week 1 - electric charge and field

8/6/2019 Week 1 - Electric Charge and Field

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WEEK 1

ELECTRIC CHARGE AND

ELECTRIC FIELD


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Lightning is a powerful natural electrostatic discharge produced during a thunderstorm.

INTRODUCTION


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Pr ope r ties of the elect r ic

cha rg eProton (positivecharge )

Charge value = 1.6 x 10 -19 CElectron (negativecharge )

Charge value = -1.6 x 10 -19 C

Total chargeNumber of charges

Charge for 1 proton/electron


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Lets t ry..For each of these pair of charges, select whether the force is attractive or repulsive.

+ -

+-+ +

- -

attraction

repulsion

attraction

repulsion

What we conclude about the type of charge between li ke charges and unli ke charges?Answer


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Lets t ry..For each of these pair of charges, select whether the force is attractive or repulsive.

+ -

+-+ +

- -

attraction

repulsion

attraction

repulsion

Answer : The force between li ke charges is repulsive and the force between unli ke charges isattractive.


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Coulomb s LawExperiment shows that the electric force betweentwo charges is proportional to the product of thecharges and inversely proportional to the squaredistance between them.

2

21

r

QQk F !

o

k TQ4

1! k is proportionality constant = 9.0 x 10 9 N.m 2 / C2

Yes! We can calculate the forces value by using


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Ex e r cise 1

Tw o isolated small objects have charges of 0.04 QC and 0.06 QC are 5cm apart. What is the vector componentof electrostatic force acting on each object?

r = 0.05 m

Q1 = + 0.04 C Q2 = -0.06 C

Solution


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Step 3

Write the vector component

F12Q1 F12 = 0.00864 N

the force direction isto + x

Hence, the vectorcomponent:- F12 = 0.00864

iStep 4

Calculate Force on Q 2

Q2 = -0.06 CF21

IF12I = IF21IF21 = -0.00864 i

i

previous


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Ex e r cise 2

- + +

Calculate the resultant force on the charge Q 3 due toother t w o charges located as sho w n in figure above.Identify the direction of resultant/ net force on Q 3 .

Q1 = -4.2 C Q2 = + 1.3 C Q3 = + 1.1 C

1.0 cm2.0 cm

Solution


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Step 3


Step 4

F31 F3

-103.95 i 1 8. 7 i

Total up all the forces

Fnet = F 31 + F3= (-103.95 i) + ( 1 8. 7 i)= 4.75 i

Hence, the M AG NITU D E of total Force = 4.75 N

the D IR ECT ION of total Force= + x direction

+

previous


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Three point charges arearranged as sho w n inFigure belo w . Find the

magnitude and directionof the net force on chargeQ A

+

+-

4 m

3 m

Q A

=+

12 C

Q C=+ 20 CQ B=-16 C

Ex e r cise 3

Solution


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Step 1D raw the direction of the electric force on Q A due to Q B and Q C

4 m

3 m

Q A=+ 12 C

Q C=+ 20 CQ B=-16 C

FAC

+

+-

FA B

Click on charge Q C and Q Bto observe the direction of force experience by chargeQA

next Question


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Q A=+ 12 C

FAC

+

FAB

Step 2

Force Calculati on Meth od Value

FAC

FAB

2k

A C Q Q

r

2k

A BQ Qr

9 6 6

2

9 0 ( 2 1 0 )(2 0 1 0 )4

v v v

9 6 6

29 10 (12 10 )(16 10 )

5

v v v

Find the magnitude of force onQ A due to QB and QC

0.135 N

0.069 N

next previous


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Step 3Write the vector component

Q A

FAC = 0.135 N

+

FAB = 0.069 N

Magnitu de X - compo nent Y - compo nent

FAC= 0.135 N

FAB = 0.069 N

Total up

53.13 o

0.135 cos 90 o= 0 i 0.135 sin 90 o = 0.135 j

0.069 cos 233.13 o= -0.04 i 0.069 sin 233.13 o = - 0.055 j

= - 0.04 i = 0.08 j

next previous


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Step 4

Find the magnitude from the vector component calculated

Hence,

2 2(0.04) (0.08)

Vector component = - 0.04 i + 0.08 j

Magnitude =

= 0.089 N

previousnext


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E lect r ic field lines

Positive charge Negative charge

Increase charge

An electric field is a region where a charge experience a force.


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+


What is the effect of the increasing the charge of the field lines?Answer


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-


Increase charge

What is the difference between the electric field due to positive and negative charge?Answer


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-


Increase charge

Answer: The directions of the field lines are opposite. Electric field experienced by thepositive charge is radially outward while electric field experienced by the negative charge isradially inward.


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-


Answer: The directions of the field lines are opposite. Electric field experienced by thepositive charge is radially outward while electric field experienced by the negative charge isradially inward.


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E lect r ic field

Direction of electric field around the charge Q can be visualized by placing smapositive charge into the field and displaying the force vector acting on each.

-

Positive charge

Negative chargeQ

A

B

C

+

+

+


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E lect r ic field st r en g thElectric field strength Eat a point is defined asthe force acting on aunit positive charge atthat point.

+ +

r

2

21

r

QQk !

Q1 Q2

Derivation

Force on test charge Q 2 at a distance rfrom a point charge Q 1 is

2

21

r

QQk !

Electric field strength

2

2

21

2

1

Qr Qk Q

Q F

E !!

2

1

r

Qk E !Hence,


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2r kQ

E p !

Step 1D raw the direction of the electric field on P due to Q

P _Q = -3.0 x 10 -6 C

Imagine there is a + vecharge here

+EP

StepCalculate the magnitude of the electric field on P

P+

EP +

9 x 10 9 Nm 2 / C2 Q = -3.0 x 10 -6 C

0.3 m

Ep = 3.0 x 105 N/C

Questionnext


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Step 3


PEP ++

The irectionof E is to -X Ep = 3. x 1

5

/E p = -3. x 1 5 i /Hence, the vector

component isF inal Answer

Magnitude = 3.0 x 105

N/C anddirection = to X axis (to the left)

previous


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S ummary The force between li ke charges is repulsive and the forcebetween unli ke charges is attractive.Coulomb s law states that the electric force between twocharges is proportional to the product of the charges and

inversely proportional to the square distance betweenthem.

Electric field strength E at a point is defined as the forceacting on a unit positive charge at that point.

2

21

r

QQk !

2r

Qk

q F

E !!


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S ummary

Electric field is a vector quantity. Its directionis the direction of force on a positive charge.

Unit for electric field is N/C or V/m.

week 1 - electric charge and field

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