C16-Electrostatic-Student1

16.1 Coulomb’s Law16.2 Electric Field16.3 Charge in a uniform electric field16.4 Electric Potential

UNIT 16 : ELECTROSTATICS(4 hours)

Electrostatics - the study of electric charges at rest, the

forces between them and the electric fields associated with

them.1

2

16.1 Coulomb’s Law (1 hour)SUBTOPIC :

LEARNING OUTCOMES :

a) State Coulomb’s law,

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

At the end of this lesson, the students should be able to :

2 24 o

Qq kQqF

r r

16.1 Coulomb’s Law

• Coulomb’s law states that the force,F between two point charges q1 and q2 is directly proportional to the product of their magnitudes and inversely proportional to the square of the distance separating them, r.

• In equation form :

21qqF 2

1

rF

attractive force repulsive force

electrostatic force Coulomb’s law equation

• Unit : N(Newton)• Vector quantity 3

221

r

qkqF

229 C m N 10 x 9.0

constant) sCoulomb' or

constantality proportion (called

k

k04

1

air) or (vacuum

space free ofty permittivi :0

)( 021212 m N C 8.85x10

++r

--2q

1q F12 F21

++r

++2q1qF12 F21

attractive force repulsive force

F12 : the force on charge q1 due to charge q2.F21 : the force on charge q2 due to charge q1.


4

Example of force diagram 5

r

F

0

F

2r

10

21qkqM Gradient,

Graphs above show the variation of electrostatic force with the distance between two charges.

221

r

qkqF

• The sign of the charge can be ignored when substituting into the Coulomb’s law equation.• The sign of the charges is important in distinguishing the direction of the electric force.

Note:


6

Example 16.116.1 Coulomb’s Law

--

cm 4.0

++ 2q1q

Two point charges, q1=-20 nC and q2=90 nC, are separated

by a distance of 4.0 cm as shown in figure below.

Find the magnitude and direction of

a) the electric force on q1 due to q2.

b) the electric force on q2 due to q1.

c) the electric force on each charge.

(Given Coulomb’s constant, k = 9.0 x 109 N m2 C-2)

7


--cm 0.4

++ 2q1q 12F

21F

q1= 2.0 x 10-8 C, q2= 9.0 x 10-8 C, r = 4.0 x 10-2 m

Solution 16.1

a)

b)

8

16.1 Coulomb’s LawExample 16.2

Three point charges lie along the x-axis as shown in figure below. C 42qC 21q

--++ ++C 63q

cm 3.0 cm 5.0

Calculate the magnitude and direction of the net electric force exerted on q2.(Given Coulomb’s constant, k = 9.0 x 109 N m2 C-2)

9

16.1 Coulomb’s LawSolution 16.2

C 42qC 21q--++ ++

C 63q

cm 3.0 cm 5.0

r21= 3.0 x 10-2 m, r23= 5.0 x 10-2 m

23F

21F

10

16.1 Coulomb’s LawExample 16.3

2q1q++

3q

--

--

12r13r

Figure below shows the three point charges are placed in the shape of triangular.

Determine the magnitude and direction of the resultant

electric force exerted on q1.

Given q1=-1.2 C, q2 =+3.7 C, q3 =-2.3 C, r12 =15 cm,

r13=10 cm, =32 and k = 9.0 x 109 N m2 C-2.

11

16.1 Coulomb’s LawSolution 16.3

q1= 1.2x10-6 C, q2= 3.7x10-6 C, q3= 2.3x10-6 C, r12= 15x10-2 m, r13= 10x10-2 m

582q1q ++

3q

--

--

12r13r

13F

12F2

12

2112

r

qkqF

N 1.7812F

22

669

)(15x10

))(3.7x10)(1.2x10(9.0x10

12F

213

3113

r

qkqF

N 2.4813F

N 3.0958 cos 13121 FFF x

N 2.1058 sin 131 0 FF y

2y1

2

x11 FFF

N 3.781F

x

y

F

F

1

1tan 34.2 12

N 3.781F

34.2

2q1q++

3q

--

--

12r13r


13

16.1 Coulomb’s LawExercise1. Two point charges of -1.0 x 10 -6 C and +2.0 x 10 -6 C are separated by a distance of 0.30 m. What is the electrostatics force on each particle ? (0.20 N , directed to one another)

