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