Electrostatics –Solving Problemsfolk.uio.no/ravi/cutn/elec_mag/5_problem.pdf · • Same thing as...
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Electrostatics – Solving ProblemsElectrostatics – Solving Problems
folk.uio.no/ravi/EMT2013/
P.Ravindran, PHY041: Electricity & Magnetism 11 January 2013: Electrostatics‐Problems1
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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
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Experiments Results
F
Line Fr‐2
r
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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
P.Ravindran, PHY041: Electricity & Magnetism 11 January 2013: Electrostatics‐Problems4
21qq
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(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
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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
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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).
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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
P.Ravindran, PHY041: Electricity & Magnetism 11 January 2013: Electrostatics‐Problems8
1085.844
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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
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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.
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Example 3Example 3
The following three charges are arranged asshown. Determine the net force acting on thegcharge on the far right (q3 = charge 3).
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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 .
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•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.
P.Ravindran, PHY041: Electricity & Magnetism 11 January 2013: Electrostatics‐Problems13
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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.
P.Ravindran, PHY041: Electricity & Magnetism 11 January 2013: Electrostatics‐Problems14
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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
P.Ravindran, PHY041: Electricity & Magnetism 11 January 2013: Electrostatics‐Problems15
431 1112p p
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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
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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
P.Ravindran, PHY041: Electricity & Magnetism 11 January 2013: Electrostatics‐Problems17
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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
P.Ravindran, PHY041: Electricity & Magnetism 11 January 2013: Electrostatics‐Problems18
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FFFF
1413121 FFFF
2122aqqKF a
213 22qqKF
23 2a
22 qqKF214 a
KF
19
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20
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21
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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?
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23
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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?
P.Ravindran, PHY041: Electricity & Magnetism 11 January 2013: Electrostatics‐Problems24
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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
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• 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
P.Ravindran, PHY041: Electricity & Magnetism 11 January 2013: Electrostatics‐Problems26
d = 30cm
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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
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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
P.Ravindran, PHY041: Electricity & Magnetism 11 January 2013: Electrostatics‐Problems28
Ex=E cos f = 6 (‐4)/5 =‐24/5 = ‐4.8 N/C
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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
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+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
P.Ravindran, PHY041: Electricity & Magnetism 11 January 2013: Electrostatics‐Problems30
E2 E123 4 a
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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
P.Ravindran, PHY041: Electricity & Magnetism 11 January 2013: Electrostatics‐Problems31
xEtan = ‐ 38.6o
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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.
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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
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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
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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.
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• 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.
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• 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
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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
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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
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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
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VB-VA = WAB / qo
EdWVV ABAB
qo
AB
qkV r
kV
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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
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-6 (0,3)
+ 2 (4,0)(0,0)
P
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• Vp = V1 + V2
21 qqkV
21 rr
kV
voltV 366
9 103.65106
4102109
54
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(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.
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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
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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
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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
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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
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-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
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JU 1091.0
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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
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Ac
At point a 10cm10cm
AaAb
+ 2q2+ 2q1
4cm6cm4cm
99 VVa 900
04.01012
06.01012109
999
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c
At point bAc
10cm10cm
AaAb
+ 2q2+ 2q1
4cm6cm4cm 4cm6cm4cm
VV 19301012101210999
9
VVb 193014.004.0
109 9
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At point cAcp
1010
b
10cm10cm
AaAb
+ 2q2+ 2q1
4cm6cm4cm
10121012 999
VVc 014.01012
1.01012109 9
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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
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VB-VA = WAB / qo
EdWVV ABAB
qoqkV r
kV qq 21
rqqkU 21
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r
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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.
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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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