2. Calculate the net electrostatics force on charge i ) q1 ii) q2 iii) q3 C 2 2qC 5 1q

--++ ++C 3 - 3q

cm 4.0 cm 2.0

(28.7 N , to the left , 116.25 N to the right , 87.55 N to the right)

3. Calculate the net electrostatics force on charge q2 .Cq 22 Cq 51

++

Cq 33

++

--

4m

3m (4.11 x10-3 N , 18.4 o below the x-axis)

14


4. What is the electrostatics force on charge q3 ?

Cq 33

Cq 5.21

++

Cq 5.22

++

++

(0,+0.30 m)

(0,-0.30 m)

(0.43 N in the +x direction)

q

l

++ ++

++++

w

q

qq

5. Four identical point charges (q = +10.0 C) are located on the corners of a rectangle as shown in figure below.

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.

(40.9 N at 263)15

16

16.2 Electric field (1 hour)

SUBTOPIC :

LEARNING OUTCOMES :

a) Define electric field.

b) Define and use electric field strength, .

c) Sketch the electric field lines of isolated point charge,

two charges and uniformly charged parallel plates.

d) Sketch the electric field strength diagram and

determine electric field strength for a system of

charges.


o

FE

q

��

16.2 Electric field

Definition of electric field :

A ………….....of space around isolated charge where an …..……..………..……. is experienced if a ………………………………. 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).

•The direction of electric field is tangent to the electric field line at each point.

17

16.2 Electric field

• Figures below show the electric field patterns around the charge.

+q+q-q-q

a. Single positive charge b. Single negative charge

Field directionField direction

the lines point radially ……….……… from the charge

the lines point radially ………………… toward the charge

18

16.2 Electric field

c. Two equal point charges of opposite sign, +q and -q

+q+q -q-q

Field direction

the lines are curved and they are directed from the

positive charge to the negative charge.

d. Two equal positive charges, +q and + q

+q+q +q+qXX

Field Field directiondirection

(point X is neutral point )

Is defined as a point (region) where the resultant electric force is zero.

It lies along the vertical dash line. 19

e. Two opposite unequal charges, +2q and -q 16.2 Electric field

+2q+2q-q-q

Field direction• note that twice as many lines leave +2q as there are lines entering –q.

• number of lines is proportional to magnitude of charge.

f. Two opposite charged parallel metal plates

The electric field lines are perpendicular to the surface of the metal plates.

The lines go directly from positive plate to the negative plate.

The field lines are parallel and equally spaced in the central region.Thus, in the central region, the electric field has the same magnitude at all points (uniform) except near the edges.

20

•The field lines indicate the direction of the electric field (the field points in the direction ………………..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 area perpendicular to the lines. The closer the lines, the ……………………… the field.

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

•The field lines never …………. because the electric field does not have two values at the same point.

The properties of electric field lines:16.2 Electric field

21

16.2 Electric field

• The electric field strength at a point is defined as the electric force (electrostatic) per unit positive test charge that acts at that point in the same direction as the force.

Electric field strength

forceelectric the of magnitude :F

where

charge test of magnitude :q0

strength fieldelectric the of magnitude :E

• It is a vector quantity.• The units of electric field strength is N C-1 or V m-1.• The direction of the electric field strength, E depends on the sign of isolated charge.

22

A positive isolated point charge.

A negative isolated point charge.

qq)( veq0

EF

r

In the calculation of magnitude E, substitute the MAGNITUDE of the charge only.

qq )( veq0

F

E

r

0

20

qr

kqq

E

20

r

kqqF

2r

kqE

0q

FE From

and

thus

charge point isolated and point

the between distance :r

16.2 Electric field

23

Example 16.516.2 Electric field

++1 5 0.q C

2.0 cm

A

Calculate the electric field strength at point A, 2.0 cm from a point charge q1.

Solution

(Given Coulomb’s constant, k = 9.0 x 109 N m2 C-2)

24


--1 5 0.q C

2.0 cm

A

Calculate the electric field strength at point A, 2.0 cm from a point charge q1.

