electrostatics –solving problemsfolk.uio.no/ravi/cutn/elec_mag/5_problem.pdf · • same thing as...

Electrostatics – Solving ProblemsElectrostatics – Solving Problems

folk.uio.no/ravi/EMT2013/

P.Ravindran, PHY041: Electricity & Magnetism 11 January 2013: Electrostatics‐Problems1

Coulomb's Torsion BalanceCoulomb s Torsion Balance

This dial allows you to adjust and measure the torque in the fibre and thus the force restraining the charge

This scale allows you to read the

2

separation of the charges

Experiments Results

F

Line Fr‐2

r


Experiments show that an electric force has the following properties:

(1) The force is inversely proportional to the square of separation, r2, between the two charged particles. 1g p

( ) h f i l h d f2

1r

F

(2) The force is proportional to the product of charge q1 and the charge q2 on the particles.

21qqF


21qq

(3) The force is attractive if the charges are of opposite sign and repulsive if the charges have the same sign.g

221

rqqF 2r


Coulomb’s LawThe electrostatic force of a charged particle exerts onanother is proportional to the product of the chargesand inversely proportional to the square of thedistance between them.

221

rqqKF

r


221 qqKF

2r

• where K is the coulomb constant = 9 109

N.m2/C2.N.m /C .

• The above equation is called Coulomb’s law• The above equation is called Coulomb’s law, which is used to calculate the force between electric charges In that equation F is measured inelectric charges. In that equation F is measured in Newton (N), q is measured in unit of coulomb (C) and r in meter (m)and r in meter (m).

7

Permittivity constant of free space

• The constant K can be written as

1

41

K

• where is known as the Permittivityconstant of free spaceconstant of free space.

• = 8.85 x 10-12 C2/N.m2

22912

/.109108584

14

1 CmNK


1085.844

Example 1Example 1

Calculate the value of two equal charges if they repel one another with a force of 0.1N when psituated 50cm apart in a vacuum.

qqS l ti2

21

rqqKF Solution

2

29

)50(1091.0 q

)5.0(

q = 1.7x10‐6C = 1.7C


Example 2One charge of 2.0 C is 1.5m away from a –3.0 C charge. Determine the force they exert on each g yother.


Example 3Example 3

The following three charges are arranged asshown. Determine the net force acting on thegcharge on the far right (q3 = charge 3).


Step 1: Calculate the force that charge 1 exerts on charge 3exerts on charge 3...

It does NOT matter that there is another chargein between these two… ignore it! It will notgeffect the calculations that we are doing forthese two. Notice that the total distancet ese two. Not ce t at t e tota d sta cebetween charge 1 and 3 is 3.1 m , since weneed to add 1.4 m and 1.7 m .need to add 1.4 m and 1.7 m .


•The negative sign just tells us the charges are opposite, so the force is attractive. Charge 1 is pulling charge 3 to the left, and vice versa Do not automatically treat a negative answer asvice versa. Do not automatically treat a negative answer asmeaning “to the left” in this formula!!! Since all I care about is what is happening to charge 3what is happening to charge 3,

•all I really need to know from this is that charge 3 feels a pullall I really need to know from this is that charge 3 feels a pull towards the left of 4.9e-2 N.


St 2 C l l t th f th t h 2 t• Step 2: Calculate the force that charge 2 exerts on charge 3...

• Same thing as above only now we are dealing with two• Same thing as above, only now we are dealing with two negative charges, so the force will be repulsive.

The positive sign tells you that the charges are either both negative orboth positive, so the force is repulsive. I know that charge 2 is pushingcharge 3 to the right with a force of 2 5e 1 Ncharge 3 to the right with a force of 2.5e‐1 N.

Step 3: Add you values to find the net force.


M l i l Ch i 2 Di iMultiple Charges in 2 Dimensions

Q2

41F 12F‐

+

F

Q1

Q313F ‐ Q3

Force on charge is vector sum

+

Q4 431 FFFF 1112 of forces from all charges

Principle of superposition


431 1112p p

Example 4p• Two equal positive charges q=2x10-6C interact

with a third charge Q=4x10-6C Find thewith a third charge Q=4x10 6C. Find the magnitude and direction of the resultant force

Qon Q.

