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22: Chapter Electric Fields
Problem 1 In Fig. 22-30 the electric field lines on the left have twice the separation as those on the right. (a) If the magnitude of the field at A is 40 N/C, what force acts on a proton at A? (b) What is the magnitude of the field at B?
6.40 10-18 N. (b) 20 N/C. Figure 22-30 Solution (a) If E is the magnitude of the electric field at A then the magnitude of the force on the proton is F = qE, (q is the magnitude of the charge on proton) F = (1.60 10-19 C)*(40 N/C) F = 6.40 10-18 N (b) Note that the field lines at B are about twice as far apart as at A. Since the density of the field lines represents the magnitude of the electric field this means that the magnitude of the field at B is about half of that at A or 20 N/C. Problem 3 What is the magnitude of a point charge whose electric field 50 cm away has the magnitude of 2.0 N/C? Answer 56 pC Solution The magnitude of the electric field due to a point charge q at a point at a distance r from the point charge is 1 q E= 4 o r 2 Therefore the magnitude of the charge is q = 4o r E =
2
B A
( 0.50 m )2 (2.0 N / C )
N m2 8.99 10 C2
9
= 56 pC
22-2
Chapter 22: Electric Fields
Problem 8 (a) In Fig. 22-31, two point charges q1 = -5q and q2 = +2q are fixed to the x axis. (a) As a multiple of distance L, at what coordinate on the axis is the net electric field of the particles zero? (b) Sketch the electric field lines. y q1 q2
L Figure 22-31
x
1.72L to the right of q2 . Solution (a) We can set up the equations for this problem if we will first determine possible regions in which the electric field could be zero. Referring to Diagram 8a, note first of all that the field cannot be zero accept along the line that runs through the two point charges. Furthermore by looking at the directions and relative sizes of the electric fields due to the two charges individually in three intervals along this line we see that any points where the field is zero must lie to the right of q2 . E q1 E q2 y q1 E q1 E q2 q2 E q1 E q2 x
L Diagram 8a
Therefore if the field is zero it is at a point (x,0) where x > L. The total electric field at such a point is 1 q1 1 q2 E tot = r+ r 3 1 4o r1 4 o r23 2 where r1 is the vector from q1 to the point (x,0) and r2 is the vector from q2 to the point (x,0). Which means that r1 = xi and r2 = (x - L)i. Therefore if the total electric field is zero the previous equation for Etot becomes
0= 1 -5q 1 2q xi + 3 ( x - L)i 3 4 o x 4 o ( x - L ) -5 2 + 2 x ( x - L )2
or
0=
or
3x2 10Lx + 5L2 = 0.
Chapter 22: Electric Fields
22-3
Solving this quadratic equation for x yields x = 0.61L or x = 2.72L. According to our previous analysis x must be greater than L therefore we find that the field will be zero at a point 1.72L to the right of the q2 . (b) A sketch of the field lines is shown in Diagram 8b. Notice in the sketch how the field is more strongly affected by the -5q charge. For quite a distance from the -5q charge the field still looks like that due to a single point charge as if the +2q charge were not present. In the vicinity of the +2q charge however the field deviates from the radial field lines much sooner.
-5q
+1q
Diagram 8b Problem 13 In Fig. 22-36, the three particles are fixed in place and have charges q1 = q2 = +e and q3 = +2e. Distance a = 6.00 m. What are the (a) magnitude and (b) direction of the net electric field at point P due to the particles? 160 N/C; (b) 45 3 a Figure 22-36 Solution Here is Fig 22-33 redrawn with vectors from each point charge to point P. a a r1 = i - j 2 2 a a r2 = - i + j 2 2 a a r3 = i + j 2 2 The net electric field at point P is y (0, a) 1 r1 r3 3 P (a/2, a/2) r2 2
y 1
a
P 2 x
(a, 0)
x
22-4
1 q1 1 q2 1 q3 r + r + r3 3 1 3 2 4 0 r1 4 0 r2 4 0 r33 1 q1 q q3 r + 2 r2 + 3 r3 3 1 3 4 0 r1 r2 r3 1 e 3 2 ( r1 + r2 + 2r3 ) 4 0 a 2 2
