Download Lecture 4 Electric Field – Chapter 23

Lecture 4
Electric Field – Chapter 23
Electric Field (8)
• Electric field lines:
– Point away from
positive and towards
negative
– Tangent to the field
line is the direction of
the E field at that
point
– # lines is proportional
to magnitude of the
charge
Electric Field (9)
• Electric field, E, is the force per unit test charge
in N/C
r
r
F
E =
q0
• For a point charge
r
q0 q
F =k 2
r
so
r
q
E = k 2
r
Electric Field (10)
• Direction of E = direction of F
• E points towards (away from) a negative
(positive) point charge
• Superposition of electric fields
r r r
r
E = E1 + E2 + ... + En
Checkpoint #2 – Rank magnitude of net E
Checkpoint #2 – Rank magnitude of net E
• a) E x = k 2q + k 3q = k 5q iˆ
d2
d2
d2
E=
5q ˆ
Ey = k 2 j
d
50 q
E +E =k
d2
2
x
2
y
• Do this for the rest and find
All Equal
Electric Field (11)
• Electric dipole – 2 equal
magnitude, opposite
charged particles
separated by distance d
• What’s the electric field at
point P due to the dipole?
Electric Field (12)
• E is on z-axis so
E = E z = E+ − E−
• giving
• where
q
q
E =k 2 −k 2
r+
r−
d
r+ = z −
2
d
r− = z +
2
Electric Field (13)
• Substituting and rearranging gives
−2
−2

kq 
d 
d  

E = 2  1 −  −  1 +  
z  2 z 
 2 z  
• Assuming z>>d then expand using binomial
theorem ignoring higher order terms d/z<<1
kq  d

  d
E = 2 1 + + ...  − 1 − + ... 
z 
z
z
 

Electric Field (14)
• Approximate E field for a
dipole is
2 kqd
E =
3
z
• Define electric dipole
moment, p, as
where direction of p is from the
negative to positive end
• E field along dipole axis at
large distances (z>>d) is
r
p = qd
2kp
E= 3
z
Electric Field (15)
• What happens when a
dipole is put in an electric
field?
• Net force, from uniform E,
is zero
• But force on charged ends
produces a net torque
about its center of mass
Electric Force (16)
r
r
• Definition of torque τ = r × F = rF sin φ
r
• For dipole rewrite it as
τ = xFsinθ +(d − x)Fsinθ
• Substitute F and d to get
r r
τ = p×E
r
Electric Field (17)
• Torque acting on a dipole tends to rotate p
into the direction of E
• Associate potential energy, U, with the
orientation of an electric dipole in an E
field
• Dipole has least U when p is lined up
with E
Electric Field (18)
• Remember
θ
θ
90
90
U = −W = − ∫ τ d θ =
∫ pE sin θ d θ
• Potential energy of a dipole
r r
U = − pE cos θ = − p • E
• U is least (greatest) when p and E are in same
(opposite) directions
Checkpoint #5
• Rank a) magnitude of torque and b) U , greatest
to least
τ = pE sin θ
• a) Magnitudes are same
U = − pE cosθ
• U greatest at θ=180
• b) 1 & 3 tie, then 2 &4
Electric Field (19)
• E field from a continuous line of charge
• Use calculus and a charge density instead of total
charge, Q
• Linear charge density
λ = Q / Length
• Surface charge density
σ = Q / Area
• Volume charge density
ρ = Q / Volume
Electric Field (20)
• Ring of radius R and positive
charge density λ
• Use
q
E = k 2
r
• Divide ring into diff. elements
of charge so
dq = λds
Electric Field (21)
• Differential dE at P is
dq
λ ds
dE = k 2 = k 2
r
r
• From trig
r =z +R
2
2
2
• Look for symmetry
– All ⊥ cancel and  point upward
Electric Field (22)
• Parallel component dE is
dE cosθ = k
(z
λds
2
+R
2
)
cosθ
• Use trig to rewrite cos θ
z
z
cosθ = = 2
1/ 2
2
r (z + R )
Electric Field (23)
• Substituting
dE cosθ = k
(z
zλ
2
+R
)
2 3/ 2
ds
• Integrate around the ring
E = ∫ dE cosθ =
kzλ
2πr
ds
∫
(z + R )
2
2 3/ 2
0
Electric Field (24)
• Finally get
E=
• Replace λ with
kzλ (2πr )
(z
2
+R
λ = q / (2πr )
• Charge ring has E of
E=
(z
kqz
2
)
2 3/ 2
+R
)
2 3/ 2
Electric Field (25)
• Charge ring has E of
E=
(z
kqz
2
+R
)
2 3/ 2
• Check z >>R then
kq
E = 2
z
• From far away ring looks like
point charge
Electric Field (26)
• Also do this for charged disk
• But now surface charge
dq = σA = σ (2πrdr )
• Integrate to get
σ 
z
1 −
E=
2
2
2ε 0 
z +R



Electric Field (27)
• Charge disk of radius R
σ
E =
2ε 0

 1 −



2
2
z +R 
z
• Let R→∞ then get
σ
E =
2ε 0
• Acts as infinite sheet of a nonconductor with uniform charge