16

66 )102)(104(qQ

229

21 29.0)5.0(

)102)(104(109 QqQq FNrqQKF

NFFx 23.05.04.029.0cos

NFFy 17.05.03.029.0sin

5.0

46.023.02x NF

0y

x

F


Example 5Example 5

• In figure what is the q - qresultant force on the charge in the lower left

q

2 3

gcorner of the square? Assume that q=110‐7 C 1 4

F13

Assume that q=110 C and a = 5cm - 2 q2 q

1 4

F14

F12


FFFF

1413121 FFFF

2122aqqKF a

213 22qqKF

23 2a

22 qqKF214 a

KF

19

Equilibrium

•ExampleT fi d h 1 C d 3 CTwo fixed charges, 1C and -3C are separated by 10cm as shown in figure below (a) where may a third charge be located so that no force acts on it?


Example• Two charges are located on the positive x-axis of a coordinate

system, as shown in figure below. Charge q1=2nC is 2cmfrom the origin, and charge q2=-3nC is 4cm from the origin.What is the total force exerted by these two charges on acharge q3=5nC located at the origin?charge q3=5nC located at the origin?


NF 42

999

13 1056.0)040(

)105)(102)(109(

213 )04.0(

NF 4999

10373)105)(103)(109(

NF 232 1037.3

)02.0(

FFF

NF

FFF444

3

32313

1081.21037.31056.0

25

• In figure shown, locate the point at -5q 2qIn figure shown, locate the point at which the electric field is zero? Assume a = 50cm

- +q q

a

-5q 2qV S P- +q qV S P

da+d

a

E2E11 2

a d

E1 = E2

22 )(5

41

)5.0(2

41

dq

dq


d = 30cm

We have q1=10 nC at the origin, q

2= 15 nC at x=4 m.

What is E at y=3 m and x=0y

P

x

3P

xq1=10 nc q2 =15 nc

4

Find x and y components of electric field due to both charges and add them up

27

R ll E k / 2

Field due to q1

Recall E =kq/r2and k=8.99 x 109 N.m2/C2

yE q1

E = 1010 N.m2/C2 10 X10‐9 C/(3m)2 = 11 N/Cin the y direction.

x3 5

xq1=10 nc q2 =15 nc4Ey= 11 N/C

Ex= 0

Field due to q2E = 1010 N.m2/C2 15 X10‐9 C/(5m)2 = 6 N/C

at some angle φResolve into x and y components

Ey= 11 + 3.6 = 14.6 N/CE 4 8 N/C

Ey=E sin f = 6 * 3/5 =18/5 = 3.6 N/C

E =E cos f = 6 * ( 4)/5 = 24/5 = 4 8 N/C

Ex= ‐4.8 N/C

E Ex2 Ey

2Magnitude


Ex=E cos f = 6 (‐4)/5 =‐24/5 = ‐4.8 N/C

E

Ey= 11 + 3.6 = 14.6 N/C

3Ex= -4.8 N/C

xq1=10 nc q2 =15 nc4

Magnitude of electric field

E E 2 E 2

Using unit vector notation we canalso write the electric field vector as:

E 14.6 2 4.8 2 15.4N /C

E Ex2 Ey

2

E 4.8 i

14.6 j

φ1 = tan‐1 Ey/Ex= tan‐1 (14.6/‐4.8)= 72.8 deg


+q +q1

2

What is the electric field in the lower left corner of the square as shown in figure? Assume that q = 1x10-7C and a = 5cm. -2q

P 3

+1EEEE

+q +q12

321 EEEE p

11 qE

P E3E2x 3

21 4 aE

21 qE

-2qP

E

3

E2y

322 24 a

32

41 qE


E2 E123 4 a

E l t th l f E E & E• Evaluate the value of E1, E2, & E3– E1 = 3.6x105 N/C,

E 1 8 x 105 N/C– E2 = 1.8 x 105 N/C,– E3 = 7.2 x 105 N/C

W fi d th t E d l i t t tWe find the vector E2 need analysis to two components

E2x = E2 cos45 E E i 45 E2y = E2 sin45

Ex = E3 ‐ E2cos45 = 7.2x105 ‐ 1.8x105 cos45 = 6x105N/C Ey = ‐E1 ‐ E2sin45 = ‐3.6x105‐ 1.8 x105 sin45 = ‐ 4.8x105 N/C

22yx EEE = 7.7 x 105 N/C

yE1tan


xEtan = ‐ 38.6o

Examplep

• A particle having a charge q=310-9C moves from• A particle having a charge q=310-9C moves from point a to point b along a straight line, a total distance d=0.5m. The electric field is uniform along this line, in the direction from a to b, with gmagnitude E=200N/C. Determine the force on q, the work done on it by the electric field, and thethe work done on it by the electric field, and the potential difference Va-Vb.


