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082-Engineering-Mathematics-Anthony-Croft-Robert-Davison-Martin-Hargreaves-James-Flint-Edisi-5-2017

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7.5 The scalar product 245

A

+

+

+

+

+

+

+

+

+

+

X

E

ds

E

E

B

Figure7.27

Two charged platessituated in avacuum.

InFigure7.27theelectricfieldvectorEintheregionofspacebetweenthecharged

plateshasadirectionperpendiculartotheplatespointingfromAtoB.Themagnitude

of the electric field vector is constant in this region if end effects are ignored. If

a charged particle is required to move against an electric field, then work must be

donetoachievethis.Forexample,totransportapositivelychargedparticlefromthe

surfaceofplateBtothesurfaceofplateAwouldrequireworktobedone.Thiswould

lead toan increase inpotential of the charged particle.

If s represents the displacement andV the potential it is conventional to write δs

to represent a very small change in displacement, and δV to represent a very small

change inpotential.

Ifaunitpositivechargeismovedasmalldisplacement δsinanelectricfield(Figure

7.27) then the change inpotential δV isgiven by

δV =−E·δs (7.1)

This is an example of the use of a scalar product. Notice that although E and δs are

vector quantities the change inpotential, δV, isascalar.

ConsideragainthechargedplatesofFigure7.27.Ifaunitchargeismovedasmall

displacement along the plane X,then δs isperpendicular toE. So,

δV = −E·δs = −|E‖δs|cosθ

With θ = 90 ◦ ,wefind δV = 0.ThesurfaceXisknownasanequipotentialsurface

because movement of a charged particle along this surface does not give rise to a

change in its potential. Movement of a charge in a direction parallel to the electric

field gives risetothe maximum change inpotential, as forthis case θ = 0 ◦ .

EXERCISES7.5

1 Ifa=3i−7jandb=2i+4jfinda·b,b·a,a·a

andb·b.

2 Ifa=4i+2j−k,b=3i−3j+3kand

c=2i−j−k,find

(a) a·a (b) a·b

(c) a·c (d) b·c

3 Evaluate (−13i −5j) · (−3i +4j).

4 Find the angle between the vectorsp = 7i +3j +2k

andq=2i−j+k.

5 Find the angle between the vectors 7i +j and 4j −k.

6 Findtheanglebetweenthevectors4i −2jand3i −3j.

7 Ifa=7i+8jandb=5ifinda·ˆb.

⎛ ⎞ ⎛ ⎞

3 5

8 Ifr 1 = ⎝1⎠and r 2 = ⎝1⎠ findr 1 ·r 1 ,r 1 ·r 2

2 0

andr 2 ·r 2 .

9 Given thatp = 2qsimplify p ·q, (p +5q) ·q and

(q−p)·p.

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