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Physics 014 Physics 014
Chapter 22: Electric Chapter 22: Electric FieldsFields
Q: How does particle 1 “know” of the presence of particle 2? That is, since the particles do not touch, how can particle 2 push on particle 1—how can there be such an action at a distance?
The Electric Field
?
Ans: particle 2 pushes on particle 1 not by touching it as you would push on a coffee mug by making contact. Instead, particle 2 pushes by means of the electric field it has set up.
The electric field E at any point is defined in terms of the electrostatic force F that would be exerted on a positive test charge q0 placed there:
Electric Field
2
||||
R
qkE
Since E is the force per unit charge, it is measured in units of N/C.
Electric Field Lines
• The field lines of a positive charge radiate outward in all directions– In three dimensions, the distribution is
spherical
• The field lines of a negative charge radiate inward in all directions
F kq qo
r2
8.99 109 N m2 C2 0.80 10 6C 15 10 6C
0.20m 2 2.7N
CN104.3C100.80
N 7.2 66-
oq
FE
Example 1. The isolated point charge of q=+15μC is in a vacuum. A test charge qo=+15μC is 0.20m to the right. Find the electric field at point p?
Electric Field Lines
• Field lines: useful way to visualize electric field E
• Field lines start at a positive charge, end at negative charge
• E at any point in space is tangential to field line
• Field lines are closer where E is stronger
Example: a negative point charge — note spherical symmetry
SuperpositionSuperposition• Question: How do we figure out the field due to
several point charges?
• Answer: consider one charge at a time, calculate the field (a vector!) produced by each charge, and then add all the vectors! (“superposition”)
• Useful to look out for SYMMETRY to simplify calculations!
2ˆi
e ii i
qk
r E r
Example 2
• 4 charges are placed at the corners of a square as shown.
• What is the direction of the
electric field at the center of the square?
(a) Field is ZERO!(b) Along +y(c) Along +x
-q
-2q
+2q
+q
y
x
Total electric field
Electric Field of a DipoleElectric Field of a Dipole
• Electric dipole: two point charges +q and –q separated by a distance d
• Common arrangement in Nature: molecules, antennae, …
• Define “dipole moment” vector p: from –q to +q, with magnitude qd
Electric Field ON axis of dipoleElectric Field ON axis of dipole
EEE :ionSuperposit
2
2
a
x
kqE 2
2
a
x
kqE
22
2
1
2
1
ax
ax
kqE 222
4
2
a
x
xakq
Pa x
-q +qa x
-q +q
Electric Field ON axis of dipoleElectric Field ON axis of dipole
222
222
4
2
4
2
a
x
kpx
ax
xakqE
What if x>> a? (i.e. very far away)
p = qa“dipole moment”-- VECTOR - +
34
22
x
kp
x
kpxE
E~p/r3 is actually true for ANY point far from a dipole (not just on axis)
3r
pE
Continuous Charge DistributionContinuous Charge Distribution
• Thus far, we have only dealt with discrete, point charges.
• Imagine instead that a charge Q is smeared out over a:– LINE– AREA– VOLUME
• How to compute the electric field E??
Q
Q
Q
Q
Electric Field – Continuous Charge Distribution, equations
• For a single charge
• For individual charge elements
• For continuous charge distribution
2ˆ
e
qkr
E r
2 20ˆ ˆlim
i
ie i eq
i i
q dqk k
r r
E r r
2
||||
R
qkE
Charge DensityCharge Density
• Useful idea: charge density
• Line of charge: charge per unit length =
• Sheet of charge: charge per unit area =
• Volume of charge: charge per unit volume =
Q/L
Q/A
Q/V
• Volume charge density: when a charge is distributed evenly throughout a volume ρ = Q / V with units C/m3
• Surface charge density: when a charge is distributed evenly over a surface area σ = Q / A with units C/m2
• Linear charge density: when a charge is distributed along a line λ = Q / ℓ with units C/m
Charge DensityCharge Density
Computing electric fieldComputing electric field of continuous charge distribution of continuous charge distribution
• Approach: divide the continuous charge distribution into infinitesimally small elements
• Treat each element as a POINT charge & compute its electric field
• Sum (integrate) over all elements• Always look for symmetry to
simplify life!
Example: Field on Bisector of Charged RodExample: Field on Bisector of Charged Rod
• Uniform line of charge +Q spread over length L
• What is the direction of the electric field at a point P on the perpendicular bisector?
(a) Field is 0.
(b) Along +y
(c) Along +x
• Choose symmetrically located elements of length dx
• x components of E cancel
q
L
a
P
o
y
x
dx dx
Example 3:Example 3: Find the electric field due to a Find the electric field due to a Line of Charge at point P? Line of Charge at point P?
• Uniform line of charge, length L, total charge Q
• Compute explicitly the magnitude of E at point P on perpendicular bisector
• Showed earlier that the net field at P is in the y direction -- let’s now compute this!
Q
L
a
P
o
y
x
Distance 22 xad
2
)(
d
dqkdE
LqCharge per unit length
2/322 )(
)(cos
xa
adxkdEdEy
Q
L
a
P
ox
dE
dx
d
2/122 )(cos
xa
a
- - - - - - electric field due to a Line of Charge at P? electric field due to a Line of Charge at P?