The force is in the same direction as the electric field since the charge is positive; the magnitude of the force is given bymagnitude of the force is given by

F =qE = 310-9 200 = 60010-9NTh k d b thi f iThe work done by this force is

W =Fd = 60010-9 0.5 = 30010-9JThe potential difference is the work per unit charge, which ischarge, which is

Va-Vb = W/q = 100VOOr

Va-Vb = Ed = 200 0.5 = 100V


Electric flux. (a) Calculate the electric flux( )through the rectangle in the figure (a). Therectangle is 10cm by 20cm and the electricfield is uniform with magnitude 200N/C.(b) What is the flux in figure if the angle is(b) What is the flux in figure if the angle is30 degrees?

Th l t i fl i The electric flux is

E E A

cosEA

So when (a) =0, we obtaincosE EA EA 2 2200 / 0.1 0.2 4.0 N mN C m C

And when (b) =30 degrees, we obtain

30EA 2 2200 / 0 1 0 2 30 3 5NN C C

34

cos30E EA 2 2200 / 0.1 0.2 cos30 3.5N mN C m C

To calculate the electric flux due to a point charge we consider an imaginaryTo calculate the electric flux due to a point charge we consider an imaginaryclosed spherical surface with the point charge in the center, this surface is calledgaussian surface. Then the flux is given by

AdE

. cosdAE ( = 0)=

= dAr

q24

= 22

44

rr

q

h h fl h h h i l i f i i l h h

=q

Note that the net flux through a spherical gaussian surface is proportional to the chargeq inside the surface.


• Consider several closed surfaces as shown in• Consider several closed surfaces as shown infigure surrounding a charge Q as in the figurebelow. The flux that passes through surfaces S1,S2 and S3 all has a value q/ Therefore weS2 and S3 all has a value q/. Therefore weconclude that the net flux through any closedsurface is independent of the shape of thesurface.surface.

• Consider a point charge located outside a closedsurface as shown in figure We can see that thesurface as shown in figure. We can see that thenumber of electric field lines entering thesurface equal the number leaving the surface.Therefore the net electric flux in this case isTherefore the net electric flux in this case iszero, because the surface surrounds no electriccharge.


• In figure two equal and opposite charges of 2Q and -2Q what is the flux In figure two equal and opposite charges of 2Q and 2Q what is the flux for the surfaces S1, S2, S3 and S4.

SolutionSolution

• For S1 the flux = zero 2QS• For S2 the flux = zero

• For S3 the flux = +2Q/ o• For S4 the flux = ‐2Q/ o

S1

S

S3

For S4 the flux 2Q/ o

-2Q

S2

S4


ExampleExample

• What must the magnitude of an isolated positive charge be for the electric potential at p g p10 cm from the charge to be +100V?

1rqV

41

CrVq 9122 101110109841004 CrVq 101.11.0109.841004


ExampleExample

• What is the potential at the center of the squareQq1 Q2q2Aa the center of the square shown in figure? Assume that 1 +110-8C 2

q2Aa

that q1= +110-8C, q2= -210-8C, q3=+310-8C,

4 +2 10 8C d

PAa

Aa

q4=+210-8C, and a = 1m.

Qq4 Qq3Aa


SolutionSolution

Qq1 Q2q2Aa q2Aa

rqqqq

VVn

n4321

41

PAa

AaThe distance r for each charge from P is 0 71mThe distance r for each charge from P is 0.71m

Qq4 Qq3Aa

VV 50071.0

10)2321(109 89


VB-VA = WAB / qo

EdWVV ABAB

qo

AB

qkV r

kV

41

ExampleExample

Two charges of 2µC and -6µC are located at positions (0,0) m and (0,3) m, respectively. (i) p ( , ) ( , ) , p y ( )Find the total electric potential due to these charges at point (4 0) mcharges at point (4,0) m.(ii) How much work is required to bring a 3µC charge from infinity to the point P?

42

-6 (0,3)

+ 2 (4,0)(0,0)

P


• Vp = V1 + V2

21 qqkV

21 rr

kV

voltV 366

9 103.65106

4102109

54


(ii) the work required is given byW = q3 Vp = 3 10-6 -6 3 103 = -18 9 10-W q3 Vp 3 10 6.3 10 18.9 103 J

Th i h k i d b hThe -ve sign means that work is done by the charge for the movement from to P.


