Calculus and Coordinate systems
EE 311 - Lecture 17
1. Calculus and coordinate systems
• In electromagnetics, we will often need to perform integrals
along lines, over surfaces, or throughout volumes
• These will require us to specify small changes in lengths, areas,
or volumes in terms of our OCS coordinates
• In Cartesian coordinates this is easy because all coordinates (x,
y, and z) are lengths
2. Cartesian system
3. Cylindrical system
• However, in many OCS’s, some of the coordinates are angles
and not lengths, so we have to multiply these coordinates by
factors to convert them into lengths before they can be used in
line, surface, or volume integrals
4. Spherical system
• Consider our general OCS, with coordinates u1, u2, and u3
2
1
Vector surface area
Differentials in an OCS
• We will be interested in small vector length changes:
ˆ (h1 du1 ) + u2
ˆ (h2 du2 ) + u3
ˆ (h3 du3 )
dl = u1
q
2
2
2
which has magnitude |dl| = (h1 du1 ) + (h2 du2 ) + (h3 du3 )
• A differential volume can be written as dv = h1 h2 h3 du1 du2 du3
ˆ as its
• A differential area along a surface which has vector u1
normal can be written as ds1 = h2 h3 du2 du3
ˆ or u3
ˆ as
• Similarly, differential areas along surfaces with u2
normals are ds2 = h1 h3 du1 du3 and ds3 = h1 h2 du1 du2
• We’ll write specific examples of these as we go...
• In many cases, we will be interested in the amount of a vector
flowing through a surface
• This will require defining a vector ds as a differential area
times a unit vector normal to the surface
• The unit normal vector is usually chosen to be outward
pointing for a closed surface
• Then A · ds gives us the normal component of A at a point on
the surface times a differential area. This can then be
integrated over the surface area to get the total amount of A
“going out” of the surface.
^
n
+q
Imaginary
spherical
surface
4
3
Cartesian system and example
Cartesian system
• Coordinates in a Cartesian system are (x, y, z), and all are real
lengths. These numbers refer to the minimum distances of
planes parallel to the yz, xz, and xy planes from the origin.
Intersection of these three surfaces locates the point (x, y, z)
• Unit vectors are x̂, ŷ, and ẑ, pointing in directions of increasing
x, y, or z coordinates respectively. Note all are constant
thoughout space
• A general small vector can be written as dl = x̂dx + ŷdy + ẑdz,
dv = dx dy dz, dsx = dy dz, etc.
Example: Point P is at coodinates (x = 1, y = 3, z = 4). Point Q is
at coordinates (x = 2, y = 5, z = 6). Find a vector rP Q from point
P to point Q.
Solution:
• First find position vectors from the origin to points P and Q.
– Position vector rP = x̂ + 3ŷ + 4ẑ
• x̂ × ŷ = ẑ and so on as specified previously
• In all OCS’s, a vector from the origin to a point P is defined as
rP , a “position vector for point P”. In Cartesian coordinates
for P = (x1 , y1 , z1 ), rP = x̂x1 + ŷy1 + ẑz1
• An arbitrary vector is written as A = x̂Ax + ŷAy + ẑAz , and
dot and cross products follow rules previously given
– Position vector rQ = 2x̂ + 5ŷ + 6ẑ
• A vector from P to Q can be seen to be rQ − rP since
rP + (rQ − rP ) = rQ by the head to tail rule
• Thus our vector rP Q = rQ − rP which is
(2x̂ + 5ŷ + 6ẑ) − (x̂ + 3ŷ + 4ẑ) or = x̂ + 2ŷ + 2ẑ or 3
if we separate magnitude and direction
³
x̂+2ŷ+2ẑ
3
6
5
Geometry of previous example
z
z2
Another Cartesian example
P1(x1, y1, z1)
z1
R12
R1
O
A chest 1 m wide by 0.5 m thick by 0.25 m deep is filled with a
material whose mass density increases exponentially as we get
deeper in the box, according to
P2(x2, y2, z2)
R2
y1
x1
y2
y
ρ = e−0.1z
where ρ is the mass density in kg/m3 and z = 0 is the top of the
box. What is the total mass of material in the box?
x2
x
8
7
´
Cylindrical system
Solution
To get the total mass we need to add up all the density times the
volume throughout the box — this is a volume integral!
