Chapter 19

Electric Potential Energy and the

Electric Potential

Van de Graaff generator used as a particle accelerator. An electron is accelerated vertically starting from rest across a +100,000 V potential difference produced by a Van de Graaff. Find a) the change in electric potential energy of the electron, b) the kinetic energy gained by the electron (neglecting gravity) and c) the final speed of the electron. For fun, compare the result in a) with the change in the gravitational potential energy of the electron assuming the Van de Graaff is 1 m tall.

VB = +105 V

VA = 0 V q = -1.6 x 10-19 C , me = 9.11 x 10-31 kg vA = 0 m/s

a)  EPEB - EPEA = qVB - qVA = q(VB - VA) = (-1.6 x 10-19)(105 - 0) = -1.6 x 10-14 J EPE decreases b) Conservation of energy: KEB + EPEB = KEA + EPEA KEB = KEA - (EPEB - EPEA) = 0 - (-1.6 x 10-14) = 1.6 x 10-14 J = 105 eV = 100 KeV

electron

Van de Graaff example -- continued

c) KEB = 1/2 mevB2

vB = [2KEB/me]1/2 = [2(1.6 x 10-14)/(9.11 x 10-31)]1/2 = 1.87 x 108 m/s The speed of light is 3.0 x 108 m/s, so this calculation would tell us that the electron is traveling at ~ 2/3 of the speed of light! To do part c) correctly, we really should use Einstein’s Theory of Special Relativity (gives 1.64 x 108 m/s) Fun calculation: compare EPEB - EPEA from a) with GPEB - GPEA for electron. GPEB - GPEA = meghB - meghA = meg(hB - hA) = (9.11 x 10-31)(9.8)(1 - 0) = 8.93 x 10-30 J From a), EPEB - EPEA = -1.6 x 10-14 J (1.6 x 10-14)/(8.93 x 10-30) = 1.79 x 1015 , GPE can be neglected for an electron!!

proton+proton collisions at 7 TeV CERN Large Hadron Collider

LHC ring in Geneva area

ALICE setup

ITS TPC

TRD

TOF PHOS

HMPID

MUON SPEC. PMD

FMD

19.3 The Electric Potential Difference Created by Point Charges

B

o

A

oAB r

kqqrkqq

W −=

BAo

ABAB r

kqrkq

qW

VV −=−

=−

Potential of a point charge r

kqV =

A positive test charge q0 is being repelled by another positive point charge q. Since the force on q0 is not constant between A and B, integral calculus must be used to find the work done by the force acting from A to B,

The potential difference between A and B is,

B A

If rB goes to infinity, and defining VB = 0 at infinity, and letting rA -> r and VA -> V,

19.3 The Electric Potential Difference Created by Point Charges

Example 5 The Potential of a Point Charge Using a zero reference potential at infinity, determine the amount by which a point charge of 4.0x10-8C alters the electric potential at a spot 1.2m away when the charge is (a) positive and (b) negative.

19.3 The Electric Potential Difference Created by Point Charges

( )( )

V 300m 2.1

C100.4CmN1099.8 8229

+=

×+⋅×

==

−

rkqV

(a)

(b)

V 300−=V

A positive point charge raises the potential.

A negative point charge lowers the potential.

19.3 The Electric Potential Difference Created by Point Charges

Example 6 The Total Electric Potential At locations A and B, find the total electric potential.

When two or more charges are present, the potential due to all of the charges is obtained by adding together the individual potentials.

19.3 The Electric Potential Difference Created by Point Charges

( )( ) ( )( ) V 240m 60.0

C100.8CmN1099.8m 20.0

C100.8CmN1099.8 82298229

+=×−⋅×

+×+⋅×

=−−

AV

( )( ) ( )( ) V 0m 40.0

C100.8CmN1099.8m 40.0

C100.8CmN1099.8 82298229

=×−⋅×

+×+⋅×

=−−

BV

-9 -9

-9 -9

19.3 The Electric Potential Difference Created by Point Charges

Conceptual Example 7 Where is the Potential Zero? Two point charges are fixed in place. The positive charge is +2q and the negative charge is –q. On the line that passes through the charges, how many places are there at which the total potential is zero?

19.4 Equipotential Surfaces and Their Relation to the Electric Field

An equipotential surface is a surface on which the electric potential is the same everywhere.

rkqV =

The net electric force does no work on a charge as it moves on an equipotential surface.

Equipotential surfaces for a point charge (concentric spheres centered on the charge)

19.4 Equipotential Surfaces and Their Relation to the Electric Field

The electric field created by any charge or group of charges is everywhere perpendicular to the associated equipotential surfaces and points in the direction of decreasing potential.

In equilibrium, conductors are always equipotential surfaces.

19.4 Equipotential Surfaces and Their Relation to the Electric Field

Lines of force and equipotential surfaces for a charge dipole.

19.4 Equipotential Surfaces and Their Relation to the Electric Field

sVEΔ

Δ−=

The electric field can be shown to be related to the electric potential. If you move a distance Δs and the electric potential changes by an amount ΔV, the component of the electric field in the direction of Δs is the negative of the ratio of ΔV to Δs, or more compactly,

ΔV/Δs is called the potential gradient.

19.4 Equipotential Surfaces and Their Relation to the Electric Field

Example 9 The Electric Field and Potential Are Related The plates of the capacitor are separated by a distance of 0.032 m, and the potential difference between them is VB-VA=-64V. Between the two equipotential surfaces shown in color, there is a potential difference of -3.0V. Find the spacing between the two colored surfaces.

19.4 Equipotential Surfaces and Their Relation to the Electric Field

mV100.2m 0.032V 64 3×=

−=

Δ

Δ−=sVE

m105.1mV100.2

V 0.3 33

−×=×

−−=

Δ−=ΔEVs

The gradient equation can be simplified for a parallel plate capacitor with voltage V across it and with plates separated by distance d as: E = V/d

+