Electric Potential EnergyElectric Potential Energy

• The definition of the electric potentialenergy of a system of charges is the workenergy of a system of charges is the work required to bring them from infinity to that

fi ticonfiguration. q1 q2q2

r

46

To workout the electric potential energy for a system of charges, assume a charge q2 at infinity and at rest as shown in figure. If

q2 is moved from infinity to a distance r from another charge q1, then the work required is given by

• W=Vq2

qq1 q2

qkV 1

rr

21qqkWU 12r

kWU

qqrqqkU 21

P.Ravindran, PHY041: Electricity & Magnetism 11 January 2013: Electrostatics‐Problems

T l l h i l f• To calculate the potential energy for systems containing more than two charges we compute the potential energy for every pair of charges separately and to add the results algebraically.separately and to add the results algebraically.

jiqqkU ijr

kU

48

ExampleExample

• Three charges are held -4q

fixed as shown in figure. What is the potential penergy? Assume that q=110-7C and a=10cm

AaAa

q 110 C and a 10cm.Aa + 2q+ 1q qq


-4q

• U=U12+U13+U23aa AaAa

Aa + 2q+ 1q

a

qqa

qqa

qqkU )2)(4()2)(()4)((

aqkU

210

a

JU 3279

109)101)(10(109


JU 1091.0

Example• Point charge of +1210-9C and -1210-9C are placed 10cm

part as shown in figure. Compute the potential at point a, b, and cand c.

• Compute the potential energy of a point charge +410-9C if it placed at points a b and c

Ac

placed at points a, b, and c.

10cm10cm

ab

i

i

nn r

qkVV

AaAb

+ 2q2+ 2q1

4cm6cm4cm

52

Ac

At point a 10cm10cm

AaAb

+ 2q2+ 2q1

4cm6cm4cm

99 VVa 900

04.01012

06.01012109

999

53

c

At point bAc

10cm10cm

AaAb

+ 2q2+ 2q1

4cm6cm4cm 4cm6cm4cm

VV 19301012101210999

9

VVb 193014.004.0

109 9


At point cAcp

1010

b

10cm10cm

AaAb

+ 2q2+ 2q1

4cm6cm4cm

10121012 999

VVc 014.01012

1.01012109 9


We need to use the following equation at each g qpoint to calculate the potential energy,

U = qVqAt point aUa = qVa = 410-9(-900) = -3610-7JUa qVa 410 9( 900) 3610 7JAt point bUb Vb 410 91930 +7710 7JUb = qVb = 410-91930 = +7710-7JAt point c Uc = qVc = 410-90 = 0


VB-VA = WAB / qo

EdWVV ABAB

qoqkV r

kV qq 21

rqqkU 21


r

ExampleExample

• In the rectangle shown in figure, q1 = -5x10-6C and q2 = 2x10-6C calculate the work required toand q2 2x10 C calculate the work required to move a charge q3 = 3x10-6C from B to A along the diagonal of the rectanglethe diagonal of the rectangle.


Course Text Book

Physics for scientists and engineering with modern physics By R A SerwayPhysics for scientists and engineering with modern physics. By R. A. Serway,

Other Recommended Resources: Borowitz and Beiser “Essentials of physics”. Addison‐Wesley Publishing Co., 1971.Borowitz and Beiser Essentials of physics . Addison Wesley Publishing Co., 1971. Halliday, D. and Resnick, R. “Physics (part two)”. John Wiley & Sons, Inc., 1978. Kubala, T.S., “Electricity 2: Devices, Circuits and Materials”, 2001 Nelkon, M. and Parker, P. “Advanced level physics”. Heinemann Educational Books p y

Ltd., 1982. Ryan, C.W., “Basic Electricity : A Self‐Teaching Guide”, 1986 Sears, F.W., Zemansky, M.W. and Young, H.D. “University physics” Addison‐Wesley

Publishing Co 1982Publishing Co., 1982. Weidner, R.T. and Sells, R.L. “Elementary physics: classical and modern”. Allyn and

Bacon, Inc., 1973. Valkenburgh, N.V., “Basic Electricity: Complete Course”, 1993Valkenburgh, N.V., Basic Electricity: Complete Course , 1993

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electrostatics –solving problemsfolk.uio.no/ravi/cutn/elec_mag/5_problem.pdf · • same thing as...

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