lecture 12 electrostatic field and potential gradient · lecture 12 slide 2. conservative property...
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Lecture 12
Electrostatic Field and Potential Gradient
Sections: 4.5, 4.6Homework: See homework file
LECTURE 12 slide 2
Conservative Property of Potential of Point Charge – 1
• the potential of a single point charge at the origin depends solely on the radial distance to the observation point A (see L11)
14A
A
QVrπε
= ⋅
• the potential difference VAB between points A and B depends solely on the their radial distances from the origin
21 1
44
B
A rd
r
r
B
AB rA BA r
Q QV d dr rr πεπε
= ⋅ = ⋅ = −
∫ ∫
aE L a L
o angular positions, θ and ϕ, of observation points do not matter
+ AAr
E
BBr
o path of integration does not matter –integrand has only r component and rdependence
the only component that matters
LECTURE 12 slide 3
Conservative Property of Potential of Point Charge – 2
1 1 04AA
A Ac
QV dr rπε
= ⋅ = − =
∫ E L
• if path of integration is closed – potential difference is zero
+ A B≡Ar
E
c
vector field such that its closed-path integral is zero for any closed contour is called conservative
LECTURE 12 slide 4
A point charge Q = 5·10−9 C is positioned at the origin (in vacuum). The following three points are given in RCS: A(5,0,0) m, B(0,3,4) m, and C(10,0,0) m.
Q1: Find the absolute potential at the point A.
9
0
1 9.0 104πε
≈ ×
AV =0
14 A
QRπε⋅ ≈
Q2: Find the potential difference between points A and B.
Q3: Find the potential difference between points A and C.
ABV =
ACV =
LECTURE 12 slide 5
Superposition of Potential
• potential of discrete (point) charges
0 1
1( )4 | |
Nn
nn
QVπε =
=′−∑r
r r
• this is an algebraic superposition
• potential due to distributed charge
0
1 ( )( ) , V4 | |
vP
v
V V dvρπε ′
′′= =
′−∫∫∫rr
r r
0
1 ( )( ) , V4 | |
s
s
V dsρπε ′
′′=
′−∫∫rr
r r 0
1 ( )( ) , V4 | |
l
L
V dlρπε ′
′′=
′−∫rr
r r
x
y
z
rP
′rQ
v′
LECTURE 12 slide 6
Superposition of Potential: Example
Find the potential VP at P(0,0,z) due to a disk of radius a charged uniformly with surface charge density ρs. The disk is centered at the origin and lies in the z = 0 plane.
LECTURE 12 slide 7
Conservative Property of Electrostatic Potential
• conservative property of potential follows from superposition and conservative property of potential of point charge
if work along a closed path is zero for a single point charge, it will be zero for any collection of charges
• electrostatic potential taken on a closed integration path is zero
0AAc
V d= ⋅ =∫ E L
• it follows that neither absolute potential nor voltage depend on the path taken
#1 #2 #1 #2 0AA AB BA AB ABV V V V V= + = − =#1 #2AB ABV V⇒ =
x
y
z
A
Bpath #1
path #2
LECTURE 12 slide 8
Conservative Property and KVL
• Kirchhoff’s voltage law in circuits is a direct consequence of the conservative property of the electrostatic field
0 0nnC
d V⋅ = ⇒ =∑∫ E L
• along any closed contour C of a circuit the sum of the branch voltages is zero
• Kirchhoff’s voltage law is valid only if the quasi-staticassumption holds – wavelength is much larger than the size of the circuit or device (frequency is sufficiently low)
• in general, KVL does not hold in high-frequency electronics!
LECTURE 12 slide 9
Gradient of Electrostatic Potential – 1 • consider a sufficiently small line element ΔLAB along which E is
constant
( )() )( x x y
AAB A B B AB
y zz xy z
V V V V VE zE xV E y
V= ⋅ = − = − − = −∆⇒ −∆ = +
∆∆ + ∆+ ∆ +⋅
La aa a a
Ea
x y zV E x E y E z⇒ −∆ = ∆ + ∆ + ∆
• for ΔLAB→0( )x y zdV E dx E dy E dz= − + +
• on the other handV V VdV dx dy dzx y z
∂ ∂ ∂= + +∂ ∂ ∂
, , x y zV V VE E Ex y z
∂ ∂ ∂⇒ = − = − = −
∂ ∂ ∂
AB x y zx y z∆ = ∆ + ∆ + ∆L a a a
d
BAB A
VdV = ⋅∫ E L
differential voltage
LECTURE 12 slide 10
, ,x xy z y zV V VE E Ex y
V V Vz x y z
∂ ∂ ∂= − = − = − ⇒ = −
∂ ∂ ∂ ∂ ∂ ∂
+ + ∂ ∂ ∂ a a aE
• remember the del vector operator from L10
x y zx y z∂ ∂ ∂
∇ = + +∂ ∂ ∂
a a a
, V/mV⇒ = −∇E gradV V∇ ≡
the E field equals the gradient of the potential with a minus sign
the gradient of V
Find the E field if the potential is given as( , , ) , VV x y z x y z= + +
Gradient of Electrostatic Potential – 2
LECTURE 12 slide 11
The distance between the plates of a parallel-plate capacitor is 10 mm: the left plate is at x = 0 and the right plate is at x = 10 mm. The left plate is at potential VL = 0 V and the right one is at potential VR = 10 V. (a) Find V(x) inside the capacitor bearing in mind that E is constant. (b) Find Einside the capacitor.