Z Z Z
dV ρ =
in Cartesian coordinates.
Z
Z 0.5m
Z 0.25m
dy
dx
−o.25m
Z Z Z
0m
dz e−0.1z = (0.5)
Z
• Resulting coordinates are (r, φ, z), where r is the radius of the
cylinder, φ is the xz plane rotation angle, and z is the distance
of the plane parallel to the xy plane from the origin
0m
z
dz e− 10
−0.25m
−0.25m
−0.5m
dx dy dz ρ
• In cylindrical coordinates, we locate points at the intersection
of a cylinder and two planes. The cylinder’s axis is the z axis,
one of the planes is parallel to the xy plane, and the other
plane is the xz plane rotated by angle φ about the z axis
• Note here that r and
z are lengths, but φ is an angle
z
¡
¢
¡
z ¢
0.025
= (0.5) −10e− 10 |z=0
z=−0.25 = −5 1 − e
zr dr dφ
dsz = ^
dz
dr
r dφ
which is 0.13 kg!
d = r dr dφ dz
dz
Note we had to integrate here because the mass density was not
constant (i.e. total mass 6= ρ times total volume!)
^ dr dz
dsφ = φ
^ r dφ dz
dsr = r
O
y
φ
r
x
dr
9
Vectors in cylindrical coordinates
• There are three unit vectors and each corresponds to a
coordinate: r̂, φ̂, and ẑ
• These are defined so that r̂ × φ̂ = ẑ, and so on; Note ẑ is the
same as in Cartesian coordinates
• r̂ at point P points in the direction of increasing r, that is,
along a line between the origin and the projection of point P
into the xy plane; only has x̂ and ŷ components
• φ̂ at point P points in the direction of increasing φ; also only
has x̂ and ŷ components and is perpendicular to r̂
y
φ^
φ
r
^y
-φ
φ^
x
φ ^r
^x
11
r dφ
10
Facts about cylindrical vectors
• Based on these definitions, a position vector to point
P = (r, φ, z) in cylindrical coordinates (remember a position
vector is a vector from the origin to point P ) is always
rP = rr̂ + z ẑ regardless of the value of φ!
• How can this be true? The above equation says the position
vector to points at different φ’s is the same!
• Reason: (VERY IMPORTANT) r̂ and φ̂ are not constant in
space! Note they change direction depending on the point at
which they are defined
• Thus, for example, r̂ defined at point P is not necessarily the
same as r̂ defined at point Q!
• For this reason, if you ever see r̂ or φ̂ inside an integral, replace
them with their Cartesian representations: r̂ = x̂ cos φ + ŷ sin φ,
φ̂ = −x̂ sin φ + ŷ cos φ
12
Cylindrical vectors and differentials
• The Cartesian representations of r̂ and φ̂ can be found by
looking at the geometry in the xy plane
• Integrals can then be done, because x̂ and ŷ DO NOT VARY
in space and can therefore be factored out of integrals
• Non-constant unit vectors are somewhat confusing, but in
many cases they are useful for describing physical problems
because the quantities we are interested in point only for
example cylindrically outward (i.e. in the r̂ direction, with r̂
varying in space)
Cylindrical example
Paint is distributed over the surface of a cylinder with radius 2 m
and length 3 m in an unusual way, such that the mass of paint per
unit area is given by cos2 φ kg/m2 . Find the total mass of paint on
the cylinder.
z
r=2
z=3
c
d
• To integrate over the volume of a cylinder, use dv = r dr dφ dz;
to integrate over the surface of a cylinder use dsr = r dφ dz.