Solution

25


++ -- 2q1q

2 0 . cm 3 0 . cm

A

Two point charges, q1=1.0 C and q2=-4.0 C, are placed 2.0 cm and 3.0 cm from the point A respectively as shown in figure above.Finda) the magnitude and direction of the electric field

intensity at point A.b) the resultant electric force exerted on q0=4.0 C if it is

placed at point A. (Given Coulomb’s constant, k = 9.0 x 109 N m2 C-2)

26

Solution 16.616.2 Electric field

++ -- 2q1q

2 0 . cm 3 0 . cmA 1AE

2AE

The electric field strength at point A due to the charges is given by

a)

27


++ -- 2q1q

2 0 . cm 3 0 . cmA

b)

28


-- -- 2q1q

2 0 . cm 3 0 . cm

A

Two point charges, q1=-1.0 C and q2=-4.0 C, are placed 2.0 cm and 3.0 cm from the point A respectively as shown in figure above.Finda) the magnitude and direction of the electric field

intensity at point A.b) the resultant electric force exerted on q0=4.0 C if it is

placed at point A. (Given Coulomb’s constant, k = 9.0 x 109 N m2 C-2)

29


-- -- 2q1q

2 0 . cm 3 0 . cmA

1AE

2AE

91

1 2 2 21

9 0 10 1 0

2 0 10

( . )( . )

( . )A

kqE

r

13 -12 3 10 N C.

Direction : to the left (q1)

92

2 2 2 22

9 0 10 4 0

3 0 10

( . )( . )

( . )A

kqE

r

13 -14 0 10 N C.

Direction : to the right (q2)

The electric field strength at point A due to the charges is given by

2A1AA EEE

13 132 3 10 4 0 10. .AE

13 -11 7 10 N C.AE

a)

Direction : to the right (q2)30


-- -- 2q1q

2 0 . cm 3 0 . cmA

b)

0

AA q

FE

A0A EqF

136 8 10 N. 134 0 1 7 10( . )( . )AF Direction : to the right (q2)

31


++ -- 2 5 00 C.q

A--

1 8 00 C.q

3 4 00 C.q

Three charges are placed on three corners of a square, as shown above. Each side of the square is 30.0 cm. Calculate the electric field strength at point A. What would be the force on a 6.00 µC charge placed at the point A?

32

Solution16.816.2 Electric field

++ -- 2 5 00 C.q

A--

1 8 00 C.q

3 4 00 C.q

30.0 cm

30.0 cm

30.0 cm

42.4 cm

EA1

EA3

EA2

9 61

1 2 2 21

9 0 10 8 00 10

42 4 10

( . )( . )

( . )A

kqE

r

5 -14 00 10 N C.

9 62

2 2 2 22

9 0 10 5 00 10

30 0 10

( . )( . )

( . )A

kqE

r

5 -15 00 10 N C.

9 63

3 2 2 23

9 0 10 4 00 10

30 0 10

( . )( . )

( . )A

kqE

r

5 -14 00 10 N C. 33


++ -- 2 5 00 C.q

A--

1 8 00 C.q

3 4 00 C.q

30.0 cm

30.0 cm

30.0 cm

42.4 cm

EA1

EA3

EA2

5 -11 4 00 10 N C.AE

5 -12 5 00 10 N C.AE

5 -13 4 00 10 N C.AE

45o

1 3

5

45

= -1.17 10 N/C

cos oAX A AE E E

2 1

5

45

= 2.17 10 N/C

sin oAY A AE E E

2 2

52 46 10 N/C.

AX AYE E E

E

o =61.7

tan AY

AX

E

E

o=61.7

AXE

AYEE

34

Exercise 16.2 Electric field

1. Determine a) the electric field strength at a point X at a distance 20 cm from a point charge Q = + 6µC. (1.4 x 10 6 N/C)b) the electric force that acts on a point charge q= -0.20 µC placed at

point X. (0.28 N towards Q)

2. ++ -- 2q1q

20 cm

X

Two point charges, q1 = +2.0 C and q2 = -3.0 C, are separeted by a distance of 40 cm, as shown in figure above. Determinea)The resultant electric field strength at point X.

