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Chapter 19

Electric and the used as a particle accelerator. An 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 of the electron, b) the gained by the electron (neglecting ) and c) the final of the electron. For fun, compare the result in a) with the change in the energy of the electron assuming the Van de Graaff is 1 m tall.

a) EPEB - EPEA = qVB - qVA = q(VB - VA)

= (-1.6 x 10-19)(105 - 0) = -1.6 x 10-14 J VB = EPE decreases +105 V

b) : KEB + EPEB = KEA + EPEA

-14 KEB = KEA - (EPEB - EPEA) = 0 - (-1.6 x 10 )

= 1.6 x 10-14 J = 105 eV = 100 KeV

-19 -31 VA = 0 V q = -1.6 x 10 C , me = 9.11 x 10 kg v = 0 m/s electron A Van de Graaff example -- continued

2 c) KEB = 1/2 mevB

1/2 -14 -31 1/2 vB = [2KEB/me] = [2(1.6 x 10 )/(9.11 x 10 )]

= 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

-14 From a), EPEB - EPEA = -1.6 x 10 J

(1.6 x 10-14)/(8.93 x 10-30) = 1.79 x 1015 , GPE can be neglected for an electron!!

LHC ring in Geneva area proton+proton collisions at 7 TeV CERN Large Hadron Collider ALICE setup

HMPID TRD TPC PMD MUON SPEC. ITS

TOF PHOS

FMD 19.3 The Electric Potential Difference Created by Point Charges

A positive test charge q0 is being repelled by another positive point charge q. Since the

on q0 is not constant between A and B, must be used to find the done by the force acting from A to B,

kqqo kqqo WAB = − rA rB The potential difference between A and B is,

−WAB kq kq VB −VA = = − qo rAB rAB

If rB goes to infinity, and defining VB = 0 at infinity, and letting rA -> r and VA -> V, kq Potential of a V = point charge r 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

(a)

kq V = = r (8.99×109 N ⋅m2 C2 )(+ 4.0×10−8 C) 1.2 m = +300 V A positive point charge raises the potential.

(b)

V = −300 V A negative point charge lowers the potential. 19.3 The Electric Potential Difference Created by Point Charges

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

Example 6 The Total Electric Potential

At locations A and B, find the total electric potential. 19.3 The Electric Potential Difference Created by Point Charges

8.99×109 N ⋅m2 C2 + 8.0×10−-98 C 8.99×109 N ⋅m2 C2 −8.0×10−-98 C V = ( )( )+ ( )( )= +240 V A 0.20 m 0.60 m

8.99×109 N ⋅m2 C2 + 8.0×10−-98 C 8.99×109 N ⋅m2 C2 −8.0×10-9−8 C V = ( )( )+ ( )( )= 0 V B 0.40 m 0.40 m 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 Surfaces and Their Relation to the

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

Equipotential surfaces for kq a point charge (concentric V = r spheres centered on the charge)

The net electric force does no work on a charge as it moves on an equipotential surface. 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 . 19.4 Equipotential Surfaces and Their Relation to the Electric Field

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 E = − Δs

ΔV/Δs is called the potential . 19.4 Equipotential Surfaces and Their Relation to the Electric Field

Example 9 The Electric Field and Potential Are Related

The plates of the 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

ΔV − 64 V E = − = + = 2.0×103 V m Δs 0.032 m

ΔV − 3.0 V Δs = − = − =1.5×10−3 m E 2.0×103 V m

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

E = V/d