Ch02 - Coulomb's Law and Electric Field Intensity
Ch02 - Coulomb's Law and Electric Field Intensity
Ch02 - Coulomb's Law and Electric Field Intensity
Q1Q2
F=
4 0R 2
Q1 and Q2: positive or negative quantities of charge (coulomb)
R: separation (meters)
0: permittivity of free space = 8.854 x 10-12 F/m
= 1/36 x 10-9 F/m
1/40 9 x 109
In vector form:
If Q1 is in r1 and Q2 is in r2, then R12 = r2 - r1 is the directed line
segment from Q1 to Q2.
z
F2 = force experienced by Q2:
F2
Q1Q2
R12
F2 =
a
12
Q
2
1
4 0R12
Q2
r2
F1
r1
F1 = F2 =
Q1Q2
4 0R12
a =
2 12
r2 r1
| r2 r1 |
Q2Q1
4 0R 21
a21
Ft =
Qi Q t
4 0Pit
Qt
ait
Ft
Qi
a
=
2 it
Q t 4 0Pit
Ft
Pit
Qi
The right side of the equation is a vector field called the electric field
intensity due to Qi.
The unit of electric field intensity E is N/C, or V/m
Q
E=
a
2 r
4 0r
z
E=
Q
a
2 m
4 0 | r rm |
y
rm
x
Qi
a
2 i
i=1 4 0 | r ri |
E(r ) =
Example:
A 1uC charge is at the origin and a 2uC charge is at P(0,10,0). What
is the electric field at Q(6,4,5)? If a 0.5 uC charge is placed at Q,
how much force would it experience?
ROQ
1 uC
ROQ = <6,4,5>
RPQ
P
2 uC
R OQ = 6 2 + 4 2 + 52 = 77 = 8.775
R PQ = 6 2 + (6) 2 + 52 = 97 = 9.849
aOQ =
aPQ =
EQO
R OQ
R OQ
R PQ
R PQ
=
=
6a x + 4 a y + 5z
8.77
6a x 6a y + 5z
9.85
1 10 6
=
(0.684a x + 0.456a y + 0.570az )
4 0 (77)
2 10 6
=
(0.609a x 0.609a y + 0.508az )
4 0 (97)
EQ
EOQ
FQ
1 uC
-0.5 uC
EPQ
P
2 uC
r
r F
v
v
Since E = , then F = QE :
Q
v
FQ = 0.5 10 6 (192.75a x 59.68a y + 160.71az )
= 96.38a x + 29.84a y 80.36az uN
Q = dQ = v dv
vol
vol
Q
r r'
v (r' )v
r r'
=
4 0 | r r'|2 | r r'| 4 0 | r r'|2 | r r'|
r - r' E
r'
r = location of Q
To get the field at the point, add the
contribution of all Qs
origin
v (r' )v
r r'
E =
4 0 | r r'|2 | r r'|
N
v (r' )v
r r'
2
i=1 4 0 | r r'| | r r'|
E(r ) =
E(r ) =
dE2
dE1
dQ2 = L dL
z
(0, 0, z)
r = a
r = z az
dQ = L dz
R = r - r = a - z az
a zaz
L dz
dE =
4 0 (2 + z2 ) 2 + z2
R = r - r'
r'
r
dE
x
dEz
dE
dE =
L dz
2
2 3/2
4 0 ( + z )
dE =
L dz
4 0 (2 + z 2 ) 3 / 2
L dz
a
2
2 3/2
4 0 ( + z )
E = E =
L ( sec 2 )d
E=
a
2
2
2
3/2
/ 2 4 0 ( + tan )
/2
tan = z/
= L
4 0
L / 2 sec 2
sec 2
da =
da
3
2
3/2
4 0 / 2 sec
/ 2 (1 + tan )
/2
L / 2
L
L
/2
=
cos
a
=
a
sin
=
a
/
2
4 0 / 2
4 0
20
L
a
20
Note:
1. The electric field due to an infinite line of charge is directed radially
outward or into to the line charge.
2. The electric field varies inversely with the distance from the line
charge
3. If the line charge density is positive, the electric field emanates
from the charge. If the line charge density is negative, the electric
field converges to the line charge.
s
y
(x, 0, 0)
Ex
dy
Therefore:
y
(x, 0, 0)
dE x =
=
S dy
2
20 x + y
S dy
2
20 x + y
dEx
cos ax
x
2
x +y
S xdy
ax
2
2
20 ( x + y )
ax
S x
1
Ex = E =
ax 2
dy
2
20
x + y
xdy
dE x = S 2
ax
2
20 x + y
Note:
du
1
1 u
=
tan
+C
2
2
a
a
a +u
y = +
S x 1
S
1 y
E=
tan
ax =
( ) a x
20 x
x y =
20 2
2
E=
S
ax
20
S
ax
20
S
E=
aN
2 0
Notes:
1. aN is a unit vector perpendicular and pointing away from the
surface
2. The electric field intensity is constant.
3. If the surface charge density is positive, the electric field
emanates from the sheet of charge. If the line charge density is
negative, the electric field is into to the surface charge.
Example:
A line charge with charge density equal to 10 nC/m is at x = 4, z = 3. A
sheet of charge with surface charge density equal to -1 nC/m2 is at the xyplane. What is the electric field at P(2,3,5)?
EL
EL
Side view:
z
ES
x
x
ES
EL
z
D
ES
D = 2a x + 2az
D = (2) 2 + 2 2 = 2.828
2a x + 2a z
aD =
= 0.707a x + 0.707az
2.828
x
10 10 9
EL =
(0.707a x + 0.707az ) = 44.939a x + 44.939az V/m
20 (2.828)
1 10 9
ES =
az = 56.472 az V/m
2 0
The arrows show the direction of the field at every point along the
line, and the separation of the lines is inversely proportional to the
strength of the field.
The lines are called streamlines.
dy
dx
Ey
E
y
Ex
x
2
x
dx
5e
sin 2y
Ey
Ex
dy
:
dx
dx = tan 2 ydy
dx = tan 2ydy
1
x = ln(sec 2 y) + C
2
2 x + C' = ln(sec 2y )
e 2 x + C' = K ' e 2 x = sec 2y
Ke 2 x = cos 2 y
To solve for K, use the fact that the streamline passes through P (0.5,
/10, 0):
Ke 2 x = cos 2 y
Ke2(0.5) = cos /5
K = 0.298
Therefore, the equation of the streamline through P is
0.298 e2x = cos 2y