(1.13 x 103 kN C-1 towards q2)b) The electric force that acts on a point charge q = 0.50 µC placed at X. (0.57 N)

20 cm

35

Exercise3. Find the magnitude of the electric field at point P due to the four

point charges as shown in the figure below if q=1 nC and d=1 cm.

4. Find the magnitude and direction of the electric field at the centre of the square in figure below if q=1.0x10-8 C and a= 5cm.

(Given 0=8.85 x 10-12 C2 N-1 m-2)(HRW. pg. 540.11)

Ans. : zero.

(Given 0=8.85 x 10-12 C2 N-1 m-2) (HRW. pg. 540.13)

Ans. : 1.02x105 N C-1, upwards.

16.2 Electric field

36

37

16.3 Charge in a uniform electric field (1 hour)

SUBTOPIC :

LEARNING OUTCOMES :

a) Explain quantitatively with the aid of a diagram the

motion of a charge in a uniform electric field.


Cases : 1. stationary charge 2. charge moving perpendicularly to the field 3. charge moving parallel to the field 4. charge in dynamic equilibrium

16.3 Charge in a uniform electric field

• Figure a and Figure b show a particle with positive charge q is held stationary and moves at constant speed respectively, in a uniform electric field, E .• The forces acted on the particle are electrostatic force (upwards) and weight (downwards).• For the particle in static equilibrium (Figure a) and dynamic equilibrium : moves horizontally in straight line (Figure b), electrostatic force = weight FE= W qE = mg

Case 1 : Stationary charge

Figure a : Case 1Figure a : Case 1

E

eF

a

0,0 avW E

eF

0aconstant,,ov

W

vov =0, a=0

Figure b : Case 4Figure b : Case 4

Case 4 : Charge in dynamic equilibrium (moves in straight line, consider the weigh, W of the particle)

38

E

y

0v

v

x

0q

FE

-

vy

vxθ

• When the electron enters the electric field , E the only force that acts on the electron is electrostatic force , FE=qE.

Case 2 : Charge moving perpendicularly to the field


39


• This causes the electron of mass m to accelerate downwards with an acceleration a.

0y

q Ea

m

• Since the horizontal component of the velocity of the electron remains unchanged as vo , the path of the electron in the uniform electric field is a parabola.

• The time taken for the electron to tranverse the electric field is given by

o

xt

v From x oR S v t

40


• The vertical component of the velocity vy , when the electron emerges from the electric field is given by

0

y y y

oy

o

oy

o

v u a t

q E xv

m v

q Exv

mv

• After emerging from the electric field, the electron travels with constant velocity v, where

2 2x yv v v

41


• The direction of the velocity v is at an angle

1 to the horizontal.tan y

x

v

v

• The position of the electron at time t is

0xs v t

2

2

1

21

2

y y y

y y

s u t a t

s a t

or 2 2

2

2 where 0

2

y y y y y

yy

y

v u a s u

vs

a

42

Example 16.916.3 Charge in a uniform electric field

+

-

20 mmuo = 1.5 x 107m/s

60 mm

An electron travelling at speed of 1.5 x 107 m/s enters the space between two parallel metal plates 60 mm long. The electric field between the plates is 4.0 x 103 V/m. a)Sketch the path of the electron in between plates, and after emerging from the space between the plates.b)Find the magnitude and direction of the acceleration of the electron in between the plates.c)Calculate the vertical and horizontal components of the electron velocity when it emerges from the space between the plates. d)Find the angle of deflection of the electron beam. 43

Solution 16.916.3 Charge in a uniform electric field

+

-

20 mmuo = 1.5 x 107m/s

60 mm

a) DIY

b)

44


c) Time taken for electron to travel through the space between

the plates is

Vertical component of velocity,

Horizontal component of velocity,

d) Angle of deflection,

45

Case 3 : Charge moving parallel to the field Consider 2 cases as shown in figure 1 and figure 2

Figure 1Figure 1

E

eF

a

Figure 2Figure 2

E

eFa

0v0v


46

Figure 1

• A particle with positive charge q moves with initial velocity vo towards negative plate (upwards).• For a positive charge, its acceleration a is in the direction of the electric field is given by

