5_problem.pdf

Electrostatics – Solving Problems Electrostatics – Solving Problems folk.uio.no/ravi/EMT2013/

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

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Coulomb'ss Torsion Balance Coulomb 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  separation of the charges 2

Experiments Results

F Line Fr‐2

r

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

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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 g particles. p charged 1

F

r

2

( ) The (2) h fforce iis proportionall to the h product d off charge q1 and the charge q2 on the particles.

F  q1 q 2 P.Ravindran,  PHY041: Electricity & Magnetism 11 January 2013: Electrostatics‐Problems

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

q1 q 2 F r2

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

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Coulomb’s Law The electrostatic force of a charged particle exerts on another is proportional to the product of the charges and inversely proportional to the square of the distance between them.

q1 q 2 F  K r2

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

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q1 q 2 F  K r2 • where K is the coulomb constant = 9  109 N.m2/C2. • The above equation is called Coulomb’s law, law which is used to calculate the force between electric charges. charges In that equation F is measured in Newton (N), q is measured in unit of coulomb (C) and r in meter (m). (m) 7

Permittivity constant of free space • The constant K can be written as K

1 4 

• where is known as the Permittivity constant of free space. space • = 8.85 x 10-12 C2/N.m2 1

1 9 2 2  9 10 . / K   N m C 4  4  8.85  10  12 P.Ravindran,  PHY041: Electricity & Magnetism 11 January 2013: Electrostatics‐Problems

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Example 1 Calculate the value of two equal charges if they p one another with a force of 0.1N when repel situated 50cm apart in a vacuum. q1 q 2 S l ti Solution F  K r2 9  10 9  q 2 0 .1  (0.5) 2

q = 1.7x10‐6C = 1.7C  P.Ravindran,  PHY041: Electricity & Magnetism 11 January 2013: Electrostatics‐Problems

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

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

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Example 3 The following three charges are arranged as shown. Determine the net force actingg on the charge on the far right (q3 = charge 3).

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

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 Step 1: Calculate the force that charge 1 exerts on charge 3... 3  It does NOT matter that there is another charge g it! It will not in between these two… ignore effect the calculations that we are doing for tthese ese two. Not Notice ce tthat at tthee tota total d distance sta ce between charge 1 and 3 is 3.1 m , since we need to add 1.4 m and 1.7 m .

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

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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. versa Do not automatically treat a negative answer as meaning “to the left” in this formula!!! Since all I care about is what is happening to charge 3, 3 •all all 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‐Problems

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• St Step 2: 2 Calculate C l l t the th force f that th t charge h 2 exerts t on charge 3... • Same thing as above, 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 or

both positive, so the force is repulsive. I know that charge 2 is pushing charge 3 to the right with a force of 2.5e‐1 2 5e 1 N. N

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

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

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M l i l Charges Multiple Ch in i 2 Dimensions Di i Q2

F41 Q1

‐

F12 +

F13 Principle of  superposition p p

‐

Q3

Force on charge is vector sum  of forces from all charges + Q4

F1  F12  F13  F14

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

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Example p 4 • Two equal positive charges q=2x10-6C interact with a third charge Q=4x10-66C. C Find the magnitude and direction of the resultant force on Q. Q

16

FQq 1

6 6 qQ ( 4  10 )( 2  10 )  K 2  9  10 9  0 .29 N  FQq 2 2 r ( 0 .5)

 0 .4  F x  F cos   0 .29    0 .23 N  0 .5   0 .3  F y   F sin    0 .29     0 .17 N  0 .5 

F F

x

 2  0 .23  0 .46 N

y

0

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

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Example 5 • In figure what is the  resultant force on the  g charge in the lower left  corner of the square?   Assume that q=110‐7 C  Assume that q=110 C and a = 5cm

q

-q 3

2

F13 1 2q

4

F14

-2q

F12

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

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    F1  F12  F13  F14 2qq F12  K 2 a 2qq F133  K 2 2 2a

2q 2q F14  K a2 19

20

21

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

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

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23

Example • Two charges are located on the positive x-axis of a coordinate system, as shown in figure below. Charge q1=2nC is 2cm from the origin, and charge q2=-3nC is 4cm from the origin. What is the total force exerted by these two charges on a charge q3=5nC located at the origin?