To integrate over a circle use dsz = r dφ dr
n^ = ^r
2
0
b
y
π/2
π/3
a
x
13
14
Solution
We want to find the total mass of paint on the surface of the
cylinder, so we need to perform an integral over the surface of the
cylinder. In cylindrical coordinates, a cylinder (aligned with the z
axis) is normal to r̂ everywhere, so we are talking about an integral
dsr = r dφ dz. We can write the total mass of paint as
Z Z
Z 3m Z 2π
dφ rρpaint
dz
dsr ρpaint =
0m
=
Z
3m
0m
Z
2π
2
dφ (2m) cos φ = 6
0
Z
2π
2
dφ cos φ
• In spherical coordinates, our 3 surfaces are a sphere, a cone,
and a plane: the sphere is centered on the origin & has radius R
• The cone is aligned with the z axis, has its tip on the origin,
and opens at half-angle θ
• The plane is the xz plane rotated about the z axis by angle φ
• Coordinates of a point P are (R, θ, φ). R is a length while θ
z
and φ are angles
R sin θ dφ
0
=6
Z
2π
dφ
0
which is 6π kg!
0
IV. Spherical coordinate system
d = R2 sin θ dR dθ dφ
1 + cos 2φ
2
dR
R
θ dθ
Cylindrical coordinates will be useful for problems that involve
circles or cylinders (note a line is a cylinder with radius 0)
y
φ
16
15
Rdθ
x
dφ
Vectors in spherical coordinates
Spherical unit vectors and differentials
• The three unit vectors are R̂, θ̂ and φ̂.
• In spherical coordinates, all three unit vectors vary in space
• Arranged so that R̂ × θ̂ = φ̂, etc
• Thus it is wise if you see them in an integral to replace them
with their Cartesian representations:
• R̂ at point P is a unit vector in the direction of a line from the
origin outward to point P
• θ̂ at point P is a unit vector in the direction of increasing θ at
point P
R̂
= x̂ sin θ cos φ + ŷ sin θ sin φ + ẑ cos θ
θ̂
= x̂ cos θ cos φ + ŷ cos θ sin φ − ẑ sin θ
φ̂ = −x̂ sin φ + ŷ cos φ
• φ̂ at point P is a unit vector in the direction of increasing φ at
point P (note only has x̂ and ŷ components)
• To integrate over the volume of a sphere use
dv = R2 sin θ dR dθ dφ; to integrate over the surface of a
sphere use dsR = R2 sin θ dθ dφ
• A position vector in spherical coordinates at point
P = (R, θ, φ) is always rP = RR̂ regardless of θ and φ
17
18
Spherical coordinate example
Calculate the volume of a sphere of radius R0 meters.
EE 311 - Lecture 18
Solution: Clearly this will involve a volume integral, with the total
volume given by
Z Z Z
V =
dv
1. Conversion between systems
Using our spherical dv we find
Z Z Z
V =
R2 sin θ dR dθ dφ
=
Z
0
R0
dR
Z
0
as expected.
π
dθ
Z
2π
2
dφ R sin θ = 2π
ÃZ
3. Line integrals
R0
dR R
0
0
2. Vector integrals
2
! µZ
π
dθ sin θ
0
¶
4. Flux integrals
5. Gradient
¢
= (2π) R03 /3 (2) = 4πR03 /3
¡
Spherical coordinates will be useful in problems involving spheres
(note a point is a sphere with radius zero!)
20
19
Point coordinate conversion equations
Conversion between coordinate systems
• Given the coordinates of a point in one system or the
components of a vector, it is sometimes necessary to convert to
a different coordinate system
• Converting vectors is more complicated than converting
coordinates, but aside from the conversions of R̂, θ̂ and φ̂ or r̂
and φ̂ to Cartesian coordinates mentioned earlier, we will
rarely need to do this
• We will often need however to convert point coordinates, so we
need to know how to do this
• Essentially just derive relationships using geometry
21
For Cartesian to cylindrical:
p
r = x2 + y 2 φ = tan−1 (y/x)
z=z
For cylindrical to Cartesian:
x = r cos φ
y = r sin φ
z=z
Ãp
!