• Since the direction of electric force is upward (same direction and parallel to the direction of motion of particle), the particle is then accelerated in straight line towards negative plate with speed v, where v > vo .


qEa

m

47


tm

qEvv

atvv

o

o

,• The velocity v of the particle after time t is given by

• The displacement s after time t is given by

2

2

2

1

,2

1

tm

qEtvs

attvs

o

o

• The displacement s in terms of velocity v is given by

mqE

vv

a

vvs

asvv

oo

o

22

,22222

22

48

Figure 2

• A particle with negative charge q moves with initial velocity vo towards positive plate (downwards).• For a negative charge, its acceleration is in the direction opposite the electric field is given by

• Since the direction of electric force is downwards, (same direction and parallel to the direction of motion of particle), the particle is then accelerated in straight line towards positive plate with speed v, where v > vo


qEa

m

49


tm

qEvv

atvv

o

o

,• The velocity v of the particle after time t is given by

• The displacement s after time t is given by

2

2

2

1

,2

1

tm

qEtvs

attvs

o

o

• The displacement s in terms of velocity v is given by

mqE

vv

a

vvs

asvv

oo

o

22

,22222

22

50


A pair of flat parallel metal plates A and B, are separated by 1.0 cm and the electric field strength that exists between the plates is 1.0 kV m-1. An electron emerges from plate A at a speed 5.0 x 106 m s-1 and moves towards B in a linear path which is perpendicular to B. Determinea) the acceleration of the electronb) the speed of the electron when it reaches Bc) the time taken by the electron to travel to B.

51


s= 1.0 x 10 -2 m , vo = 5.0 x 106 m s-1, q = 1.6 x 10 -16 C, me = 9.11 x 10 -31 kg

52


An electron is released from rest and allowed to accelerate in a straight line in a uniform electric field of strength 3.0 kV m-1. Determinea)its accelerationb)its speed after 3.0 s. Solution q = 1.6 x 10 -16 C,

me = 9.11 x 10 -31 kg

53

54

16.4 Electric Potential (1 hour)SUBTOPIC :

LEARNING OUTCOMES :

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.

e) Use for uniform E.

f) Deduce the change in potential energy, between two points in

electric field.

g) Calculate potential energy of a system of point charges.


QV k

r

AB A B

BAAB

V V V

WV

q

VE

d

U

U q V

1 3 2 31 2

12 13 23

q q q qq qU k

r r r

16.4 Electric Potential

55

• The electric potential V at a point in an electric field is the ……… done to bring a unit positive charge from infinity to that point. (energy required to bring 1 C of positive charge from infinity to that point)

• The electric potential at infinity is considered zero.

• Scalar quantity.

• Its unit is volt (V) or J/C.

• Its formula is

QV k

r

0V


56

+

r

A•Q

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

A

A

QV k

rQ

V kr

-

r

A•Q

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

A

A

QV k

rQ

V kr

Case A Case B


57

+

r

A•Q

+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

02

but E

kQqF

r

The electric potential V at a point in an electric field is the work done to bring a unit positive charge from infinity to that point.

0

rA

WV

q

oA

o

A

kQqV

q r

QV k

r

Case A


58

-

r

A•Q

+qo

r

FE

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

02

but E

kQqF

r

The electric potential V at a point in an electric field is the work done to bring a unit positive charge from infinity to that point.

0

rA

WV

q

oA

o

A

kQqV

q r

QV k

r

Case B


59

Example 16.12

++q A

m10

Figure above shows a point A at distance 10 m from the positive point charge, q = 5C.

Calculate the electric potential at point A and describe the meaning of the answer.(Given Coulomb’s constant, k = 9.0 x 109 N m2 C-2)

Meaning :


60

Example 16.13

+ +

•

Q1 =2µC Q2 =2µC

5 cm3 cm

A

Calculate the electric potential at point A .


61

Example 16.14

+ -

•

Q1 =2µC Q2 =2µC

5 cm3 cm

A

Calculate the electric potential at point A .


62

Example 16.15

+

+

+

-

o

+0.01 µC +0.02 µC

+0.01 µC -0.02 µC

10 cm

10 cm

Calculate the electric potential at point O .

