chapter1 electrostatic 2016 reviewed

8/18/2019 Chapter1 Electrostatic 2016 Reviewed

1/98

Chapter 1 : ELECTROSTATICS{ 4 Hours }

Charles Coulomb

Electrostatic - electricity at

rest.


2/98

1.1 Coulomb’s Law.1.2 Electric feld.1.3 Electric Potential

1.4 Charge in a uniorm electric feld

ELECTROSTATICS

The stud o

electriccharges at rest!the orcesbetween themand the electricfelds associatedwith them.


3/98

Electric charges , Q

There are two kinds of charges in nature – positive andnegative charge.

Charges of opposite sign attract one another – attractive

force.

Charges of the same sign repel one another – repulsive

force.

1. !ntroduction


4/98

"rinciple of conservation of charges state the

total charge in an isolated system is constant

#conserved$.

Charge is %uanti&ed.

Electric charge e'ists as discrete (packets)and written as

neQ =

n * positive integer num+er, 1, , ,

e * 1./ 0 1 –1 C


5/98

Charge, Q is a scalar %uantity.

The 2.!. unit of charge is coulom+ #C$.

1 Coulom+ is defined as the total chargetransferred +y a current of one ampere in one

second.


6/98

Q1 Q

QQ1

1.1 Coulom+3s 4aw

#i$#i$ inversely proportionalinversel

y proportional to theto the s%uare of thes%uare of the

separation, r separation, r between the two charges, and ;between the two charges, and ;

#ii$#ii$ directly proportionaldirectl

y proportional to theto the product of theproduct of the

magnitudes of the chargesmagnitudes of the charges, Q, Q11 and Qand Q22..

Coulom+3s 4awCoulom+3s 4aw states that thestates that the electrostatic force, 5electrostatic force, 5

between two charges separated by a distance, r, isbetween two charges separated by a distance, r, is


7/98

where *

k 6 Coulom+ constant which has the value of

× 1 7 m C –

Q1 6 magnitude of charge Q1

Q 6 magnitude of charge Q

r 6 distance +etween the two charges.

8athematically9

2

21

r QQk F =


8/98

Coulom+ constant k is given +y

9 2 -219.0 10 Nm C

4k

ο πε

= = ×

where * :o 6 permittivity of free space

# ;.;< × 1 –1 5 m –1 $


9/98

5igures #a$ and #+$ show the variation of

electrostatic force with the distance +etween

two charges.

"radient o thegra#h$%


10/98

The electrostatic force between two charges is

attractive if the charges are of opposite sign and

repulsive if the charges have the same sign.

=QQ11 QQ22

5511 5511

The notation F12 denotes the force exerted on

charge 1 by charge 2 and F21 is the force exerted on

charge 2 by charge 1.

= =QQ22QQ1155

11 5511

This electrostatic force is directed along the line

joining the charges.


11/98

2ince electric forces o+ey 7ewton3s Third

4aw, therefore the forces 51 and 51 are e%ual

in magnitude +ut opposite in direction.

>ence, it can +e written as

51 6 – 51

7ote *

The sign of the charge can +e ignored when

su+stituting into the Coulom+3s law e%uation.

The sign of the charges is important in

distinguishing the direction of the electric force

when we draw the electric force vector.


12/98

The electrostatic force

is a vector %uantityand has a direction as

well as magnitude.

?hen adding

electrostatic forces,

must take into

account the directions

of all forces, using

vector components as

needed.


13/98

E'ample 1 *

Three point charges are firmly held on astraight line of @ cm in length as shown in the

figure +elow.

5ind the resultant electric force acting on 9

#a$ charge Q#+$ charge Q1

Q1 6 =1 AC Q 6 =< AC Q 6 -; AC

cm cm


14/98

2olution *

55 5511

Q 6 =< AC Q 6 -; ACQ1 6 =1 AC

The force acting on Q due to Q1 is repulsive+ecause Q1 and Q have the same sign,

therefore the direction of 51 is to the right.

The force acting on Q due to Q is attractive

+ecause Q and Q have the opposite sign,

therefore the direction of 5 is also to the

right.

2tep 1 * Braw the electric force vectors


15/98

The magnitudes of 51 and 5 are given +y *

2

2121

r

QQk F =

2tep * se Coulom+ e%uation, find the

magnitude of each of the electric

forces.

N 1125=

2

3223

r QQk F =

2

669

2302.0

)108)(105()109(

−− ×××= F

N 900=

2

669

2102.0

)105)(1010()109(

−− ×××= F


16/98


17/98

#+$

51 51

Q 6 = < AC Q 6 - ; ACQ1 6 = 1 AC

The force acting on Q1 due to Q is repulsive+ecause Q1 and Q have the same sign,

therefore the direction of 51 is to the left.

The force acting on Q1 due to Q is attractive

+ecause Q1 and Q have the opposite sign,

therefore the direction of 51 is to the right.

2tep 1 * Braw the electric force vectors


18/98

The magnitudes of 51

and 51

are given +y 9

2

2112

r

QQk F =

2

669

12 02.0

)105)(1010(

)109(

−− ××

×= F

2tep * 5ind magnitude of electric force

1125 N=

2

31

13 r

QQ

k F =

2

669

1304.0

)108)(1010()109(

−− ×××= F

450 N=


19/98

Therefore, the resultant electric force acting on

charge Q1 is 9

5T 6 51 = 51

6 #– 11


20/98

E'ample *

5igure shows three point charges that lie in the

', y plane in a vacuum. 5ind the electrostaticforce on %1

&2

&1 &3


21/98

2olution

+

&1

Step 1 ** Braw the electric force vectors

F12

cos 73

F12

sin 73


22/98

The magnitude of the forces are

2

21

1212r

QQk F =

2

669

)15.0(

)106)(104()109(

−− ××

×=12 9.6 N F =

2tep * 5ind magnitude of electric force

2

1

13

3

13

r

QQk F =

2

669

)10.0(

)105)(104()109(

−− ×××=

13

18 N F =


23/98

5orce ' - comp y - comp

51 =./ cosD

6 =.; 7

=./ sinD

6 =. 7

51 = 1; 7 75 5' 6 = 1 7 5y 6 =. 7

The electrostatic force acting on %1 *

22 y x F F F += 22 2.921 += 23 N=

x

y

F

F =θ tan

21

2.9= xabove24 +°=θ

2tep * dds as vector #consider the direction$


24/98

C charge lies on the straight line +etweena C charge and a 1 C charge. Theseparation +etween the C and 1 C is @ cm.

(a) Draw the position of the three charges andshow the forces acting on the −2 µC charge.

(b) Calculate the distance of µC fro! −2 µCwhere net force on −2 µC is "ero.

E'ample E'ample


25/98

2olution*

C103 61−×=Q

C101 62−×=Q

C102 6−×−=q

m104 2−×=d x xd −

#a$

#+$ 7ett force acting on % is &ero, 21 F F = 222211 r qkQr qkQ = ( )22 62 6 104 101103 x x −× ×=× − −−


26/98

== – –#;$ small point charge of mass ; g and charge of =./

C is hung +y a thin wire of negligi+le mass. charge of. C is held 1


27/98

1.2 Electric 'ield

(n electric feld is a region in which anelectric orce will act on a test charge whichis #laced in that region.

l i i ld h


28/98

)efnition * The electric orce! ' acting on atest charge that is #laced in the electric feldregion di+ided b the magnitude charge othe test charge! &,.

o

F E

q=

r

E is a +ector &uantit. - unit or E is / C01 or m01Electric feld #atterns can be re#resented belectric feld lines which is drawn #ointing inthe direction o the E +ector at an oint.

Electric 'ield -trength! E


29/98

(1) #hese electric filed lines ne$er cross each

other and the nu!ber of lines deter!ine the

strength of the electric field.

(2) % is large when the field lines are close

together and s!all when they are far apart.

() & positi$e and a negati$e charge can produce

electric field.

(') #he direction of % for a positi$e point charge

is outward fro! the charge in all direction

( di!ension ).() #he direction of % for a negati$e point charge

is toward the charge in all direction (

di!ension)


30/98

%lectric field lines (patterns) for two eual and

opposite point charges.

(*) #he nu!ber of electric field lines entering or

lea$ing a charge is proportional to the !agnitude

of the charge.


31/98

Electric feld lines #atterns or two e&ual#ositi+e #oint charges.

Electric feld lines #atterns or a #ointcharge 2& and a second #oint charge 5&.


32/98

Electric feld lines #atterns or twoo##osite charged #arallel metal #lates

The electric feld lines are #er#endicular tothe surace o the metal #lates.

The lines go directlrom #ositi+e #late tothe negati+e #late.


33/98

The feld lines are #arallel and e&ualls#aced in the central region ar rom the

edges but ringe outward near the edges.

n the central region! the electric feld hasthe same magnitude at all #oints.


34/98

Electric feld strength! E or a#oint charge.

'rom defnition *

(2)

o

F E

q

=r

L

2(1)

oQq F k

r

=r

L

(ccording to Coulomb’s Law 6

Consider a test charge! &, located at a

distance r rom a #oint charge 7!


35/98

where

8 5 Coulomb constant 9.:;1:9 / m2 C52 7 5 #oint charge that #roduce

electric feldr 5 distance a #oint rom the#oint charge


36/98

/otice that E is in+ersel #ro#ortional to r2

2

1

r E ∝

The relationshi# between E and r can beshown in the gra#h below.

E

r

The strength o E will decrease when the

distance rom the charge increase.


37/98

E'ample @

Betermine#a$ the electric field strength at a point G at a

distance of cm from a point charge Q

6 =; AC.

#+$ the electric force that acts on a point charge% 6 –1 AC placed at point G.

2 l i


38/98

2olution*2olution*

#a$ 5rom#a$ 5rom 2r

Qk E =

2

69

2.0

)108()109(

−××=

E 6 1.; × 1/ 7 CH1

#+$ Inowing that#+$ Inowing that

o

F

E q=

r

6 #1 × 1 H/$ #1.; × 1/$

5 6 1.; 7 towards Q

+7 $ = μC >

2: cm

o F q E =

–

& o$ 5 1μC'e

E


39/98


40/98

E'ample <Two point charges, Q1 6 =D AC and Q 6 -< AC are

separated +y a distance of . m +etween each other

as in figure +elow. Betermine the resultant Eproduced +y these two charges at point ".

. m

.@ m

"

Q1 Q

2olution *

5irst, we have to draw the vector diagram for E

produced +y Q1 and Q at point ".

E1 is produced +y Q1 and E is produced +y Q.

++ −−


41/98

+. !+. !

+.' !+.' !

"

QQ11 QQ22

E1

E2

>ow we draw the electric

field line, E that e'ist at point

" J

θ

θ

−

+. !+. !

se this concept *E is outward for = charge

E is inward for H charge

Kector E is draw along the line that

Loining point " and the charge.

Th it d f th l t i fi ld i t "


42/98

The magnitude of the electric fields on point ",

11 2

1

kQ E

r

=

( ) ( )( )

9 6

5 -1

1 2

9 10 7 103.93 10 N C

0.4 E

−× ×= = ×

( ) ( )

( )

9 6

2 2

9 10 5 10

0.5 E

−× ×=

5 -1

1.80 10 N C= ×

2

2 2

2

kQ E

r =

and


43/98

""

EE11

EE22

MM

Fiven that 9

tan M 6 .@ N .

M 6


44/98

Therefore, the magnitude of the resultant E is 9

∑∑ += 22 y x E E E

5 2 5 2(1.08 10 ) (2.49 10 ) E = × + ×5 -1

2.714 10 N C E = ×

5 11.08 10 x E N C

−= ×∑5 1

2.49 10 y E N C −= ×∑

Birection of the resultant E is given +y 9

x

y

E

E =θ tan

5

5

2.49 10

1.08 10

×= ÷×

a+ove the positive '-a'is


45/98

#$ #wo point charges, Q1 − .+ µC and Q2 − .+ µC,are placed 12 c! and + c! fro! the point -

respecti$ely as shown in figure below.

Deter!ine

(a) the !agnitude and direction of the electric

field intensity at -,

-- -- -


46/98

(b) the nett electric force eerted on q0

/1µC if it is placed at -,(c) the distance of a point fro! Q1 where theelectric field intensity is "ero.

&nswer 0

(a) %- 1. 1+* 3 C1 ; towards Q1

(b) 1. 3 ; towards Q1

(c) +.1 !

2 l ti2 l ti2


47/98

2olution *2olution *

a.

4y applying the euation of electric field intensity,

thus

--

-

2

(1) raw the !

vectors

(2) Find

magnitude !

--


48/98

Direction 0 to the left (towards Q1)

Direction 0 to the right (towards Q2)



49/98


Therefore the electric field intensity at " is given +y Birection * to the left #towardsBirection * to the left #towards QQ11$$

(") #dd ! asvectors

---

--

()()


50/98


+$ Fiven

5rom the definition of electric field intensity,

The nett electric force e'erted on qo is given+y

Birection * to the left #towardsBirection * to the left #towards QQ11$$

()()



51/98


c.

The electric field at is &ero, hence

---- & To get electric field at #$ero means !1# and !2#must have samemagnitude but pointin opposite direction

13ElectricPotentialV


52/98

1.3 Electric Potential, V

Electric potential,V is defined as the work done +ye'ternal force per unit test charge to +ring that test

charge from infinity #R$ to a point #r $ in an electric fieldproduced +y a source point charge, Q.

Electric potential, V is a scalar %uantity

2! nit is S C –1

or Kolt #K$.

oqW V r ∞=

E%uipotential 4ines and 2urfaces


53/98

E%uipotential 4ines and 2urfaces

The +lue lines represent the e%uipotential surfaces #lines$.

Ped line with arrow head represents electric field lines.

E%uipotential line or surfaceE%uipotential line or surface is a line or surface on whichis a line or surface on which

all pointsall points on it are at theon it are at the same potentialsame potential..


54/98

(1) #he electric field at e$ery point on aneuipotential surface is perpendicular to the

surface.

The important properties of e%uipotential surfaces are

(2) #he electric field points in the direction

of decreasing electric potential.

() #he surfaces are closer together where the

electric field is stronger, and farther apartwhere the field is wea5er


55/98

(') 7o two euipotential surfaces can intersect

each other. &n euipotential surface is

nor!al to the electric field. 6f twoeuipotential surfaces intersect each other

then at the point of intersection there will +e

two directions of electric field, which is

i!possible.

Electric field lines and cross sections of


56/98

equipotential surfaces for

(a) A point charge

(b) A uniform field

E%uipotential

surfaces

E%uipotential lines


57/98

V can also +e written in terms of k, Q and r which is

given +y *

r Qk V =

7ote*The total electric potential at a point in space is e%ual

to the alge+raic sum of the constituent potentials at

that point.!n the calculation of U and V, the sign of the charge

must +e su+stituted in the related e%uations.


58/98


59/98

E'ample D *

point charge %1 6 =


60/98

V V at " due to each charge can +e calculated from 9at " due to each charge can +e calculated from 9

r

Qk V =

1

11

r

Qk V =

V 4

69

10125.1

0.4

100.5)109(

×=

×+×=

−

The totalThe total V V is the scalar sum of these two potentials.is the scalar sum of these two potentials.

2

22

r

Qk V =


61/98

69

4

2.0 10(9 10 )

5.0

0.36 10 V

− − ×= × ÷

= − ×Therefore,Therefore,

V V p p 66 V V 11 + V + V 2 2

6 #=1.1


62/98

Two point charges, Q16 @ C and Q26 C are

separated +y a distance of 1< cm as shown in 5igure

+elow.

Calculate

a. the electric potential at point and descri+e the

meaning of the answer,

+. the electric potential at point U.

nswer * #a$

#+$

-- -- &

4



63/98


a. 7i$en

#he electric potential at point & is

-- --

&



64/98

b. 7i$en

#herefore the electric potential at 4 is

-- --

4

Electric "otential Bifference, VK


65/98

"otential difference +etween two points U in an

electric field is the work done # or re%uired $ in +ringing apositive test charge from point to point U in an electric

field per unit charge.

A B BA

W V

q

→∆ =

BAV ∆

where * VV BA = V B – V A ( * V final – V initial


66/98

BA B A ( final initial

The electric field is a conservative field. The work done

to +ring a charge from one point to another point in an

electric field is independent of the path.

!f the value of workwork is positivepositive – work is re%uiredwork is re%uired #we#we

need e'ternal force to move the charge$.need e'ternal force to move the charge$.

!f the value of work donework done is negativenegative – no work re%uired

# or work done +y the electric force itself, no e'ternal# or work done +y the electric force itself, no e'ternal

force is needed to move the charge$.force is needed to move the charge$.

CT!7 W!n the calculation of U, Wand V , the sign of thesign of the

charge mustcharge must +e su+stitutedin the related e%uations.


67/98

7TE*


68/98

0 A DW → =

0 A C W → = A B BAW q V → = ∆

0 EF V ∆ =

7TE*

8The potential difference +etween any two points

on an e%uipotential surface is &ero.8>ence, no work is re%uired to move a charge at

constant speed along an e%uipotential surface.

E'ample ; *


69/98

E'ample ; *

"oints and U are at distances of . cm and . cm

respectively from a point charge Q 6 –1 AC.

Betermine 9

#a$ the electric potential at and U,#+$ the work re%uired in moving a point charge % 6

=. AC, from to U.

2olution *

69

7

100 10(9 10 )

0.02

4.5 10 V

− − ×= × ÷

= − ×

A

A

kQ

V r =#a$

–

1μC

& 4

kQV =


70/98

B

B

V r

=6

9

7

100 10(9 10 )

0.033.0 10 V

− − ×= × ÷

= − ×

#+$ the work re%uired is given +y *

AB BAW q V = ∆

6 7

( )

( 2 10 )'( 3.0) ( 4.5) 10

30

A Bq V V −

= −= + × − − − ×= 8 ?ork is done +y e'ternal force, ? positive

8 "otential energy, increase

& –

1μC

4

9o%

544? " EGEPC!2E


71/98

#1$ Two points, 2 and T are located around a point charge of

=


72/98

(1) a. 7i$en

#he electric potential difference between S and # is gi$en by

()()S # ==

a.


73/98

+. 5rom *

8 ?ork is done +y the system, ? negative8 "otential energy, decreases

#$ t t h = C i l d f i t


74/98

#$ test charge q% 6=. C is placed cm from a point

charge Q. < Loule of work is done +y e'ternal force

re%uired in +ringing the test charge q% to a distance 1<

cm from the charge Q.

Betermine

a. the potential difference +etween point 1< cm and

cm from the point charge, Q,# $

+. the value of charge Q,

# $

c. the magnitude of the electric field strength at point

1 cm from the charge Q.

# $

#Fiven k 6 ×11

7 m,

C ,

$

ns*

ns*

ns*

Change in Electric "otential Energy, V +etween points

i l t i fi ld


75/98

in electric field

?hen a charged particle moves in an electric field, the

field e'erts a force that can do work on the particle.?hen a charged particle moves from a point where the

potential energy is to a point where it is U , the

change in potential energy is V 6 U – and the work,

? U done +y the e'ternal force to move the charge is

%& &%W U = ∆

%& & %( )W U U = −

& % & %( )q V V U U − = −

&% &%q V U ∆ = ∆

7ote* U * 5inal 9 * !nitial


76/98

!f the electric potential at a point in an electric field

is V , then a charge % placed at that point will have

electric potential energy, of *

qV U =

7ote*The total electric potential at a point in space is e%ual

to the alge+raic sum of the constituent potentials at

that point.!n the calculation of U and V, the sign of the charge

must +e su+stituted in the related e%uations.

E'ample 1


77/98

E'ample 1

?hen an alpha particle which has charge e

moves +etween points with potential

difference of 1 K, the change in potential

energy is

, q V ∆ = ∆)1000(2e=19

2(1.6 10 )(1000)−

= ×16

, 3.2 10 −∆ = ×

Electric Potential Energy, U of two point charges


78/98

r

qqk U

21 =

wherek – Coulom+ constant

%1 – charge 1

% – charge

r – distance +etween %1 %

8 is =ve if the charges %1 % have the same sign, –ve if

they have opposite sign.

8 is proportional to 1

r


79/98

Electric Potential Energy, U of a system of pointcharges


80/98

charges

The potential energy associated with thecharge %1 at point a is the sum of the potential

energy of interaction for each pair of charges.


81/98

++=23

32

13

31

12

21

r

qq

r

qq

r

qqk U

( Algebraic sum with the sign of charge included )

Example 9


82/98

Take distance d = 14.0 cm & charge q =150 nC

What is the electric potential energy of

this system of charges ?

Solution nclude sign of


83/98

Solution

++=23

32

13

31

12

21

r

qq

r

qq

r

qqk U

( 4 )( 2 ) ( 4 )( 5) ( 5 )( 2)q q q qk

d d d

+ − + + + − = + +

−++

−= d d d

q

k

22210208

+=

d

qk

22

9 29 2(150 10 )8.99 10

0.14

− + ×= ×

32.9 10 U −= ×

nclude sign of

charge in the

substitution.

c.


84/98

The total electric potential energy for the system of three

charges is given +y

-

====

--

&ubstitute


85/98

...sign ofcharge (' or ) ineuation

1.@ Charge in a uniform electric field.


86/98

uniform electric fielduniform electric field is represented +y a set ofis represented +y a set ofelectric field lineselectric field lines which arewhich are straight, parallel to eachstraight, parallel to each

other and e%ually spaced.other and e%ually spaced.

!t can +e!t can +e produced +yproduced +y two flattwo flat parallel metal platesparallel metal plates

which is charged, one is positive and one is negativewhich is charged, one is positive and one is negative

and is separated +y a distance as shown in the figureand is separated +y a distance as shown in the figure

a+ove.a+ove.TheThe directiondirection of a uniform E is from theof a uniform E is from the positivepositive plateplate

toto thethe negativenegative plate.plate.

- - - - - - - - - - - - - - - - - - -

EE

Consider a stationary particle of chargecharge qq and massmass mm is

placed in a uniform electric fielduniform electric field EE


87/98

==

placed in a uniform electric fielduniform electric field E E ,

--

5igure #a$

5igure #+$

The electric forceelectric force FeFe e'erted on the charge is given +y

e F qE =2ince only electric forceelectric force e'erted on the particle, thus

this force contri+utes the nett force,nett force, F F and causes the

particle to accelerateaccelerate.

*f the mass of thecharged particle is

too small (eg+

proton or

electron), weight

can be negligible.

!f the mass of

the chargedparticle is too

small

#eg* proton or

electron$,

weight can +e

negligi+le.

ccording to 7ewton3s second law, then the magnitudemagnitude

of the accelerationof the acceleration of the particle is


88/98

net F ma=qE ma=

- a is the acceleration of the charged particle

E m

qa =

where - m is mass of the charged particle

Eg* !f charged particle is electron,m 6 .11 × 1 –1 kg

To find other kinematics %uantities such as v, s and t, we

can always apply the e%uations of linear #straight line $

motion. at uv +=asuv 2

22 +=2

2

1at ut s

+=

of the accelerationof the acceleration of the particle is

Consider an electronelectron #ee$ with mass,mass, !!ee enters a uniform

E


89/98

electric field, E perpendicularly with an initial velocityinitial velocity " " , theupward electric forceelectric force will cause the electron to move along acause the electron to move along a

para+olic pathpara+olic path towards the upper plate as shown in 5igure #c$

qa E

m

= ÷

qE ma=net F ma

=*t experiencesacceleration ,a along yaxis only.

Inowing that 9

--

--

--

?e can determine the velocity in the ' and y component.


90/98

con!tant x xv u= =

at uv y y +=

y y p

!n y-component, uy 6 , therefore 9

y

qE

v at t m= =The final position Xcoordinate 9 #',y$Y of the electron at

a certain time can +e determined from 9

t u s x x =2

2

1at t u s

y y +=

E'ample /*


91/98

uniform electric field e'ists in the space +etween two

identical parallel charged metal plates. The plates are

1. cm apart. n electron is released from rest at thenegatively charged plate. !t arrives at the positively

charged plate . ns later. Betermine 9

#a$ the electric field strength

#+$ the speed of the electron when itarrives at the positively charged plate.

#given 9 me 6 .11 × 1 –1 kg$

2olution *

#a$ 5rom 5 6 mea and 5 6 %E

(1)em a

E q

= L

5rom kinematics e%uation 9


92/98

2

2

1at ut s +=

2

2

10

2

2 (2)

s at

sat

= +

= L

2

2em s E qt =

Fiven that u 6 #from ret$, so 9

#$ !nto #1$ *

14

2919

31

1085.2

)102)(106.1(

)01.02)(1011.9(

−

−−

−

×=

××××

=

C N

#+$ 5rom v - u ' at


93/98

#+$ 5rom v u at

t t

sv

+= 22

0

t

sv

2=

9102

)01.0(2−×

=v

7 -11 10 m != ×


94/98

The graph is a straight line with negative constantdi t th


95/98

gradient, thus

5or uniform5or uniform E E suchsuch

as in capacitor.as in capacitor.

The negative sign indicates that the valueof electric potential decreases in thedirection of electric field.

E'ample D*


96/98

charged particle of .; ' 1-1; C enters the region

+etween two parallel plates as shown in 5!FPE.

Calculate the electric field strength in the region

+etween the plates.

2olution *

1 mm K

14

3103

1010

300 −− ×=×

== Vmd

V E

544? " EGEPC!2E


97/98

Two charged parallel plates are separated +y a

distance of


98/98

&ext Chapter'C:&-#% 20

C it d di l t i

chapter1 electrostatic 2016 reviewed

Documents