Chapter 22: Electric Fields
E=
E=
E=
E=
1 e a a a a a a i - j + - i + j + 2 i + j 3 2 2 4 0 a 2 2 2 2 2 2 2
1 2 2e 1 2 2e ( ai + aj) = ( i + j) 3 4 0 a 4 0 a 2
E=
2 2 2 1.6 10 -19 C N 9 N-m E = 8.99 10 ( i + j) = 113 ( i + j) 2 2 C C 6 10 -6 m
(
(
)
)
The magnitude of the field is 160 N/C and the direction as a standard angle is 45. Problem 23 Figure 22-43 shows two parallel nonconducting rings arranged with their central axes along a common line. Ring 1 has uniform charge q1 and radius R; ring 2 had uniform charge q2 and the same radius R. The rings are separated by a distance 3R. The net electric field point P on the common line, at distance R from ring 1, is zero. What is the ratio q1 /q2 ?
q1
Ring 1 P
Ring 2 q2
R R d Figure 22-43
Answer 0.506 Solution The magnitude of the electric field on the axis of a uniformly charged ring is given by qz E= 3/ 2 . If the field at point P is zero then the magnitudes of the fields due to 4 o z 2 + R 2 Ring 1 and Ring 2 are equal.
(
)
Chapter 22: Electric Fields
q1R q2 ( 2R )
2
22-5
q1 2q2
4 o R 2 + R
(
2 3/ 2
)
=
4 o ( 2R ) + R 2
(
)
3/2
or
(1 + 1)
3/2
=
( 4 + 1)3/2
Problem 25 In Fig. 22-45, two curved plastic rods, one of charge +q and the other of charge -q, form a circle of radius R in an xy plane. An x-axis passes through their connecting points, and charge is distributed uniformly on both rods. What are the (a) magnitude and (b) direction relative to the positive direction of the x axis) of the electric field E produced at P, the center of the circle? y +q P -q Figure 22-45 Answer 1 4q , in the negative Y direction 4o R 2 Solution Diagram 22-25 shows the electric field contribution, dE, due to a piece of the ring with charge dq. The piece of the ring is located at an angle measured from the x-axis and subtends an angle d about the origin. y +q dq r -q P dE x x
Diagram 22-25 Following the form of the point charge we can write dE = where r = - (R cos() i + R sin() j) is directed from dq to the origin and 1 dq r 4 o R 3
22-6 dq = Therefore q R d, R
Chapter 22: Electric Fields where the + sign applies for between 0 and and the - sign applies for between and 2.
q d 1 ( -R cos( )i - Rsin() j) dE = 4o R3
Integrating this with the + sign from 0 to and adding it to the integral with the - sign from to 2 yields 1 4q E=- j 4o R 2 Problem 26 Charge is uniformly distributed around a ring of radius R = 2.40 cm, and the resulting electric field magnitude E is measured along the ring's central axis (perpendicular to the plane of the ring). At what distance from the ring's center is E maximum? Answer R z= 2 Solution Given a uniformly charged ring of radius R, the electric field on the axis of the ring at a distance z from the plane of the ring is E= 1 qz 4 o ( z2 + r 2 )3 / 2
We can find the value of z for which E has a maximum value by differentiating E with respect to z, setting the derivative equal to zero and solving for z: dE = dz q 4 o ( z2 + r 2 )
3/ 2
- qz4o
( 4 o ) 2 (z 2 + r 2 )
1/ 2 3 2z(z 2 + r2 ) 2 3
Setting the derivative to zero (the numerator must be zero) we get q 4 o (z 2 + r 2 ) or z2 + R2 = 3z2 z= R 2
3/ 2 1/ 2 3 = qz4 o 2z( z2 + r 2 ) 2
or
Chapter 22: Electric Fields
22-7
Problem 27 In Fig. 22-46, a non-conducting rod of length L = 8.15 cm has charge q = -4.23 fC uniformly distributed along its length. (a) What is the linear charge density of the rod? What are the (b) magnitude and (c) direction (relative to the positive direction of the x axis) of the electric field produced at point P a distance a = 12.0 cm from the rod? What is the electric field magnitude produced at distance a = 50 m by (d) the rod and (e) a particle of charge q = -4.23 fC that replaces the rod.. -q L Figure 22-46 51.9 fC/m; (b) 1.57 10-3 N/C; (c) 180; (d) 1.52 10-8 N/C; (e) 1.52 10-8 N/C. Solution (a) The linear charge density, , is constant since the charge is uniformly distributed -q -4.23 fC fC = = -51.9 and is therefore equal to the average charge density, . L 8.15cm m y dq r P x L Diagram 22-27 (b,c) Referring to Diagram 22-27, the charge element, dq, at position (x, 0) has a contribution to the E-field given by dE = 1 dq r 4o r 3 L+a a P x