VA = 2546 V


63

• The potential difference between points A and B, VAB is given by

q

WVVV BA

BAAB

electric potential at point B:BV

electric potential at point A:AV

work done in bringing a charge from point B to point A:BAW

a charge:q

Potential Difference

q

WV BA

AB

or

ABBA qVW

In calculation ifi)W positive : work done by electric fieldii)W negative : work done on electric field


64

Example 16.16

A point charge of q = +50 µC moves from point A to point B in an electric field. The work done by the field is 10 µJ. Determine the electric p.d crossed by the charge.


65

Example 16.17

A point charge of q =-2.0 µC travels from point X to point Y in an electric field, crossing a potential rise of 200 V. a)Determine the work done in transferring the charge.b)Is the work done on or by the electric field ?c)An electric potential of -20 V exists at point Y. Determine the potential at point X.

a)

Solution

b) c)


66

Example 16.18Two point charges q1=+2.40 nC and q2=-6.50 nC are 0.100 m apart. Point A is midway between them, point B is 0.080 m from q1 and 0.060 m from q2 as shown in figure below.

1q++ --

2q

B

A

m060.0 m080.0

m050.0 m050.0Find

a) the electric potential at point A,

b) the electric potential at point B,

c) the work done by the electric field on a charge of 2.5 nC

that travels from point B to point A.


67

Solution 16.18q1=+2.40 nC , q2=-6.50 nC

1q++ --

2q

B

A

m060.0 m080.0

m050.0 m050.0

a) 1 2A A AV V V

b) 1 2B B BV V V

( )BA A BW q V V c)


68

Example 16.19a) What is the electric potential at the point

i) 10 cm and ii) 50 cm from a point charge of 2 µC ?b) Find the work done to move a charge of 0.05 µC from a point 50 cm from the point charge 2 µC to a point 10 cm from the point charge. State whether the work done by or on the electric field.

9 65

10 2

9 0 10 2 101 8 10 V

10 10( )

( . )( ).

( )cm

kqV

r

9 6

450 2

9 0 10 2 103 6 10 V

50 10( )

( . )( ).

( )cm

kqV

r

50 10 10 50

-6 5 4 3 =-0.05 10 1 8 10 3 6 10 7 2 10 J

( )

. . .

cm cm cm cmW q V V

Work done on the electric field.

a)

b)

16.4 Electric PotentialExercise1. What is the electric potential 11.0 cm from a 4.25 µC point charge ?

(3.47 x 10 5 V)2. What is the electric potential 2.0 x 10 -10 m from a proton (charge +e) ? (7.2 V)3. The electric potential and the electric field strength at a point in an

electric field produced by a point charge Q are + 500 V and 150 V/m respectively. Determinea) the distance of the point from the point charge (3.33 m)b) the charge Q. (1.85 x 10 -7 C)

4. A negative point charge of 0.75 mC travelling from point X to point Y in an electric field, experiences a pd drop of 200 V.a) Determine the potential at point Y if a potential of +20 V exists at point X. (+220 V)b) Determine the work done in this charge transfer. Is the work done on or by the field ? (-0.15 J, work done on the field)

5. A rectangle ABCD has length AB = 4.0 m and breadth AD = 3.0 m.Point charges Q1 = + 0.08µC and Q2 = - 0.03 µC are placed at A and C respectively. Determinea) the electric potential VB-VD across B and D (-72.8 V)b) the work required in moving a point charge q = =0.50 µC from D to B.

(+36.4 J)

69


70

Relation Between V and E

• The relationship between electric field intensity E at a point in a uniform electric field and electric potential V is given by

d

VE

d

V(uniform) E

• Unit ; Vm-1

• Vector quantity : directed towards negative plate

Example 16.20

Two parallel plates are charged to a voltage of 50 V. If the separation between the plates is 0.050 m, calculate the electric field strength between them.


Exercise

1. How strong is the electric field between twp parallel plates 5.0 mm apart if the potential difference between them is 110 V ? (22 kV m-1)

2. An electric field of 800 V/m is desired between two parallel plates 6.0 cm apart. How large a voltage should be applied ? (48 V)

3. The electric field between two parallel plates connected to a 45-V battery is 600 V/m. How far apart are the plates ? (75 mm)

71

72

Equipotential Lines and Surfaces

• The electric potential can be represented graphically by drawing equipotential lines or in three dimensions, equipotential ……………………...