For Cartesian to spherical:
R=
p
x2 + y 2 + z 2
−1
θ = tan
x2 + y 2
z
φ = tan−1 (y/x)
For spherical to Cartesian:
x = R sin θ cos φ y = R sin θ sin φ z = R cos θ
22
Vector integrals
Point coordinate conversion example
Point P is located at coordinates (R = 2, θ = 45◦ , φ = 45◦ ) in a
spherical coordinate system. Find the Cartesian coordinates of
point P .
Solution: Use the spherical to Cartesian equations:
x = R sin θ cos φ = 2 sin 45◦ cos 45◦ = 1
= R sin θ sin φ = 2 sin 45◦ sin 45◦ = 1
√
z = R cos θ = 2 cos 45◦ = 2
√
so P is (x = 1, y = 1, z = 2) in Cartesian coordinates.
y
There are several types of integrals we can imagine involving vector
functions, for example:
R
[1] V F dv, which adds up a vector function throughout a volume,
including directions. Result is a vector.
R
[2] C V dl, which adds up a scalar along a curve C, including
direction of curve. Result is a vector.
R
[3] C F · dl, which adds up the component of a vector along the
direction of curve C. Result is a scalar.
R
[4] S A · ds, which adds up the normal component of a vector over
surface S. Result is a scalar.
24
23
Line integrals
[2] and [3] are both “line integrals”, and require us to specify (i.e.
know the equation of) the curve C that we are integrating over
R
Example of [2]: find C V dl with V = xy over two different paths
in the xy plane between the origin and the point (x = 1, y = 1).
First path: along the curve y = x3 . Second path: up y axis then
over to (1, 1).
Integrating a vector through a volume
[1] is easy, just represent F in terms of its 3 scalar components
(Cartesian vectors will work best!) and do each individually
Example of [1] with F = x̂ + 2ŷ + 3xyẑ:
Z 1 Z 2 Z 3
dx
dy
dz (x̂ + 2ŷ + 3xy ẑ)
0
= x̂
Z
0
1
dx
Z
0
2
dy
0
Z
3
dz+2ŷ
0
Solution: First note in Cartesian coordinates for a path in the xy
plane, dl = x̂dx + ŷdy
0
Z
0
1
dx
Z
0
2
dy
Z
3
dz+3ẑ
0
= 6x̂ + 12ŷ + 9ẑ
Z
0
1
dx
Z
0
2
dy
Z
3
1
dz xy
0
0.8
y
0.6
Path 1
0.4
3
y=x
0.2
Path 2
25
26
0
0
0.2
0.4
0.6
0.8
1
x
Integral over path 1
R
R
• We want to perform C1 xy dl = C1 xy (x̂dx + ŷdy) from (0, 0)
to (1, 1) on the curve C1 : y = x3
R
R
• Note we can’t just separate this into one dx and one dy
because x and y are changing simultaneously as we move along
C1
³
´
R
dy
• The trick is to re-write this as C1 xy dx x̂ + ŷ dx
because we
dy
= 3x2
know y = x3 on C1 , so dy = 3x2 dx and thus dx
R
• Plugging this in we get C1 xy(x̂ + ŷ3x2 )dx
R1
R1
= x̂ 0 dx xy + 3ŷ 0 dx x3 y
R1
R1
• Now use y = x3 on C1 to get x̂ 0 dx x4 + 3ŷ 0 dx x6
= 51 x̂ + 37 ŷ
Integral over path 2
• Path two has two legs: on the first, x = 0 and y varies from 0
to 1. One the second, y = 1 and x varies from 0 to 1
R1
R1
R
• This gives C2 xy dl = ŷ 0 dy y (0) + x̂ 0 dx x (1)
• The answer is thus 21 x̂
• Note in this example we got two different answers for the
integral depending which path we chose! This is called a “path
dependent” integral
28
27
Example of line integral [3]
Flux integrals
R
Find the integral C F · dl, where F = xy (x̂ + ŷ) over the same two
paths as in the previous example.
