For electromagnetics we deal with quantities that vary in both space and time Scalar and vector fields are a function of four variables

(t, u1, u2, u3) Need a method to describe the space rate of change of a scalar field at a given time Partial derivatives with respect to the three space coordinates are involved The rate of change can be different in different directions so need a vector to define the space rate of change of the scalar field at a given point in space and a given time

2(

Gradient of a Scalar Field (2.6)

Scalar V has(a(constant(value((V1) over(a(surface( V1

A(second(surface(where(the(value(of V is(also(constant((V1+dV)

V1+ dV dV(is(a(small(change(in(V

P1

P2 dn"

P3 d��

α"

P1(is(a(point(on(surface(V1 P2(is(a(point(on(surface(V1+ dV along(the(normal(vector(dn

P3(is(a(second(point(on(surface(V1+ dV close(to(P2(along(another(vector(d�(≠(dn

For(the(same(change(dV(in(V(the(space(rate(of(change,(dV/d�,(is(greatest(along(dn(since(dn(is(the(shortest(distance(between(the(two(surfaces(

The(magnitude(of(dV/d�(depends(on(the(direcCon(of(d�,(therefore(dV/d�(is(a(direcConal(derivaCve(

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We define the vector that represents both the magnitude and the direction of the maximum rate of increase of a scalar as the gradient of that scalar

gradV = andVdn

To(reduce(the(amount(of(wriCng(the(del(operator(is(used(and(this(is(represented(by(the(symbol((∇

∇V = andVdn

So(far(it(is(assumed(that(V(increases,(dV(is(posiCve,(if(V(decreases(then((((((((((will(be(negaCve(in(the(an(direcCon.(

∇V

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The directional derivative along d� is

dVd�

=dVdn

dnd�

=dVdncosα

=dVdnan ⋅a � = ∇V( ) ⋅a �

In(words(this(is(that(the(space(rate(of(change(of(V(in(the(a�(direcCon(is(equal(to(the(component((projecCon)(of(the(gradient(of(V(in(that(direcCon(

dV = ∇V( ) ⋅d�

This(expression(for(dV(is(the(total(differenCal(in(V(when(moving(from(point(P1(to(point(P3

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Can be generalised to express the change in terms of a coordinate system

where d�1, d�2 and d�3 are the components of the vector differential displacement d� in a chosen coordinate system. In terms of a general coordinate system (u1,u2,u3)

dV =dVd�1

d�1 +dVd�2

d�2 +dVd�3

d�3

d� = au1d�1 + au2d�2 + au3d�3= au1 h1du1( ) + au2 h2du2( )+ au3 h3du3( )

We can write dV as the dot product of two vectors

dV = au1dVd�1

+ au2dVd�2

+ au3dVd�3

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%&⋅ au1d�1 + au2d�2 + au3d�3( )

= au1dVd�1

+ au2dVd�2

+ au3dVd�3

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%&⋅d�

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We can rewrite ∇V

∇V = au1∂V∂�1

+ au2∂V∂�2

+ au3∂V∂�3

∇V = au1∂Vh1∂u1

+ au2∂Vh2∂u2

+ au3∂Vh3∂u3

We can use the above equation to calculate the gradient of a scalar if the scalar is a function of space coordinates Using Cartesian coordinates [(u1, u2, u3) = (x, y, z) and h1 = h2 = h3 =1]

∇V = a x∂V∂x

+ ay∂V∂y

+ az∂V∂z

∇V = a x∂∂x+ ay

∂∂y+ az

∂∂z

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'(V

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It is useful to think of in Cartesian coordinates as a vector differential operator

∇

∇ ≡ a x∂∂x+ ay

∂∂y+ az

∂∂z

∇ ≡ au1∂

h1∂u1+ au2

∂h2∂u2

+ au3∂

h3∂u3

In a general orthogonal coordinate system it can be defined as

Need to take care using this last equation as differentiating a vector may lead to a new vector in a different direction.

∂ar∂φ

= aφ∂aφ∂φ

= −ar

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Example

The electrostatic field intensity E is derivable as the negative gradient of a scalar electric potential.

E = −∇V

Determine E at the point (1, 1, 0) if

V = Eoe−x sin π y

4

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Solution

where

E = − a x∂∂x+ a y

∂∂y+ az

∂∂z

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'(Eoe

−x sin π y4

= a x sinπ y4− a y

π4cosπ y

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&'Eoe

−x

so

E 1,1, 0( ) = a x − a yπ4

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&'Eo

2= aEE

E = Eo121+ π

2

16!

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%& aE =

1

1+ π 2

16( )a x + a y

π4

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