# • Electric field due to point charge • Chapter 27 (The Electric Field

```Lecture 18
• Electric field due to point charge
• Chapter 27 (The Electric Field)
Ē due to configuration of (source) charges (today)
parallel plate capacitor (uniform Ē )
motion of (other) charges in Ē
•
•
Electric Field of Point Charge
Use Coulomb’s law:
!
&quot;
F̄on q!
1 q
Ē = q! = 4π&quot;0 r2 , away from q
field diagram (sample vectors): at
tail of vector; does not stretch...
Unit Vector Notation
• mathematical notation for “away from q’’
• î, ĵ, k̂ : magnitude 1 (no units), purely directional information
• r̂ : unit vector pointing from origin to point (“straight outward
from point” like Ē )
•
applies to q &lt; 0 (−r̂ points towards charge)
Electric Field Models
•
good approx. for small
charged object, real
wire
•
Coulomb’s law !
K=
1
4π&quot;0
• Ē due to point charge
•
limiting cases (check +
simpler formula): like Ē
point charge very far
( ! size)
&quot;
•
Electric Field of Multiple Charges
Principle of superposition
!
+F̄on q = F̄1 on q + F̄2 on q ⇒ Ēnet =
i Ē
!i
Strategy: picture..., identify P, find Ēi , i Ēi (use symmetry)
•
Electric field of a dipole (two
opposite charges separated by
small distance s): no net charge,
but does have a net Ē
(Edipole )y
=
=
=
(E+ )y + (E− )y
1
q
+
2
4π&quot;0 (y − 1/2s)
1
−q
4π&quot;0 (y + 1/2s)2
q
2ys
4π&quot;0 (y − 1/2s)2 (y + 1/2s)2
Electric Field of a Dipole
•
For y &gt;&gt; s: (Edipole )y ≈
•
dipole moment: p̄ = (qs,
Picturing Ē
•
•
•
draw vectors or lines
•
•
starts on +ve, end -ve
tangent: direction of Ē
closer together for
larger Ē
cannot cross (unique
direction at point)
1 2qs
4π&quot;0 y 3
from negative to positive)
Electric Field of a Continuous Charge Distribution
•
even if charge is discrete, consider it
continuous, describe how it’s
distributed (like density, even if atoms
•
Strategy (based on of point charge and
principle of superposition)
divide Q into point-like charges ∆Q
find Ē due to ∆Q
convert sum to integral:
∆Q → density &times;dx
(x describes shape of ∆Q)
Electric Field of a Line of Charge
•
•
•
Problem
Strategy
Solution...: Erod =
|Q|
1
√
4π&quot;0 r r 2 +(L/2)2
Infinite line of charge:Eline = limL→∞ 4π&quot;1 0 r√r2|Q|
=
2
+(L/2)
|Q|
1
4π&quot;0 rL/2
=
1 2|λ|
4π&quot;0 r
Review of Line of charge
•
!
due to segment i: Ēi
&quot;
=
x
=
1 ∆Q
cos θi
2
4!0 ri
1
r∆Q
4!0 (r2 + y 2 )3/2
i
•
•
•
Sum over segments:
!N
Ex = i (Ei )x =
1
4π&quot;0
!N
r∆Q
3/2
2
2
r +yi
(
)
Relate ∆Q to coordinate: ∆Q = λ∆y = (Q/L) ∆y
Q/L !N
r∆y
Ex = 4π&quot;0 i
3/2
2
2
(r +yi )
&quot; L/2
!N
Sum to integral: N → ∞; ∆y → dy; i → −L/2
Ex =
Q/L
4π&quot;0
! L/2
i
rdy
−L/2 (r 2 +y 2 )3/2
Ring and Disk of charge
•
•
On axis of ring: (Ering )z =
Divide disk of surface
! sum: &quot;
Ering i
z
=
zQ
1
4π&quot;0 (z 2 +R2 )3/2
Q
charge density η = A
z∆Qi
1
4π&quot;0 (z 2 +r 2 )3/2
i
(Edisk )z =
!N &quot;
i
=
Ering i
#
z
Q
πR2
=
into rings;
z
4π&quot;0
!N
i
∆Qi
3/2
(z2 +ri2 )
•
•
Disk of Charge
Relate ∆Q to coordinate: ∆Qi = η∆Ai = η2πri ∆r
2πηz !N
ri ∆r
(Edisk )z = 4π#0 i
3/2
2
2
(z +ri )
&quot;R
!N
Sum to integral: N → ∞; ∆r → dr; i → 0
(Edisk )z =
•
ηz
2&quot;0
!R
0
rdr
(z 2 +r 2 )3/2
For z &gt;&gt; R, (Edisk )z =
Q
4π&quot;0 z 2
(point charge), as expected
Plane and Sphere of Charge
•
Electrodes (charged surfaces used to steer
electrons): model as infinite plane if
distance to plane, z &lt;&lt; distance to edges
•
Outside sphere (or shell) of uniform
charge
```