• Integral [4] is a “flux integral” because it adds up the amount
of some vector flowing “through” or “out of” a surface
• Again dl³= x̂dx +
´ ŷdy, so F · dl = xy (dx + dy)
dy
= xy dx 1 + dx
R
dy
= 3x2 , so C1 F · dl
• On C1 , y = x3 and dx
¡
¢ R1
¡
¢
R1
= 0 dx (xy) 1 + 3x2 = 0 dx x4 + 3x6 = 51 +
3
7
=
• The distinction between “through” and “out of” can be made
because we can think of two types of surfaces: closed or open.
A closed surface completely encloses some volume of space (for
example a cylindrical or spherical surface) while an open
surface does not (for example a plane)
22
35
• Note we obtain a scalar result in this case!
R1
R1
R1
• On C2 , we have two legs, 0 dy (0) + 0 dx xy = 0 dx x =
1
2
• Again this is a path dependent integral
Of these two types of line integrals, we will encounter [3] much
more often. We will also usually work with line integrals that are
path independent, i.e. the same answer is obtained for an integral
between two points no matter what path is taken
29
• For closed surfaces, we take the amount of the vector flowing
“out of” the surface, meaning we add up the outward normal
component of the vector over the surface
• For an open surface we add up the amount flowing “through”
by choosing either the upward or downward pointing normal to
the surface, and integrating this component of the vector over
the surface
30
Example of a flux integral
Normal vectors
A few useful normal vectors:
• An outward vector normal to a spherical surface (centered on
the origin) is R̂
• An outward vector normal to the body of a cylinder (aligned
with the z axis) is r̂
• An outward normal on the top “end cap” of a cylinder is ẑ
• An outward normal on the bottom “end cap” is −ẑ
R
R
Note the flux integral S A · ds can also be written as S A · n̂ds,
where ds is now scalar, to emphasize we are taking the normal
component of A.
H
For closed surfaces we also sometimes write S A · ds
R
Find S (x̂ xy) · ds, where S is a closed cylindrical surface of radius
3 m and length 2 m. The cylinder is aligned with the z axis, and
the bottom of the cylinder is at z = 0.
• This flux integral will have three parts, since we need to do the
integrals over the cylinder body, top, and bottom end caps
• On the top cap, n̂ = ẑ, on the bottom cap, n̂ = −ẑ, and on the
cylinder body n̂ = r̂
• Note however that the function we are integrating here has no
ẑ component, thus (x̂ xy) · ds is zero on the top and bottom
end caps
• The integral over the body can then be written as
R 2 R 2π
dz 0 dφ (3) (xy x̂ · r̂). Note that we are using
0
dsr = r dφ dz with r = 3 here
• Now get r̂ out of the integral! Change r̂ = x̂ cos φ + ŷ sin φ to
find x̂ · r̂ = cos φ
32
31
Gradient
• Also use x = r cos φ, y = r sin φ and r = 3 for our cylinder to
R 2 R 2π
get 0 dz 0 dφ (3 cos φ) (3 sin φ) (3) cos φ
R 2 R 2π
3
2π
• This is 27 0 dz 0 dφ cos2 φ sin φ = − 54
3 cos φ|0 = 0!
A note on symmetry (IMPORTANT):
• Notice that we could have realized that we should get 0 for the
R 2π
previous integral 0 cos2 φ sin φ just by thinking about
symmetry properties of the integrand
• Now we move to talking about derivatives of scalar and vector
functions in 3d space
• First let’s consider the “derivative” of a scalar function
V (x, y, z) of 3d space. How can we do this?
• Start by thinking about a surface (called surface 1) where
V (x, y, z) has a constant value, say V0 . If V (x, y, z) is
continuous this should always be construct-able
• Since cos2 φ between π and 2π is identical to itself from 0 to π,
but sin φ becomes the negative of itself over this range, the
integration from 0 to π exactly cancels that from π to 2π
• Now think about a nearby surface (called surface 2) where
V (x, y, z) = V0 + ∆V , a slightly different value
• Thus we could get 0 as an answer without doing any
integration at all!