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

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(9  109 )( 2  109 )(5  109 ) 4 F31   0 . 56  10 N 2 (0.04) (9  109 )(3  109 )(5  109 ) 4 F32   3 . 37  10 N 2 (0.02)

F3  F31  F32  F3  0.56  104  3.37  104  2.81  104 N 25

•

-5q q -

In figure shown, locate the point at which the electric field is zero? Assume a = 50cm

V

2q q + a

-5q 1 -

S a

2q 2 +

E1

P

E2

d aa+dd

E1 = E2

1 4 

2q 1 5q  ( 0 .5  d ) 2 4  (d ) 2 d = 30cm

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

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We have q =10 nC at the origin, q = 15 nC at x=4 m.  1 2 What is E at y=3 m and x=0

y

P 3

q1=10 nc

4

x q2 =15 nc

Find x and y components of electric field due to both charges and add  them up 27

Recall E =kq/r R ll E k / 2 and k=8.99 x 109 N.m2/C2 Field due to q q1 E = 1010 N.m2/C2 10 X10‐9 C/(3m)2 = 11 N/C in the y direction.

Ey= 11 N/C

y

E

 3

q1=10 nc

5  4

x q2 =15 nc

Ex= 0 Field due to q2 E = 1010 N.m2/C2 15 X10‐9 C/(5m)2 = 6 N/C at some angle φ Resolve into  x and y components Ey=E sin f = 6 * 3/5 =18/5 = 3.6 N/C

Ey= 11 + 3.6 = 14.6 N/C Ex= ‐4.8 N/C 4 8 N/C Magnitude

E  E x2  E y2

Ex=E cos  =E cos f = 6  f = 6 * (‐4)/5 =‐24/5 = ‐4.8 N/C ( 4)/5 = 24/5 = 4 8 N/C P.Ravindran,  PHY041: Electricity & Magnetism 11 January 2013: Electrostatics‐Problems

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E

Ey= 11 + 3.6 = 14.6 N/C Ex= -4.8 N/C

 3 4

q1=10 nc

x q2 =15 nc

Magnitude of electric field

E  E x2  E y2

E

14.6  4.8  15.4N /C 2

2

Using unit vector notation we can also write the electric field vector as:    E  4.8 i  14.6 j

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

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

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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.

    E p  E 1  E 2  E3 E1  E2  E3 

1 4  1 4  1 4 

+q 1

+q 2

P

3 -2q

+q 1

+q + 2

q a2 q 2 a2 2q a2

E2x

E3

P

E2y E2

3 -2q

E1

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

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• Evaluate E l t the th value l off E1, E2, & E3 – E1 = 3.6x105 N/C, – E2 = 1.8 1 8 x 105 N/C, N/C – E3 = 7.2 x 105 N/C

W fi d th We find the vector E t E2 need analysis to two components d l i t t t  E2x = E2 cos45  E2y = E E2 sin45 i 45

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

E  Ex2  Ey2   tan1

Ey Ex

= 7.7 x 105 N/C  = ‐ 38.6o

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

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Example p • A particle having a charge q=310-9C moves from point a to point b along a straight line, a total distance d=0.5m. The electric field is uniform g this line, in the direction from a to b, with along magnitude E=200N/C. Determine the force on q, the work done on it by the electric field, and the potential difference Va-Vb.

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

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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 by F =qE = 310-9  200 = 60010-9N Th workk done The d by b this thi force f is i W =Fd = 60010-9  0.5 = 30010-9J The potential difference is the work per unit charge, which is Va-Vb = W/q = 100V Or Va-Vb = Ed = 200  0.5 = 100V P.Ravindran,  PHY041: Electricity & Magnetism 11 January 2013: Electrostatics‐Problems

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Electric flux. ((a)) Calculate the electric flux through the rectangle in the figure (a). The rectangle is 10cm by 20cm and the electric field is uniform with magnitude 200N/C. (b) What is the flux in figure if the angle is 30 degrees? Th electric The l t i flux fl is i

   E  E  A  EA cos  So when (a) =0, we obtain





 E  EA cos   EA   200 N / C   0.1  0.2m 2  4.0 N  m 2 C

And when (b) =30 degrees, we obtain





0 1 0.2 0 2m2 cos30 30  3.5N 3 5N  m2 C  E  EA cos30 30   200 N / C   0.1 34

To calculate the electric flux due to a point charge we consider an imaginary closed spherical surface with the point charge in the center, this surface is called gaussian surface. Then the flux is given by      E.dA = E  dA cos 

=

q 4  r

=

dA = 2 

( = 0) q

4  r

2 4  r 2

q



Note that h the h net flux fl through h h a spherical h i l gaussian i surface f i proportional is i l to the h charge h q inside the surface.

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

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•

Consider several closed surfaces as shown in figure surrounding a charge Q as in the figure below. The flux that passes through surfaces S1, S2 and S3 all has a value q/. Therefore we conclude that the net flux through any closed surface is independent of the shape of the surface.

•

Consider a point charge located outside a closed surface as shown in figure. figure We can see that the number of electric field lines entering the surface equal the number leaving the surface. Therefore the net electric flux in this case is zero, because the surface surrounds no electric charge.