• An equipotential surface is a surface on which all points are at the same ……………………….• ….. 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)

• An equipotential surface must be ……………………. to the electric field at any point.


73

i) An isolated charge

E

C

AB

The dashed lines represent the equipotential surface (line).

CBA VVV

)( BAABBA VVqVqW

0WBA


74

ii) A uniform electric field

The dashed lines represent the equipotential surface (line).

CBA VVV

)( BAABBA VVqVqW

0WBA

C

A

B

E

+

+

+

+

+

-

-

-

-

-



75

Electric Potential Energy, U

• The electric potential energy U of a charge q at a point in an electric field is given by

75

+

r

A•Q

q

U qV

• Scalar quantity.• Its unit is joule (J).• The sign of charges must be substituted in the calculation.

-

r

A•Q

q

Exercise

What is the potential energy of an electron that is 0.53 x 10 -10 m from a proton ? (-4.35 x 10-18 J)


The Change in Electric Potential Energy, ∆U

AB

A B

A B

U U

U U

qV qV

The difference between two electric potential energy

The difference between electric potential energy at point A and electric potential energy at point B

7676


77

Example 16.21

77

a) What is the electric potential energy of a charge 0.05µC at the pointi) 10 cm and

ii) 50 cm from a point charge of 2 µC ?b) Find the change in electric potential energy between i) and ii).

10 10

( ) ( )cm cmU qV

kQq

r

50 50

( ) ( )cm cmU qV

kQq

r

10 50

10 50

( )

( ) ( )

cm cm

cm cm

U U

U U

a)

b)


Electrical Potential Energy of A System of Point Charges

• Consider a system with three point charges, q1 , q2 and q3.• The potential energy of the system is given by

12 13 23

1 3 2 31 2

12 13 23

U U U U

q q q qq qU k

r r r

+

+

+q1

q2

q3 r12r12

r23

• The sign for the charge (+ or -) must be substituted in the calculation.

The electric potential energy of the system of

charges is the work done to move all the

charges from infinity to the points where the charges

are placed.

78


79

Example 16.22

79

+

+

-

q2 = 10µC

r12= 4.0 cm

r23 = 5.0 cm

r13 = 5.0 cm

q1 = - 10µC

q3 = 10µC

Calculate the potential energy for the system of charges shown above.

12 13 23U U U U

U

U

U = -22.5 J

16.4 Electric PotentialExercise

80

1. A point charge Q = +9.10 µC is held fixed at the origin. A second point charge with a charge of q = -0.420 µC is placed on the x -axis, 0.960 m from the origin. Calculate the electric potential energy of the pair of charges. (-3.58 x 10-2 J)

2. • • • •A D C B

20 cm 20 cm60 cm

+200 pC -100 pC

In figure above, the charge at A is +200 pC, while the charge at B is-100 pC. a)Calculate the electric potentials at points C and D. (-2.25 V, +7.88 V)b)If a charge of +500 µC is placed at points C and D, calculate the electric potential at points C and D. (- 1.13 x 10 -3 J, 3.94 x 10-3 J)c) The change in potential energy between points C and D. (-5.07 mJ)

80

16.4 Electric PotentialExercise

81

3. A point charge q1 = +2.00 nC is placed at the origin and a second point charge q2 = -3.00 nC is placed on the x -axis, at x = +20.0 cm. A third point charge q3 = 5.00 nC is to be placed on the x -axis between q1 and q2. a) What is the electric potential energy of the system of the three charges if q3 is placed at x = +10.0 cm ? (-7.20 x 10 -7 J) b) Where should q3 be placed to make the potential energy of the system equal to zero? (6.91 cm)

4. Three equal point charges q = 8.40 x 10-7 C are placed at the corner of an equilateral triangle whose side is 1.00 m. What is the potential energy of the system? ( 19.1 mJ)

81

C16-Electrostatic-Student1

Documents

Transcript of C16-Electrostatic-Student1