where r = ((L+a) - x)i , r = (L+a) - x and dq = dx = -q dx/L. Putting in these expressions yields dE = 1 -q dx ( L + a - x)i 4o L (L + a - x )3
As expected, the E-field is in the x-direction. Solving for the x-component of the field yields
L 1 -q Ex = 2 dx 4o L (L + a - x) 0
22-8
Chapter 22: Electric Fields
1 -q 1 -4.23 fC mN = = -1.57 4 o a ( L + a ) 4 o 12.0cm ( 8.15cm + 12cm ) C
Ex =
The negative sign on the x-component of the field indicates that the field is in the negative direction of the x axis. (d) The field at a = 50 m is
Ex = 1 -q 1 -4.23 fC nN = = -15.2 4 o a ( L + a ) 4 o 50m ( 8.15cm + 50m ) C
(e)
The field due to a 4.23 fC point charge at distance 50 m is
Ex = 1 -q 1 -4.23 fC nN = 2 = -15.2 2 4 o a 4 o ( 50m ) C
This result should be expected. At distance 50 m a 8.15 cm long rod will strongly resemble a point. Problem 32 In Fig. 22-51, positive charge q = 7.81 pC is spread uniformly along a thin nonconducting rod of length L = 14.5 cm. What are the (a) magnitude and (b) direction (relative to the positive direction of the x axis) of the electric field produced at point P, at distance R = 6.00 cm from the rod along its perpendicular bisector? (a) 6.22 N/C; (b) In the direction of the positive y axis.
y P
R x L Figure 22-51
Solution Referring to diagram 22-32, we can write the contribution to the E-field due to a bit of charge dq located at position (x, 0) on the axis as dE = 1 dq r 4o r 3
y dE P r = (0 - x)i + (R - 0)j dq x x Diagram 22-32 L/2
From the diagram we see that r = -xi + Rj and therefore r = (x2 + R2 )1/2. Since the charge is distributed uniformly on the rod dq = q dx/L. Substituting theses -L/2 terms into the expression for dE we get
Chapter 22: Electric Fields
22-9
dE =
1 qdx 4 o L x 2 + R 2
(
)
3/2
( -xi + Rj)
The total E-field is
E = dE =
L /2
- L /2
1 qdx 2 4 o L x + R 2
(
)
3/2
( -xi + Rj)
Or, writing the integrals for the x- and y-components
Ex = Ey =
-1 qx dx - L /2 4 2 2 o L x + R
L /2
( (
) )
3/2
L/2 -L/2
1 qR dx 4 o L x 2 + R 2
3/ 2
Integrating the integral for the x-component (use a u-substitution with u = x2 + R2 ) gives Ex = 0 which we probably should have expected from symmetry. Integrating the integral for the ycomponent (use a trig-substitution with x2 = R2 tan2 () ) gives
Ey = 1 q 1 == 4 o R L2 + 4 R 2 4 o 6.00cm 7.81pC
(14.5cm )2 + 4 ( 6.00cm )2
+ R P + +
= 6.22
N C
Problem 33 In Fig. 22-52, a "semi-infinite" nonconducting rod (that is, infinite in one direction only) has uniform linear charge density . Show that the electric field Ep at point P makes an angle of 45 with the rod and that this result is independent of the distance R. (Hint: Separately find the component of Ep parallel to the rod and the component perpendicular to the rod.) Solution Diagram 22-33 illustrates the situation with a bit of charge , dq, shown on the rod. Treating dq as a point charge we can write the electric field at point P as
E = dE = E=
+
Figure 22-52 y + (x, 0) + + + dq = dx r = -xi - Rj (0, -R) P Diagram 22-33
x
+
0
1 dx 4 o x 2 + R 2
1 ( -i - j) 4 o R
(
)
3/ 2
( -xi - Rj)
22-10
Chapter 22: Electric Fields
Equal x- and y- components mean that the field makes an angle of 45 with the rod. Problem 39 An electron is released from rest in a uniform electric field of magnitude 2.0 104 N/C. Calculate the acceleration of the electron. (Ignore gravitation.) Answer 3.51 101 5 m/s2 Solution The electron will accelerate because of the force on it due to the electric field, E. The magnitude of that force is F = qE, (q is the magnitude of the charge on the electron) from but Newton's second law a force will cause a mass to accelerate according to F = ma, (where m is the mass of the electron in this case) Equating these two expressions for F and solving for a yields