2/
2/2/322 )(
L
Ly
xa
dxakE
What is E very far away from the line (L<<a)?What is E if the line is infinitely long (L >> a)?
224
2
Laa
Lk
2/
2/222
L
Laxa
xak
a
k
La
LkEy
222
- - - - - - electric field due to a Line of Charge at P? electric field due to a Line of Charge at P?
Problem-Solving Strategy• Conceptualize
– Establish a mental representation of the problem
– Image the electric field produced by the charges or charge distribution
• Categorize– Individual charge?
– Group of individual charges?
– Continuous distribution of charges?
Problem-Solving Strategy, cont• Analyze
– Units: when using the Coulomb constant, ke, the charges must be in C and the distances in m
– Analyzing a group of individual charges: • Use the superposition principle, find the fields due to
the individual charges at the point of interest and then add them as vectors to find the resultant field
• Be careful with the manipulation of vector quantities
– Analyzing a continuous charge distribution:• The vector sums for evaluating the total electric field at some
point must be replaced with vector integrals• Divide the charge distribution into infinitesimal pieces,
calculate the vector sum by integrating over the entire charge distribution
Problem Solving Hints, final
• Analyze, cont.– Symmetry:
• Take advantage of any symmetry to simplify calculations
• Finalize– Check to see if the electric field expression is consistent with
your mental representation
– Check to see if the solution reflects any symmetry present
– Image varying parameters to see if the mathematical result changes in a reasonable way
22-4 The Electric Field Due to a Line of Charge
• A differential element of charge occupies a length ds.
•This element sets up an electric field dE at point P.
• The components perpendicular to the z axis cancel.
• The parallel components add.
Example 4 – Find the electric field of a uniformly Charged Ring at point P?
© 2014 John Wiley & Sons, Inc. All rights reserved.
22-4 The Electric Field Due to a Line of Charge
• Adding Components & replace cosθ .
• Integrating from a starting point (call it s=0) through the full circumference (s=2πR).
- - - electric field of a uniformly charged ring at P?
22-5 The Electric Field Due to a Charged Disk
• We superimpose a ring on the disk as shown
• The ring has radius r, radial width dr, and a charge element dq.
• It sets up a differential electric field dE at point P on its central axis.
• Then, we can sum all the dE contributions with
• We find
Example 5 – Find the electric field of a uniformly Charged Disc at point P?
© 2014 John Wiley & Sons, Inc. All rights reserved.
22-6 A Point Charge in an Electric Field
If a particle with charge q is placed in an external electric field E, an electrostatic force F acts on the particle:
Ink-jet printer. Drops shot from generator G receive a charge in charging unit C. An input signal from a computer controls the charge and thus the effect of field E on where the drop lands on the paper.
A Point Charge in an Electric Field
© 2014 John Wiley & Sons, Inc. All rights reserved.
22-7 A Dipole in an Electric Field
The torque on an electric dipole of dipole moment p when placed in an external electric field E is given by a cross product:
A potential energy U is associated with the orientation of the dipole moment in the field,
as given by a dot product:
(a) An electric dipole in a uniform external electric field E. Two centers of equal but opposite charge are separated by distance d. The line between them represents their rigid connection.
(b) Field E: causes a torque τ on the dipole. The direction of τ is into the page, as represented by the symbol (x-in a circle) .
A Dipole in an Electric Field
© 2014 John Wiley & Sons, Inc. All rights reserved.
22 SummaryDefinition of Electric Field• The electric field at any point
Electric Field Lines • provide a means for visualizing the directions and the magnitudes of electric fields
Field due to an Electric Dipole•The magnitude of the electric field set up by the dipole at a distant point on the dipole axis is
Eq. 22-9
Field due to a Charged Disk•The electric field magnitude at a point on the central axis through a uniformly charged disk is given by
Eq. 22-26
Field due to a Point Charge•The magnitude of the electric field E set up by a point charge q at a distance r from the charge is
Eq. 22-1
Eq. 22-3
Summary
© 2014 John Wiley & Sons, Inc. All rights reserved.
22 Summary
Force on a Point Charge in an Electric Field• When a point charge q is placed in an external electric field E
Dipole in an Electric Field•The electric field exerts a torque on a dipole
•The dipole has a potential energy U associated with its orientation in the field
Eq. 22-34
Eq. 22-38
Eq. 22-28
© 2014 John Wiley & Sons, Inc. All rights reserved.
Summary
Adding vectors: Review
• Draw vector A.
• Draw vector B starting at the tip of vector A.
• The resultant vector R = A + B is drawn from the tail of A to the tip of B.
Graphical method (triangle method):
Components of a vector
sin
cos
AA
AA
y
x
22yx AAA
x
y
A
A1tan
The x- and y-components of a vector:
The magnitude of a vector:
The angle between vector and x-axis:
A
Vector addition using unit vectors:
We want to calculate: R = A + B
From diagram: R = (Ax + Bx)i + (Ay + By)j
The components of R:Rx = Ax + Bx
Ry = Ay + By
Only add components!!!!!
2222 )()( yyxxyx BABARRR
xx
yy
x
y
BA
BA
R
Rtan
The magnitude of a R:
The angle between vector R and x-axis:
Vector addition using unit vectors: