Ch 16 student_201516

Topic --- Electrostatics

The study of electric

charges at rest, the

forces between them

and the electric fields

associated with them.

16.1 Coulumb’s Law (1 Hour)

16.2 Electric Field (1 Hour)

16.3 Electric Potential (1 Hour)

16.4 Charge In A Uniform Electric Field (1 Hour)


Elec

tro

stat

ics

Coulomb’s Law

Electric field

Electric field strength

Charge in uniform electric field

Electric potential

Potential energy

Equipotentialsurface


16.1 COULOMB’S LAW

(a) State Coulomb’s law,

(b) Sketch the force diagram and apply Coulomb’s law for a system of point charges

22

04 r

kQq

r

QqF


Two kind of charges:

(+)ve & (–)ve charges

Charges of opposite

sign attractone another –

attractive force

Charges of the same sign repel

one another –repulsive

force

16.1 COULUMB’S LAW

Figure 16.1


Principle of conservation of

charges

the total charge in an isolated

system is constant

(conserved)

electric charge can neither

be created nor destroyed

where: Q = electric chargen = positive integer number = 1, 2, …

e = fundamental amount of charge = 1.6 x 10-19 C

Charge is quantized

Electric charge exists as discrete

“packets”

Charge, Q is a scalar quantity

S.I unit: Coulumb (C)

1 C is defined as the total

charge transferred by a

current of 1 Ampere in 1

second

neQ


How many electrons in 1 C and what are the total

mass of these electrons?

http://en.wikipedia.org/wiki/Electric_charge


• states that the magnitude of the electrostatic (Coulomb or electric) force between two point charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them

• Since

hence the Coulomb’s law can be written as

= permittivity of free space (vacuum or air) = 8.85 x 10– 12 C2 N-1 m-2

Permittivity is a property of a material that is indicative of how well it supports an electric

field


2

21

r

QQF 2

21

r

QkQF

F = magnitude of electrostatic forceQ1, Q2 = magnitude of charges

r = distance between two point charges

k = electrostatic (Coulumb) constant = 9.0 x 109 N m2 C-2

04

1

k

2

21

04

1

r

QQF


What happens to the force between two charges, if

(a) the distance between them is doubled?

(b) the distance between them is cut in half?

(c) the magnitude of one charge is doubled?

(d) the magnitude of both charges is doubled?


• Graphically • 3 CASES

F12 : the force on charge q1 due to charge q2

F21 : the force on charge q2 due to charge q1

r

F

0

F

2

1

r0

Gradient,m = kQ1Q2

repulsive force

attractive force


Figure 16.2

Figure 16.3

Simulation 16.1

S_8/AF_2307.html


In electromagnetism, permittivity is the measure of how much resistance is encountered when forming an electric field in a medium. In other words, permittivity is a measure of how an electric field affects, and is affected by, a dielectric medium. Permittivity is determined by the ability of a material to polarize in response to the field, and thereby reduce the total electric field inside the material. Thus, permittivity relates to a material's ability to transmit (or "permit") an electric field.It is directly related to electric susceptibility, which is a measure of how easily a dielectric polarizes in response to an electric field.


Figure 16.4

http://www.wikipedia.org/wiki/Electromagnetism

http://www.wikipedia.org/wiki/Electric_field

http://www.wikipedia.org/wiki/Dielectric

http://www.wikipedia.org/wiki/Polarization_(electrostatics)

http://www.wikipedia.org/wiki/Polarization_density

http://www.wikipedia.org/wiki/Electric_field


Points to bring out about Coulomb’s law:

• The form is exactly the same as Newton’s law of universal gravitation; in particular, it is an inverse-square law

• This force can be attractive or repulsive

• The magnitude of the force can be calculated by this equation, and the direction should be obvious from the signs of the interacting charges

• Actually, if you include the signs of the charges

in the equation, then whenever you get a

negative answer for the force, there is an

attraction, whereas a positive answer indicates

repulsion


Coulomb’s Law VS Newton’s Law of Gravitation

Coulomb’s Law

Attractive or repulsive force

Force due to charge interaction

The force is a short-range force

Newton’s Law of Gravitation

Only attractive force

Force due to mass interaction

The force is a long-range forces

2

21e

r

QkQF 2

21g

r

mGmF



Two point charges, Q1 = 85 C and Q2 = 50 C are separated by a distance of 3.5 cm as shown in Figure 16.5.Determine the magnitude and direction of (a) the electric force that

Q2 exerts on Q1(b) the electric force that

Q1 exerts on Q2(Given Coulomb’s

constant, k = 9.0 109

N m2 C2)

-

cm 5.3

+1Q

2Q

Figure 16.5



Solution: m 105.3 C; 1050 C; 10 85 26

2

6

1

rQQ

-

cm 5.3

+1Q

2Q21F

1 chargeon

2 chargeby force :21

F

STEP 1: Draw the electric force vectors

The force acting on Q1 due to Q2 is attractive because Q1 and Q2 have the

opposite sign, therefore the direction of F21 is to the right

STEP 2: Use Coulomb’s equation to calculate electric force

22

669

21

105.3

10501085100.9

F

N 1012.3 4

21 F

Direction: to the right (towards Q2)

2

2121

r

QkQF

Since electric forces obey Newton’s Law, therefore the forces F21 and F12 are equal in magnitude but opposite in direction

F12 = – F21 F12 = 3.12 x

104 N



Three point charges lie along the x-axis as shown in Figure 16.6.

(a) Calculate the magnitude and direction of the total electrostatic force exerted on Q1

(b) Suppose the charge Q2 can be moved left or right along the line connecting the charges Q1 and Q3. Determine the distance from Q3 where Q2 experiences a nettelectrostatic force of zero

(Given permittivity of free space, 0 = 8.85 1012 C2 N1 m2)

cm 20

-+

μC 121 Q μC 202 Q

+

μC 363 Q

cm 12 Figure 16.6



Solution:

STEP 1: Draw the electric force vectors

(a) Q1 will experience of forces by Q2 and Q3

m 1012 C; 1020 C; 10 12 2

12

6

2

6

1

rQQ

m 1032 C; 10 36 2

13

6

3

rQ

12r

13r

-+1Q

2Q

+3Q

21F

31F

The force acting on Q1 due to Q2 is attractive because Q1 and Q2 have the

opposite sign, therefore the direction of F21 is to the right

The force acting on Q1 due to Q3 is repulsive because Q1 and Q3 have the

same sign, therefore the direction of F31 is to the left



STEP 2: Use Coulomb’s equation to calculate electric force

By applying the Coulomb’s law equation, thus

Direction : to the right (towards Q2)

2

120

2121

4 r

QQF

2212

66

21

10121085.84

10201012

F

N 15021 F

Direction : to the left

2212

66

31

10321085.84

10361012

F

N 9.3731 F

STEP 3: Electric force adds as vectors

+1Q

NF 15021

N.F 93731

The total electrostatic force on Q1

31211 FFF

N 1121 F9.371501 F Direction: to the right

(towards Q2)



Solution:

(b)

The nett force acting on Q2 is zero thus

m 1012 C; 1020 C; 10 12 2

12

6

2

6

1

rQQ

m 1032 C; 10 36 2

13

6

3

rQ

xr 13

13r

-+1Q

2Q

+3Q

32F

x

12F

3212 FF

2

230

32

2

120

21

44 r

QQ

r

QQ

2

6

22

6 1036

1032

1012

xx

m 203.0x OR cm 3.20



For each diagram below, draw the direction of electric force acting on Q1.



Figure 16.7 shows three point charges that lie in the x-y plane in a vacuum.

Calculate the magnitude and direction of the nettelectrostatic force on Q2.

(Given electrostatic constant, k = 9.00 109 N m2 C2)

cm 20

μC 0.61 Q

μC 0.42 Q μC 0.53 Q

cm 12

-

+ -

17

Figure 16.7



Given 0 = 8.85 1012 C2 N1 m2

1. Two point charges are placed on the x-axis as follows : Charge Q1 = +4.00 nC is located at x = 0.200 m, charge Q2 = +5.00 nCis at x = 0.300 m. Determine the magnitude and direction of the total electric force exerted by these two charges on a negative point charge Q3 = 6.00 nC that is placed at the origin. (University physics, 11th edition, Young & Freedman, Q21.20, p.829)

ANS: 2.4 N to the right

2. A point charge Q = 0.35 nC is fixed at the origin. Where a proton must be placed in order for the electric force acting on it to be exactly opposite to its weight? (Given charge of proton, Qp= 1.60 1019 C and mass of the proton, mp = 1.67 1027 kg ) (Physics, 3rd edition, J.S. Walker, Q18, p.657)

ANS: 5.55 km below Q



3. Four identical point charges (Q = +10.0 C) are located on the corners of a rectangle as shown in Figure 16.8.

The dimension of the rectangle are l = 60.0 cm and w = 15.0 cm. Calculate the magnitude and direction of the resultant electric force exerted on the charge at the lower left corner by the other three charges. (Physics for scientists and engineers, 6th edition,Serway&Jewett, Q57, p.735)

ANS: 40.9 N at 263

cml 0.60

cmw 0.15

Q

Q + +

++ Q

Q

Figure 16.8



16.2 ELECTRIC FIELD(a) Define and use electric field strength,

(b) Use for point charge

(c) Sketch the electric field strength diagram and determine electric field strength E for a system of charges

0q

FE

2r

kQE


• is defined as a region in which an electric force will act on a charge that, is place in that region/ a region of space around isolated charge where an electric force is experienced if a positive test charge placed in the region

• Electric field around charges can be represented by drawing a series of lines

• These lines are called electric field lines (lines of force)

16.2 ELECTRIC FIELD


Simulation 16.2

S-26/AF_2324.html


The field lines indicate the direction of the electric field (the field points in the direction tangent to the field line at any point)

The lines are drawn so that the magnitude of electric field is proportional to the number of lines crossing unit areaperpendicular to the lines. The closer the lines, the stronger the field

Electric field lines start on positive charges and end on negative charges, and the number of starting or ending is proportional to the magnitude of the charge

The field lines never cross because the electric field don’t have two value at the same point


• defined as the electric (electrostatic) force per unit positive charge that acts at that point in the same direction as the force

• Mathematically,

0q

FE

force electric :F

charge test :0q

strength field electric :E

It is a vector quantity. Unit: N C1 OR V m 1

EqF

0

16.2 ELECTRIC FIELD

E


• Consider a test charge, q0

located at a distance r from a point charge, Q

• A test charge is a charge small enough to leave the main charge

configuration undisturbed

• According to Coulumb’s Law

• From definition electric field:

• (1) in (2):

Q: magnitude of a point charger: distance between a point & point

charge

Q0q

F

r

2

21

r

QkQF (1)

0q

FE

(2)

2r

kQE OR 2

04 r

QE

16.2 ELECTRIC FIELD

Figure 16.11


Determine

(a) the electric field strength at a point X at a distance of 20 cm from a point charge Q = 8 C

(b) the electric force that acts on a point charge q = – 1 C placed at point X.

+20 cm

Q = 8CE

+20 cm

Q = 8C

–

q = – 1 C

FX

16.2 ELECTRIC FIELD


The direction• of the electric

field strength, E depends on the sign of the point charge only

• of the electric force, Fdepends on both signs of the point charge and the test charge

Figure 16.12

16.2 ELECTRIC FIELD

Simulation 16.3

S_31/AF_2313.html


Two point charges, Q1= 3.0 C and Q2= 5.0 C, are placed 12 cm and 30 cm from the point P respectively as shown in Figure 16.13.

Determine(a) the magnitude and direction of the electric field intensity at P,(b) the nett electric force exerted on q0= +1 C if it is placed at P,(c) the distance of a point from Q1 where the electric field intensity

is zero.(Given electrostatic constant, k = 9.00 109 N m2 C2)

cm 30

-1Q 2Q

cm 12

-P

Figure 16.13

16.2 ELECTRIC FIELD


Two point charges, Q1= 2.0 nC and Q2= +3.2 nC, are placed 3.0 cm apart as shown in Figure 16.14.

Determine the magnitude and direction of the resultant electric field intensity at point M. (Given permittivity of free space, 0 = 8.85 1012 C2 N1 m2)

1Q

2Q

cm 0.3

M+

-

cm 0.4

Figure 16.14

16.2 ELECTRIC FIELD


Given 0 = 8.85 1012 C2 N1 m2

1. Sketch an electric field lines pattern for following cases:

(a) Two equal negative charges, Q and Q.

(b) Two unequal negative charges, 2Q and Q

Q

- -Q

Q2

- -Q

16.2 ELECTRIC FIELD


2. Determine the magnitude of the electric field at point P due to the four point charges as shown in Figure 16.15 if q = 1 nC and d = 1 cm.(Fundamental of physics, 6th edition, Halliday, Resnick & Walker, Q11, p.540)

ANS: zero Figure 16.15

16.2 ELECTRIC FIELD


3. Calculate the magnitude and direction of the electric field at the centre of the square in Figure 16.16 if q =1.0 108 C and a = 5 cm. (Fundamental of physics,6th edition, Halliday, Resnick&Walker, Q13, p.540)

ANS: 1.02 105 N C1 ; upwards

Figure 16.16

16.2 ELECTRIC FIELD


16.3 ELECTRIC POTENTIAL

(a) Define electric potential

(b) Define and sketch equipotential lines and surfaces of (i) an isolated charge(ii) a uniform electric field

(c) Use for a point charge and a system of charges

(d) Calculate potential difference between two points

0q

WV

r

kQV

BAAB VVV

0

BAAB

q

WV



(e) Deduce the change in potential energy, U between

two points in electric field

(f) Calculate potential energy of a system of point

charges

VqU 0

23

32

13

31

12

21

r

QQ

r

QQ

r

QQkU


0q

WV



• is defined as the locus of points that have the same electric potential

Figure 16.17



The electric field, E at every point on an equipotential

surface is perpendicular to the surface

The electric field points in the direction of decreasing

electric potential

The surface are closertogether where the electric field is stronger, and farther

apart where the field is weaker

No two equipotential surfaces can intersect

each other

A (+)vepoint charge

A uniform alectric

field



APPLICATION: NO work is done when a charge moves from one point on an equipotential surface to another point on the same

surface (because the potential difference is zero)



• The electric potential at point A at distance r from a positive point charge Q is

• The electric potential at point A at distance r from a negative point charge Q is

+

r

A•

Q

–

r

A•

Q

r

QkV

r

QkV

A

A

)(

r

QkVA



A (+)ve point charge A (-)ve point charge


For a point charge

ORr

kQV

r

QV

04


+

r

A•

+qo

r

FEFext

extdW F dr

EdW F dr

2

1

r r r r

Er r

r

o

r

o

o

dW F dr

W kQq r dr

W kQq r

QqW k

r

0

2but

E

kQqF

r

The electric potential at

point A at distance r from a positive point

charge Q

0

r

A

WV

q

o

A

o

A

kQqV

q r

QV k

r



Figure 16.18 below shows a point A at distance 10 m from the positive point charge, q = 5C. Calculate the electric potential at point Aand describe the meaning of the answer.

Figure 16.18

V = (+)ve W = (+)vework is required (need external force to move the charge

V = (–)ve W = (–)veNo work required (work done by the electric force itself// No external force is needed to move the charge



Two point charges, Q1= 40 C and Q2= 30 C areseparated by a distance of 15 cm as shownin Figure 16.19. Calculate

(a) the electric potential at point A and describe the meaning of the answer,

(b) the electric potential at point B.(Given 0 = 8.85 1012 C2 N1 m2)

- -1Q 2Q

A

B

cm 5 cm 10

cm 13

Figure

16.19

• The total electric potential at a point in space is equal to the algebraic sum of the constituent potentials at that point

• In the calculation of U , W and V, the sign of the charge mustbe substituted in the related equations



• defined as the work done in bringing a positive test charge from a point to another point in the electric field per unit test charge

• The electric field is a conservative field The work done to bring a charge from one point to another point in an electric field is independent of the path

• If the value of work done is positive – work done on the system (need external force)

• If the value of work done is negative – work done by the system (no external force is needed)

0q

WV

initialfinal VVV



• Consider a positive test charge is moved by the external force, F from point A to point B as shown in Figure 16.20

• For this (+) charge, +Q

VB > VA

• The potential difference between points A and B, VAB

is given by

where

WBA: work done in bringing a (+)vetest charge from point A to point

B

VA: electric potential at point A

VB: electric potential at point B

0qF

Figure 16.20

0q

WV BA

AB VVV

AB





• Moving the test charge (+)vefrom A to B is harder and will require work (Need external force)

WAB = (+)ve

Work done by the test charge/ on an electric field

• V = VB – VA = (+)ve VB > VA

(From lower to higher electric potential, potential energy of the charge will increase)

• Moving the test charge (+)vefrom B to A is easier because it will naturally move from B to A in the direction of electric field. No external force needed

WBA = (–)ve

Work done on the test charge/ by an electric field

• V = VA – VB = (–)ve VA < VB

(From higher to lower electric potential, potential energy of the charge will decrease)

++ABE

+ABE

+



• Moving the test charge (+)vefrom A to B is easier because it will naturally move from A to B in the direction of electric field. No external force needed