qE = m 1.6 10 -19 C 2.0 10 4 9.11 10 -31 kg N C = 3.51 1015 m s2
a=
Problem 44 Humid air breaks down (its molecules become ionized) in an electric field of 3.0 106 N/C. In that field, what is the magnitude of the electrostatic force on (a) an electron and (b) an ion with a single electron missing? 4.8 10-13 N; (b) 4.8 10-13 N Solution (a) The magnitude of the force on a point charge is F = qE therefore the force on the electron is F = (1.6 10-19 C)*(3.0 106 N/C) = 4.8 10-13 N (b) Same answer as part (a) since the ion has the same magnitude of charge.
Problem 48 At some instant the velocity components of an electron moving between two charged parallel plates are vx = 1.5 105 m/s and vy = 3.0 103 m/s. Suppose that the electric field between the plates is given by E = (120 N/C)j. (a) What is the acceleration of the electron? (b) What will be the velocity of the electron after its x coordinate has changed by 2.0 cm?
Chapter 22: Electric Fields 2.1 101 3 m/s2 j; (b) 3.5106 m/s Solution (a) The acceleration of a point charge in an electric field is a = F/m = qE/m. The acceleration of the electron in the given field is
qE = m 1.6 10 -19 C 120 N C = 2.1 1013 m -31 9.11 10 kg s2
22-11
a=
(b) The acceleration of the electron is constant and directed in the positive y-direction. This is like an "upside-down projectile" problem from Physics I. The x-component of the velocity will not change so all we need to find is the y-component of the velocity. We can solve the following two equations for vy . and vy = vy o + ay *t x = vo *t
Solving for vy yields
vy = vyo + ax x m m 2cm m = 3 10 3 + 2.1 1013 2 = 3.5 10 6 vo s s 1.5 10 5 m s s
Problem 53 A 10.0 g block with a charge of +8.00 10-5 C is placed in electric field E = (3.0 103 )i 600j, where E is in newtons per coulomb. (a) What are the magnitude and direction of the force on the block? (b) If the object is released from rest at the origin at t = 0, what will be its coordinates at t = 3.00 s? 0.245 N, 11.3 clockwise from the +x axis. (b) x = 108 m; y = -21.6 m. Solution (a) Computing each of the components of the force separately Fx = qEx Fx = (8.0 10-5 C)*(3.0 103 N/C) = 0.24 N and the y-component Fy = qEy Fy = (8.0 10-5 C)*(-600 N/C) = -0.048 N and the z-component is zero. Therefore the force is in the xy-plane with magnitude
22-12
F = Fx2 + Fy2 =
Chapter 22: Electric Fields
( 0.24 N )2 + ( 0.048N )2
= 0.245N
and the direction in terms of the standard angle, , is given by
tan ( ) = Fy Fx = -0.048N = -0.20 0.24N
= tan-1 ( -0.20 ) = -11.3
(b) Since each of the force components is constant the acceleration will have constant components and we can compute the displacements from (x - x0 ) = vx 0t + ax t2 /2, (y - y0 ) = vy 0t + ay t2 /2 and (z - z0 ) = vz0t since Fz = 0. The initial components of the velocity and position are all zero so x = ax t2 /2 and y = ay t2 /2 and z = 0 Where and ax = Fx /m = (0.24 N)/(0.010 kg) = 24 m/s2 ay = Fy /m = (-0.048 N)/(0.010 kg) = -4.8 m/s2
Therefore and x = (24 m/s2 )*(3.0 s)2 /2 = 108 m y = (-4.8 m/s2 )*(3.0 s)2 /2 = -21.6 m.
Problem 56 An electric dipole consists of charges of +2e and -2e separated by 0.78 nm. It is in an electric field of strength 3.4 106 N/C. Calculate the magnitude of the torque on the dipole when the dipole moment is (a) parallel to, (b) perpendicular to, and (c) antiparallel to the electric field. 0; (b) 8.5 10-22 N-m; (c) 0 Solution (a) The torque on an electric dipole moment of magnitude p = qd when the angle between the dipole moment and the electric field, E, is is given by = qdE sin() When the dipole moment and electric field are parallel = 0 and the torque is
Chapter 22: Electric Fields
22-13
= qdE sin() = 2(1.6 10-19 C)*(0.78 10-9 m)*( 3.4 106 N/C) sin(0) = 0 (b) When the dipole moment and electric field are perpendicular the torque is = 2(1.6 10-19 C)*(0.78 10-9 m)*( 3.4 106 N/C) sin( /2) = 8.5 10-22 N-m (c) When the dipole moment and electric field are antiparallel the torque is = 2(1.6 10-19 C)*(0.78 10-9 m)*( 3.4 106 N/C) sin() = 0 Problem 57 An electric dipole consisting of charges of magnitude 1.5 nC separated by 6.2 m, is in an electric field of strength 1100 N/C. (a) What is the magnitude of the electric dipole moment? (b) What is the difference in potential energy corresponding to dipole orientations parallel to and antiparallel to the field? 9.30 10-15 C-m. (b) 2.05 10-11 J. Solution (a) The electric dipole moment, p, for a dipole with charges +q and -q separated by a distance d is p = qd = (1.5 10-9 C)*(6.2 10-6 m) p = 9.3 10-15 C-m. (b) The potential energy of a dipole in a field when making an angle with the field is U() = -pE cos() Therefore the difference between parallel and antiparallel is U = U(180) - U(0) = (-pE cos(180)) - (-pE cos(0)) U = 2pE = 2(9.3 10-15 C-m)*(1100 N/C) U = 2.05 10-11 J. Problem 78 Calculate the electric dipole moment of an electron and a proton 4.30 nm apart. Answer 6.88 10-28 C-m
22-14 Solution The magnitude of the electric dipole moment is
Chapter 22: Electric Fields
p = qd = (1.60 10-19 C)*( 4.30 10-12 m) = 6.88 10-28 C-m Problem xx A charged cloud system produces an electric field in the air near the Earth's surface. A particle of charge -2.0 10-9 C is acted on by a downward electrostatic force of 3.0 10-6 N when placed in the field. (a) What is the magnitude of the electric field? (b) What is the magnitude and direction of the electric force exerted on a proton placed in this field? (c) What is the gravitational force on the proton? (d) What is the ratio of the magnitude of the electrostatic force to the magnitude of the gravitational force in this case? 1.50 103 N/C. (b) 2.4 10-16 N, up. (c) 1.64 10-26 N. (d) 1.46 101 0 Solution (a) The magnitude of the electric field is defined by Thus E = F/q, (where F is the force on a charge of magnitude q) E = (3.0 10-6 N)/(2.0 10-9 C) E = 1.50 103 N/C (b) The force acting on a charge in an electric field written in vector form is F = qE From part (a) we know that the electric field has magnitude of 1.50 103 N/C. We also know that this field causes a downward force on a negative particle therefore the electric field must be directed upward. (If q is negative the F and E must be anti-parallel.) Therefore the force on the proton will be F = (+1.60 10-19 C)*(1.50 103 N/C), upward F = 2.4 10-16 N, upward. (c) The gravitational force on the proton is just its weight mg = (1.67 10-27 kg)*(9.8 m/s2 ) mg = 1.64 10-26 N (d) The gravitational force is negligible compared to the electric force qE/mg = (2.4 10-16 N)/(1.64 10-26 N) qE/mg = 1.46 101 0
Chapter 22: Electric Fields
22-15
The electric force is 14.6 billion times larger than the gravitational force. Problem xx In Fig. 23-29 three point charges are arranged in an equilateral triangle. Consider the field lines of force due to +Q and -Q and from them determine the direction of the force that acts on +q because of the presence of the other two charges. (Hint: See Fig. 235) +Q Figure 23-29 Answer To the right in the figure. Solution If you rotate Figure 23-5 ccw 90 the two charges in Figure 23-5 would correspond to the +Q and -Q of Figure 23-21. Note that all of the field lines crossing the top half of the perpendicular bisector of the line joining +Q and -Q will then be horizontal and directed toward the right.
+q
a a
a
-Q
Figure 23-5
Problem xx In Fig. 23-34, charges are placed at the vertices of an equilateral triangle. For what value of Q (both sign and magnitude) does the total electric field vanish at C, the center of the triangle?
+ + Diagram 3
22-16
Chapter 22: Electric Fields
Y 3S 6 -S/2 C
Q
S/2 X
+1.0 C
-1.0 C
Figure 23-34 Answer +1.00 C Solution You may be satisfied with a symmetry argument that says that it is intuitively obvious that the third charge must also be +1.0 C. If you would like a more methodical approach to the problem you may write the electric field vectors for each of the three charges, set their sum to zero and solve for Q. Using the coordinate system shown in Figure 23-34 you may write the electric field vector for the bottom-right charge as E1 = 1 q s 3s r , where q = +1.0 C and r1 = - i + j 3 1 4o r1 2 6
the field due to the bottom-left charge is E2 = 1 q s 3s r , where q = +1.0 C and r2 = i + j 3 2 4o r2 2 6
and the field due to the top charge is E3 = 1 Q 3s r , where r3 = - j 3 3 4 o r3 3 3s =r 3
Of course r1 = r2 = r3 =
Setting the sum of these three vectors to zero and solving for Q: E1 + E 2 + E 3 = 1 q 1 q 1 Q r+ r + r =0 3 1 3 2 3 4o r1 4 o r2 4o r3 3
Chapter 22: Electric Fields 1 (qr + qr2 + Qr3 ) = 0 4o r3 1 qr1 + qr2 + Qr3 = 0 s s 3s 3s 3s q - i + j + q i + j - Q j=0 2 6 2 6 3 q 3s 3s j-Q j=0 3 3
22-17
or Q = q = +1.0 C. Problem xx Make a quantitative plot of the electric field along the central axis of a charged ring having a diameter of 6.0 cm and a uniformly distributed charge of 1.0 10-8 C. Answer Diagram 27 Solution Given a uniformly charged ring of radius R, the electric field on the axis of the ring at a distance z from the plane of the ring is E= 1 qz 4 o ( z2 + r 2 )3 / 2
Letting R = 3.0 cm and q = 1.0 10-8 C and plotting E versus z we get the graph of Diagram 27. Note the occurrence of the maximum value of E as in Problem 29.
E (kN/C) 40
E-Field on Axis of Charged Ring
20
5
10
15
20
25
z (cm)
Diagram 27
22-18
Chapter 22: Electric Fields
Problem xx (a) What is the acceleration of an electron in a uniform electric field of 1.4 106 N/C? (b) How long would it take for the electron, starting from rest, to attain one-tenth the speed of light? (c) How far would it travel? (Use Newtonian mechanics.) 2.46 101 7 m/s2 . (b) 0.122 ns. (c) 1.83 mm. Solution (a) Working with magnitudes only, the acceleration is the force divided by the mass and the force is the product of the charge of the electron and the electric field. That is a = F/m = qE/m where q = 1.6 10-19 C, m = 9.11 10-31 kg and E is 1.4 106 N/C. Therefore a = (1.6 10-19 C)*(1.4 106 N/C)/(9.11 10-31 kg) a = 2.46 101 7 m/s2 . (b) Starting from rest the initial velocity is zero and we want a final velocity of c/10, where c = 3.0 108 m/s is the speed of light. Solving for the time t = (v - v0 )/a t = (3.0 107 m/s - 0)/2.46 101 7 m/s2 t = 1.22 10-10 s = 0.122 ns. (c) Knowing the initial velocity, the final velocity and the time we can get the displacement from (x - x0 ) = (v + v0 )t/2 (x - x0 ) = (3.0 107 m/s + 0)*(1.22 10-10 s)/2 (x - x0 ) = 1.83 10-3 m = 1.83 mm. Problem xx Sketch qualitatively the lines of force associated with a thin, circular, uniformly-charged disk of radius R. (Hint: Consider as limiting cases points very close to the disk, where the electric field is perpendicular to the surface, and points very far from it, where the electric field is like that of a point charge.) Solution We can use the hint to sketch the field lines in two steps. Diagram 3a shows a sketch of the field lines in the two limiting cases. Note that we are looking at the edge of the disk and
Chapter 22: Electric Fields
22-19
only showing the lines on one side of the disk since the other side will be symmetric. As given in the problem we sketch field lines that are perpendicular to the disk surface near the surface. Far from the surface we sketch the uniformly spaced radial field lines associated with a point charge. In Diagram 3b we join these two sets of lines smoothly.
Diagram 3a
Diagram 3b