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

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•

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

Solution • • • •

For S1 the flux  = zero For S2 the flux  = zero For S3 the flux  = +2Q/ o For S4 the flux  For S the flux  = ‐2Q/ 2Q/ o

S3

2Q

S1 S2 -2Q

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

S4

37

Example • What must the magnitude of an isolated ppositive charge g be for the electric potential p at 10 cm from the charge to be +100V?

1 q V 4 r  q  V 4  r 2  100  4  8.9  10 12  0.1  1.1  10 9 C

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

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Example Q

q1

a

q2

a

Q2

A

P

a

A

Q

q4

A

a

A

Q

• What is the potential at the center of the square shown in figure? Assume that q1= 1 +110-8C, C q2= 2 -210-8C, q3=+310-8C, q4=+210 4 +2 10-88C, C and da= 1m.

q3

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

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Solution

Q

q1

a

q2

a

Q2

A

P

1 q 1  q 2  q3  q 4  V   Vn  4 r n

a

A

A

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

Q

q4

a

A

Q

q3

9  10 9 (1  2  3  2)  10 8 V   500V 0.71

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

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VB-VA = WAB / qo

W AB VB  VA   Ed qo

q V k r 41

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

42

-6

(0,3)

(0,0) +2

(4,0) P

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

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• Vp = V1 + V2

 q1 q 2  V  k    r1 r 2  6 6   2 10 6 10  3 9  V  9  10   6.3  10 volt  4 5  

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

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(ii) the work required is given by W = q3 Vp = 3  10-6  -6 6.3 3  103 = -18 18.9 9  103J Th -ve sign The i means that h workk is i done d by b the h charge for the movement from  to P.

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

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Electric Potential Energy Electric Potential Energy • The definition of the electric potential energy of a system of charges is the work required to bring them from infinity to that configuration. fi ti q1

q2

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 q1

q1 V  k r

q2

r

q 1q 2 U W  k r12

q1q 2 U k r P.Ravindran,  PHY041: Electricity & Magnetism 11 January 2013: Electrostatics‐Problems

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

qi qj U  k rij

48

Example -4q

• Three charges are held fixed as shown in figure. p What is the potential energy? Assume that qq=110 110-7C and a a=10cm 10cm.

a

a

A

+ 1q q

A

a

A

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

+ 2q q

49

-4q

• U=U12+U13+U23 a

a

A

+ 1q

A

a

A

+ 2q

 (  q )( 4 q ) (  q )( 2 q) ( 4 q)( 2 q )  U  k    a a a 

10q 2 U  k a 7 2

9  10 (10)(1  10 ) U    9  10 3 J 0.1 9

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

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

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Example • Point charge of +1210-9C and -1210-9C are placed 10cm part as shown in figure. Compute the potential at point a, b, and c. c • Compute the potential energy of a point charge +410-9C if it placed at points a, a b, b and c. c A

qi V   Vn  k  ri n

c

10cm b

10cm

A

A

a

+ 2q 1 4cm

6cm

4cm

+ 2q 2

52

A

At point a

c

10cm b

10cm

A

A

a

+ 2q 1 4cm

6cm

4cm

+ 2q 2

9 9      12 10 12 10 9   900V Va  9  10   0.04   0.06

53

At point b A

c

10cm b

10cm

A

A

a

+ 2q 1 4cm

6cm

4cm

+ 2q 2

 12  10 9  12  10 9    1930V Vb  9  10   0.14   0.04 9

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

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At point c p

A

c

10 10cm b

10 10cm

A

A

a

+ 2q 1 4cm

6cm

4cm

+ 2q 2

9 9   12  10  12  10 9   0V Vc  9  10   0.14   0.1

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

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We need to use the following g equation q at each point to calculate the potential energy, q U = qV At point a Ua = qVa = 410 410-9(-900) 9( 900) = -3610-7J 3610 7J At point b Ub = qVb Vb = 410-91930 410 91930 = +7710-7J +7710 7J At point c Uc = qVc = 410-90 = 0

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

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VB-VA = WAB / qo

q V k r

W AB VB  VA   Ed qo

q1q 2 U k r

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

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

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

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

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Course Text Book Physics for scientists and engineering with modern physics By R A Serway Physics for scientists and engineering with modern physics. By  R. A. Serway,

Other Recommended Resources:  Borowitz and Beiser and Beiser “Essentials Essentials of physics of physics”.. Addison Addison‐Wesley Wesley Publishing Co., 1971. 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 1982 Publishing Co., 1982.  Weidner, R.T. and Sells, R.L. “Elementary physics: classical and modern”. Allyn and  Bacon, Inc., 1973. Valkenburgh, N.V., “Basic Basic Electricity: Complete Course Electricity: Complete Course”,, 1993 1993  Valkenburgh, N.V., 

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