WBA = (–)ve


• V = VB – VA = (–)ve VB < VA


• Moving the test charge (+)vefrom B to A is harder and will require work (Need external force)

WAB = (+)ve


• V = VA – VB = (+)ve VA > VB


+–ABE

–ABE

+



• Moving the test charge (+)vefrom A to B is difficult and will require work (Need external force)

WAB = (+)ve


• V = VB – VA = (+)ve VA < VB


• Moving the test charge (+)vefrom B to A is easier because it will naturally move from B to A in the direction of electric field. No external force needed

WBA = (–)ve


• V = VA – VB = (–)ve VB > VA


+–ABE

+ABE

+



• The potential difference between any two points on an equipotential surface is zero

• Hence no work is required to move a charge along an equipotential surface



Figure 16.21 show two points A and B are at a distance of 2.0 cm and 3.0 cm respectively from a point charge Q = –100 C.

Determine:(a) the electric potential at A and B(b) the work require in moving a point charge q = + 2 C,

from A to B

–100 C

A B

Figure 16.21



Two points, S and T are located around a point charge of +5.4 nC as shown in Figure 16.22. Calculate

(a) the electric potential difference between points S and T,

(b) the work done in bringing a charge of 1.5 nC from point T to point S. (Electrostatic constant, k = 9.00 109 N m2

C2)

nC 4.5

S Tcm .08

cm .06

+

Figure 16.22



Change in a potential energy, U

• From the definition of electric potential difference, V

• Therefore the change in a potential energy is given by

0

Δq

WV UW and

0

Δq

UV

VqU Δ0

12 VV 12 UU

final initial



• Consider a system of three point charges as shown in Figure 16.23.

• The total electric potential energy, U can be expressed as

1Q

2Q

3Q

12r 23r

13r

Figure 16.23

231312 UUUU

23

32

13

31

12

21

r

QkQ

r

QkQ

r

QkQU

23

32

13

31

12

21

r

QQ

r

QQ

r

QQkU



Solution:Take distance d = 14.0 cm and charge q = 150 nC. What is the electric potential energy of this system of charges?

23

32

13

31

12

21

r

QQ

r

QQ

r

QQkU



0 = 8.85 1012 C2 N1 m2; me= 9.11 1031 kg; e = 1.60 1019 C

1. At a certain distance from a point charge, the magnitude of the electric field is 500 V m1 and the electric potential is 3.00 kV. Calculate

(a) the distance to the charge.

(b) the value of the charge.

(Physics for scientists and engineers,6th

edition,Serway & Jewett, Q17, p.788)

ANS: 6.00 m; 2.00 C



2. Four point charges are located at the corners of a square that is 8.0 cm on a side. The charges, going in rotation around the square, are Q, 2Q, 3Q and 2Q, where Q = 4.8 C as shown in Figure 16.24.

Determine the electric potential at the centre of the square.

ANS: 1.53 106 V

Q Q2

Q2 - Q3

+ +

+

Figure 16.24



3. Initially two electrons are fixed in place with a separation of 2.00 m. How much work must we do to bring a third electron in from infinity to complete an equilateral triangle?

(Fundamental of physics,7th edition, Halliday, Resnick & Walker, Q79, p.653) ANS: 2.30 1022 J

4. Two point charges, Q1= +q and Q2= +2q are separated by 1.0 m as shown below.

Determine the position of a point where

(a) the nett electric field intensity is zero,

(b) the electric potential due to the two charges is zero.

(Fundamental of physics,7th edition, Halliday, Resnick& Walker, Q81, p.653) ANS: 0.41 m, U think

1Q2Q+ +

m 0.1




0q

WV

0q

WV

+

ve)(W

ve)(W

–

ve)(W

ve)(W

0

Δq

UV

0

Δq

KV

231312 UUUU

23

32

13

31

12

21

r

QQ

r

QQ

r

QQkU


16.4 CHARGE IN A UNIFORM ELECTRIC FIELD

(a) Explain quantitatively with the aid of a diagram the motion of a charge in a uniform electric field

(b) Use for uniform Ed

VE


• A uniform electric field is represented by a set of electric field lines which are straight, parallel to each other and equally spaced

• It can be produced by two flat parallel metal plates which is charged, one with positive and one is negative and is separated by a distance

• Direction of E: (+)ve plat to (–) plat (Figure 16.24)

Figure 16.24



CASE1: A a particle with positive charge q is held stationary

CASE2: A particle with positive charge moves at constant speed respectively, in a uniform electric field, E

•The forces acted on the particle are electrostatic force (upwards)and weight (downwards).

mgWandqEFe WqE,WFe

Figure 16.23



• When a test charge is placed between two oppositely charged plates there will be a push force from the positive plate and a pull force from the negative plate.

