calculate electric field from the potential
TRANSCRIPT
1/29/2020 2
Calculate Electric Field from the Potential
Electric field always points in the direction of steepest descent of V (steepest slope) and its magnitude is the slope.
Potential from a Negative
Point ChargePotential from a Positve
Point Charge
-V(r )
y
x
V(r )
y
x
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Calculating the Electric Field from the Potential Field
If we can get the potential by integrating the electric field:
We should be able to get the electric field by differentiating the potential.
b
a
ba ldEV
VE
In Cartesian coordinates:
dx
VEx
dy
VEy
dz
VEz
In the direction of
steepest descent
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Calculating the Electric Field from the Potential
Field
ˆˆ ˆ
points in the direction in which
the potential decreases most rapidly.
From this we see that
, , and .x y z
V V VE V i j k
x y z
E
V V VE E E
x y z
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total chargeQ
small piecesof charge
dq
Line of charge: = charge per unit length [C/m]
dq = dx
Surface of charge: = charge per unit area [C/m2]
dq = dA
Volume of Charge: = charge per unit volume [C/m3]
dq = dV
Charge Densities
dq rdrddz
dq r2dr sindd
Cylinder:
Sphere:
dq rddz
dq r2 sindd
Cylinder:
Sphere:
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Example: Calculating the Electric Field from the
Potential Field
What is the electric field at any point on the central
axis of a uniformly charged disk given the potential?
Given: V
20
( z2 R2 z)
VE
Ex
Ey
Ez
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Calculate Potential on the central axis of a
charged disk (another way)
V E dl
Ez
2o
1z
z2 R2
From Lecture 3:
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Calculate Potential due to an infinite sheet
++++++++++++++++
ˆˆ ˆ ˆ2 ( )
2 2
dV E dl
dV k i dx i dx j dx k
V k dx k x C
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E due to an infinite line charge
Corona discharge around a high voltage power line,
which roughly indicates the electric field lines.
August 26, 2014: Four Kentucky Firefighters Electrocuted
When Participating in ALS Ice Bucket Challenge
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V due to an infinite line charge
R
Lline charge with
charge density
Gaussian
cylinder
Remember Gauss’s Law
s
o
insiden
s
QdAEAdE
EndcapsbarrelRs
n dAEdAEdAE0
o
RBarrelR
LRLEAE
)2(
LQinside
R
kE
RE
R
o
R
2
or2
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V due to an infinite line charge (continued)
dV E dl ERˆ R dl ERdR
R
RkV
R
RR
R
Rk
R
dRk
dREVV
ref
ref
refref
ref
P
R
R
R
RRrefP
P
ref
P
ref
ln2
and 0
.ln and 0ln
because or 0
chooset can'you :Problem
ln2
2
Where we define V = 0 at R = Rref
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Equipotentials
Definition: locus of points with the same potential.
•General Property: The electric field is always
perpendicular to an equipotential surface.
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Equipotentials: Examples
infinite positive
charge sheet
electric dipolePoint charge
V(x) 2kx Vo
V(r) kq
r
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+++++++++++++++
–––––––––––––
Locally
Gauss:
at electrostatic
equilibrium
E
0
E|| 0
in electrostatic equilibrium
all of this metal is an equipotential;
i.e., it is all at the same voltage
Equipotential Lines on a Metal Surface
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Potential inside & outside a conducting sphere
The electric field is zero inside a conductor.
The electric potential is constant inside a conductor.
Vref 0
at r .
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The graph gives the electric potential V as a function of
x. Rank the five regions according to the magnitude of
the x-component of the electric field within them,
greatest first.
Quiz 3
1 2 3 4 5
V
x
(A) 4 > 2 > 3 > 1 > 5
(B) 3 > 1 > 5 > 2 = 4
(C) 2 > 4 > 1 = 3 = 5
(D) 1 = 2 > 4 = 3 > 5
(E) 1 = 3 = 5 > 2 > 4
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electric field
linesV1 100 V
V2 80 V
V3 60 V
V4 40 V
Work and Equipotential Lines
Which of the following
statements is true about the
work done by the electric
field in moving a positve
charge along the paths?
(A) III = IV > 0, I = II = 0
(B) I = II = 0, III > IV
(C) III = IV < 0, I = II = 0
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Summary
• If you know the functional behavior of the potential V at any point, you can calculate the electric field.
• The electric potential for a continuous charge distribution can be calculated by breaking the distribution into tiny pieces of dq and then integrating over the whole distribution.
• Finally no work needs to be done if you move a charge on an equipotential, since it would be moving perpendicular to the electric field.
• The charge concentrates on a conductor on surfaces with smallest radius of curvature.