• Pick a point P1 on surface 1 and follow the normal to surface 1
at this point (call it dn) until surface 2 is intersected. Call the
intersection point P2
• Thinking about the symmetry of an integral before trying to do
it can simplify things immeasurably
• We an also consider another point P3 on surface 2, obtained by
starting at P1 and following an arbitrary vector dl until surface
2 is intersected
34
33
Definition of gradient
Figure for Gradient
• Now consider the rate of change of V with distance. We can
∆V
∆V
or |dn|
. It turns out that ∆V
is always <= to |dn|
choose ∆V
|dl|
|dl|
as ∆V becomes small
Surface where V(x,y,z)=V0 +∆
P3
Vector
dl
Surface where V(x,y,z)=V0
P2
Normal
at point P1
• This means the shortest distance between surfaces 1 and 2 is
along the normals to surface 1. This makes sense because
moving from surface 1 in a direction that is tangential to
surface 1 does not change V
• Due to this unique nature of the normal direction, we can
define the “gradient” of a scalar function V (x, y, z) as
grad V = n̂
dV
dn
(1)
where |dn| has been replaced by dn.
• The gradient thus represents the maximum rate of change of V
(i.e. dV
dn ) times a unit vector in a direction normal to the
surfaces of constant V
36
35
Example of gradient
Properties of gradient
• The gradient operates on a scalar function and
produces a vector rate of change
• grad V is sometimes also written as ∇V
• The gradient points in the direction of the maximum rate of
increase of V
• We can use the gradient to find the change in V along any
other direction dl as well, by ∆V = ∇V · dl
37
The gradient in our three coordinate systems is
∂V
∂V
• Cartesian: ∇V = x̂ ∂V
∂x + ŷ ∂y + ẑ ∂z
1 ∂V
∂V
• Cylindrical: ∇V = r̂ ∂V
∂r + φ̂ r ∂φ + ẑ ∂z
1 ∂V
1
∂V
• Spherical: ∇V = R̂ ∂V
∂R + θ̂ R ∂θ + φ̂ R sin θ ∂φ
Example: Find ∇(x2 + y 2 ). Note we take gradient of a scalar!
∇(x2 + y 2 ) = x̂
∂(x2 + y 2 )
∂(x2 + y 2 )
∂(x2 + y 2 )
+ ŷ
+ ẑ
∂x
∂y
∂z
= x̂2x + ŷ2y
Note the answer is a vector! Note also that 2 (x̂x + ŷy)
= 2r (x̂ cos φ + ŷ sin φ) = 2rr̂. This makes sense because surfaces of
constant x2 + y 2 are cylinders!
38
Fundamentials of electrostatics
• Electrostatics is the study of electric forces between charges at
rest; electric forces are produced by charges
EE 311 - Lecture 19
• Charge is measured in Coulombs (C) in the MKS system, and
occurs in discrete multiples of the charge on an electron
−1.6 × 10−19 C
• Charge can be neither created or destroyed, so the “principle of
conservation of electric charge” always applies
1. Fundamentals of electrostatics
2. Coulomb’s Law
• We can define a “volume charge density” as
ρV = lim
∆v→0
3. Electric Fields
∆q
∆v
(2)
in C/m3 , where ∆q is the amount of charge contained in
volume ∆v
• Since the charge on an electron is very small, we usually regard
ρV as a continuous point function of space; can be positive or
negative!
40
39
Surface and line charges
Coulomb’s law
• We can also spread charges out over surfaces or lines
• A surface charge density has units C/m2 and is defined as
∆q
ρS = lim
∆s→0 ∆s
where ∆s is a small piece of surface area
(3)
• By point charges we mean that the objects carrying the charge
are very small compared to the distance between the charges
• A line charge density has units C/m and is defined as
∆q
ρL = lim
∆l→0 ∆l
where ∆l is a small piece of length
• Coulomb’s law describes the force the occurs between two
“point charges”
(4)
• If we call the two charges q1 and q2 at points P1 and P2 , the
(vector) force exerted by q1 on q2 is
F e21 = R̂12 k
• If we want to find the total charge, Qtot , in a volume containing
a volume charge density, we just integrate over the volume:
Z Z Z
dV ρV
(5)
Qtot =
V
• Similarly to get the total charge on a surface, integrate ρS over
the surface area, or the total charge on a line is obtained by
integrating ρL over length
q1 q2
2
R12
(6)
• In this equation, R̂12 is a unit vector from q1 to q2 , which is a
unit vector in direction rP 2 − rP 1
2
is the distance between q1 and q2 squared
• R12
• Note the force is repulsive if q1 and q2 have the same sign,
while it is attractive if q1 and q2 have opposite signs
42
41
Electric fields
Coulomb’s law and electric fields
• Finally in Coulomb’s law, the constant k is given by
1
9
−12
F/m is known as
4πǫ0 ≈ 9 × 10 m/F, where ǫ0 = 8.854 × 10
the “permittivity of free space”
• We could work out the units of k also by looking at units on
both sides of the equation: N = C 2 /m2 [k], so k must have
units N m2 /C 2 , same as m/F
• The concept of an electric field, E, allows us to talk about
electric effects by knowing only about one of the charges!
• An electric field is defined as:
lim
qtest →0+
F 2,test
qtest
• Note the units of E are N/C or Volts/m
• E is a vector function of 3d space
• Given the definition of E, the force on a stationary charge q in
an electric field is given by F = qE
• Using Coulomb’s law, we can find the electric field at point P2
produced by a point charge q1 at point P1 :
• Coulomb’s law tells us about forces between charges if we know
their quantities and locations
E=
• E is thus the force per unit charge experienced by a small
positive test charge qtest when placed in E
(7)
E(P2 ) =
F e1,test
q1
= R̂12 k 2
qtest
R12
• A new way of writing this: find field at a point whose position
vector is R generated by a point charge q at position vector R′ :
¡
¢
q R − R′
E(R) =
(9)
¯
¯3
4πǫ0 ¯R − R′ ¯
• The field points radially outward from a positive charge;
radially inward toward a negative charge
44
43
(8)
Example electric field calculation
Principle of Superposition
A 2 nC point charge is located at (x = 1 m, y = 1 m, z = 1 m) in a
Cartesian coordinate system. Find the electric field at the origin
produced by this point charge, and the force on a −1 nC point
charge placed at the origin.
Solution: Use expression for point charge field. Observation point
is the origin, so R = 0. The point charge is located at (1, 1, 1), so
√
R′ = x̂ + ŷ + ẑ. Thus R − R′ = −x̂ − ŷ − ẑ, with magnitude 3.
Thus
E(R = 0)
=
2 × 10−9 (−x̂ − ŷ − ẑ)
¡√ ¢3
4πǫ0
3
= (−3.46) (x̂ + ŷ + ẑ)
(10)
(11)
N/C. Note directed away from the positive charge!
• For example, the electric field produced by multiple point
charges can be found by adding the fields produced by each
individual point charge:
¢
¡
N
1 X qi R − R i
E tot =
(12)
¯
¯
4πǫ0 i=1 ¯R − Ri ¯3
for point charges qi at locations Ri with i = 1 to N
46
45
Basic Laws of Electrostatics
• Our study of electrostatics will proceed in terms of electric
fields only. Fields are useful because only sources, not receivers,
are needed
(13)
(14)
C
• The first of these is known as Gauss’ Law, while the second
states that the line integral of E around a closed path yields
zero.
• Both of these can be derived starting just from our Coulomb’s
law definition of the field of a single charge.
47
• This means that the solution for multiple sources is simply the
solution for each individual source added together
• If we continue this process for charge distributions rather than
a set of point charges, we can arrive at integral expressions for
the total field. We are not convering these computations here
Force on the −1 nC point charge at the origin is
qE = −10−9 (−3.46) (x̂ + ŷ + ẑ). Note attractive!
• The two basic laws (or “fundamental postulates”) of
electrostatics in free space can be expressed in terms of
integrals of E:
I
Z
ρV
Qtot
dV
E · dS =
=
ǫ
ǫ0
0
V
S
I
E · dl = 0
• The equations of electrostatics are linear, which means that
they satisfy the principle of superposition