x0LV =
10RV =V
10 mm0
LECTURE 12 slide 12
Gradient and Directional Derivative – 1
• assume a scalar field V(x,y,z)
• an infinitesimal displacement along the x-axis dx brings us to a slightly different scalar value V(x+dx,y,z)
( , , ) ( , , )x xd V V x dx y z V x y z d dx= + − =L a
( , , ) ( , , )y yd V V x y dy z V x y z d dy= + − =L a
( , , ) ( , , )z zd V V x y z dz V x y z d dz= + − =L a
• there are analogous changes in V for displacements dy and dz
• an infinitesimal displacement x y zd dx dy dz= + +L a a ainvokes all three changes at the same time
( , , ) ( , , )x y zdV d V d V d V V x dx y dy z dz V x y z= + + = + + + −
LECTURE 12 slide 13
Gradient and Directional Derivative – 2
x y zV V VdV d V d V d V dx dy dzx y z
∂ ∂ ∂= + + = + +
∂ ∂ ∂
( )x y z x y z
dV
V V VdV dx dy dzx y z
∇
∂ ∂ ∂⇒ = + + ⋅ + + ∂ ∂ ∂
L
a a a a a a
( )LdV V d V dL⇒ =∇ ⋅ = ∇ ⋅ ⇒L a
directional derivative
LdV VdL
= ∇ ⋅a
LECTURE 12 slide 14
Gradient and Directional Derivative – 3
LdV VdL
= ∇ ⋅a
max min| |, | |dV dVV V
dL dL = ∇ = − ∇
• the maximum directional derivative is the magnitude of the gradient
• the directional derivative in any direction is determined by the projection of the gradient onto this direction
• the directional derivative is a scalar which shows the rate of change of the scalar function in a specified direction aL
• the gradient is a vector which shows the direction and the magnitude of the maximum rate of ascent of a scalar function
LECTURE 12 slide 15
E-field and Potential Gradient
V= −∇E
E points in the direction of the fastest descent of V and is equal to the rate of this descent
LECTURE 12 slide 16
Gradient and Directional Derivative: Example
The potential is given by(a) Find ∇V(x,y).(b) Find E(x,y).(c) Find the directional derivative dV/dL at the point at P(1,1,0) in
the direction ( ) / 2.L x y= −a a a
1 1( , ) .V x y x y− −= +
LECTURE 12 slide 17
Equipotential Surface • equipotential surface is the geometrical place of all points with
equal potential
• in 2-D problems the equipotential surface collapses into a line –it is a perpendicular cut through an infinite cylindrical surface
http://www.falstad.com/vector3de/
LECTURE 12 slide 18
Equipotential Surface and Gradient • any direction tangential to the equipotential surface is a direction
of zero directional derivative (no ascent/descent)
• the directions normal to the equipotential surface are the directions of fastest ascent/ descent, i.e., they are aligned with ∇V
Can you estimate roughly the E field direction and magnitude from a potential map?
0V V Vτ ττ∂
= ∇ ⋅ = ⇒ ∇ ⊥∂
a a
nn
VVL∆
= −∇ ≈ −∆
E a 1 VV =
2 VV =
4 VV =
3 VV =
τana
EV∇
E
V∇
V∇E
LECTURE 12 slide 19
Unit Normal to Equipotential Surface • if a surface is defined by the equation ( , , ) 0f x y z =
its unit normal can be found at any point on this surface as
| |nff
∇= ±
∇a
• if an equipotential surface is given by 0 0( , , ) where is a constantV x y z V V=
the E-field vector at this surface is along
| |EVV
∇= −
∇a
• an equipotential surface V(x,y,z) = V0 has an equation such that 0( , , ) ( , , ) 0 or ||k f x y z V x y z V k f V f V⋅ = − = ⇒ ∇ =∇ ∇ ∇
any nonzero real constant
LECTURE 12 slide 20
An equipotential surface is defined by the equation y = 10 m, where the potential is V0 = 100 V and the field strength is |E0| = 50 V/m.(a) Find the unit normal an to the equipotential surface pointing away
from the origin. (b) Find the E vector at the surface that points away from the origin.
Equipotential Surface: Example
LECTURE 12 slide 21
Field Maps: Field Lines and Equipotential Surfaces – 1
quarter of a coaxial line
LECTURE 12 slide 22
Field Maps: Field Lines and Equipotential Surfaces – 2 charged cylinder
LECTURE 12 slide 23
Field Maps: Field Lines and Equipotential Surfaces – 3 twin-lead line
LECTURE 12 slide 24
Field Maps: Field Lines and Equipotential Surfaces – 4 parallel-plate line
LECTURE 12 slide 25
Gradient in CCS and SCS
1 2 31 2 3
V V VdV dl dl dl V dl l l
∂ ∂ ∂= + + = ∇ ⋅∂ ∂ ∂
L
1 2 31 2 3
V V VVl l l
∂ ∂ ∂⇒∇ = + +
∂ ∂ ∂a a a
1CCS:
1 1SCS:sin
z
r
V V VVz
V V VVr r r
ρ φ
θ φ
ρ ρ φ
θ θ φ
∂ ∂ ∂∇ = + +
∂ ∂ ∂∂ ∂ ∂
∇ = + +∂ ∂ ∂
a a a
a a a
to be practiced in tutorial
LECTURE 12 slide 26
You have learned:
that the electrostatic field is conservative, i.e., work does not depend on the path taken and work is zero along a closed path
what gradient is and what directional derivative is
how to apply the principle of superposition in order to find the potential of a system of charges
how to find the unit normal to a surface
that the E field is equal to the gradient of the potential V with a minus sign
that the E field is always perpendicular to an equipotential surface