• The force gets weaker as the charge gets further from the plate.

• The left charge gets a very strong push (75% of the total forces) from the nearby positive plate and a weak pull (25%) from the faraway negative plate.

• The middle charge which is halfway between the two plates gets half of the total force from the pushing positive plate and 50% of the force from the pulling negative plate.

• The right charge gets 25% push from the positive and 75% pull from the negative.



• CASE3: Consider an electron (e) with mass, me enters a uniform electric field, E perpendicularly with an initial velocity u in Figure 16.24

• The upward electric force will cause the electron to move along a parabolic path towards the upper plate as shown below

u

0q-

E

xxs

v

-

ys

Simulation 16.4

Figure 16.24


S_76/AF_2326.html


Therefore the magnitude of the electron’s acceleration is given by

ey

m

eEaa

direction: upwards

since 0xa

The components of electron’s velocity after pass through the electric field are given byx-component:

y-component:

constant uvx

0 yyyy u,tauv

tm

eEv

ey

The position of the electron is

2

2

1t

m

eEs

ey

utsx

tatus yyy2

1

and

The path makes by the electron is similar to the motion of a ball projected horizontally above the ground



Figure 16.25 shows an electron entering a charged parallel plate with a speed of 5.45 106 m s1. The electric field produces by the parallel plates has deflected the electron downward by a distance of 0.618 cm at the point where the electron exits. Determine

(a) the magnitude of the electric field,

(b) the speed of the electron when it exits the parallel plates.

(Given e = 1.60 1019 C and me = 9.11 1031 kg)

Figure 16.25


(Physics, 3rd edition, J. S. Walker, Q78, p.661)


• CASE4: Charge moving parallel to the field

• Since only electric force exerted on the particle, thus this force contributes the nett force, F and causes the particle to accelerate

• For a (+)ve charge: For a (–)ve charge:m

Eqa 0

m

Eqa 0



+a

eF

CASE1: Consider a stationary particle of charge q and mass m is placed in a uniform electric field, E in Figure 16.26

a

-

eF

qEFe The electric force Fe exerted on the charge is given by Since only electric force exerted on the particle, thus this force

contributes the nett force, F and causes the particle to accelerate From Newton’s 2nd law,

maFF e

maEq 0 m

Eqa 0

Figure 16.26

For a positive charge, its acceleration is in the direction of the electric field, for a negative charge (electron), its acceleration is in the direction opposite the electric field



0y

x

u

,u

0q-

E

x tus xx 22yx vvv

-

02

1 yyyy u,tatus

2

2

1t

m

eEs

ey

0 yyyy u,tauv

tm

eE

e

xx uv



d

VE eF

a

0vm

Eqa,EqF 0

0 maF

0 u rest,at eargch

v velocity, )(at withplatarrive

2

2

1 mvK,energykineticinGain

,q

WV

0

KW 0q

KV

FdW 0q

FdV



• Figure 16.27 show a uniform electric field between two plats

• Consider a uniform electric field is produced by a pair of flat metal plates, one at which is earthed and the other is at a potential of +Vas shown below Figure 16.27



• The V against r graph for pair of flat metal plates can be shown

The graph is a straight line with negative constant gradient, thus

V

d

)d(

)V(

rΔ

VΔE

0

0

d

VE

For uniform E such as in capacitor



Two parallel plates are separated 5.0 mm apart. The electric field strength between the plates is 1.0 104 N C1.

A small charge of +4.0 nC is moved from one conducting plate to another. Calculate

(a) the work done on the charge, and

(b) the potential difference between the plates.



0 = 8.85 1012 C2 N1 m2; me = 9.11 1031 kg; e = 1.60 1019 C

An electron beam enters at right angle into a uniform electric field between two horizontal plates separated of 5.0 cm apart. The plates are connected across a potential difference of 1000 V. The length of the plates is 10.0 cm. The beam is deflected vertically at the edge of the field by a distance of 2.0 cm. Calculate the speed of the electrons entering the field.

ANS: 2.97 107 m s1


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Ch 16 student_201516