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Vanderbilt - PHYSICS - 2049

Chapter 28: Magnetic FieldsProblem 1 An electron that has a velocity given inby v = (2.0 106 i m/s) + (3.0 106 j m/s) moves through a uniform magnetic field B = 0.030 T i - 0.15 T j. (a) Find the force on the electron due to the magnetic field. (b

Vanderbilt - PHYSICS - 2049

Chapter 29: Magnetic Fields Due to CurrentsProblem 3 A surveyor is using a magnetic compass 6.1 m below a power line in which there is a steady current of 100 A. (a) What is the magnetic field at the site due to the power line? (b) Will this interfe

Vanderbilt - PHYSICS - 2049

Chapter 30: Induction and InductanceProblem 1 In Fig. 30-37, the magnetic flux through the loop increases according to the relation B = 6t2 + 7t, where B is in milliwebers and t is in seconds. (a) What is the magnitude of the emf induced in the loop

Vanderbilt - PHYSICS - 2049

Chapter 31: Electromagnetic Oscillations and Alternating CurrentProblem 1 In a certain oscillating LC circuit, the total energy is converted from electrical energy in the capacitor to magnetic energy in the inductor in 1.50 s. What are (a) the perio

Vanderbilt - PHYSICS - 2049

Chapter 32: Maxwell's Equations; Magnetism of MatterProblem 3 A Gaussian surface in the shape of a right-circular cylinder with end caps has a radius of 12.0 cm and a length of 80 cm. Through one end there is an inward magnetic flux of 25.0 Wb. At t

Vanderbilt - PHYSICS - 2049

Chapter 33: Electromagnetic WavesProblem 3 A certain helium-neon laser emits red light in a narrow band of wavelengths centered at 632.8 nm and with a "wavelength width" (such as on the scale of Fig. 33-1) of 0.0100 nm. What is the corresponding "fr

Vanderbilt - PHYSICS - 2049

Chapter 34: ImagesProblem 1 A moth at about eye level is 10 cm in front of a plane mirror; you are behind the moth, 30 cm from the mirror. For what distance must you focus your eyes to see the image of the moth in the mirror: that is, what is the di

Vanderbilt - PHYSICS - 2049

Chapter 35: InterferenceProblem 1 The speed of yellow light (from a sodium lamp) in a certain liquid is measured to be 1.92 108 m/s. What is the index of refraction of this liquid for the light? Answer 1.56 Solution The index of refraction is equal

Vanderbilt - PHYSICS - 2049

Chapter 36: DiffractionProblem 3 Light of wavelength 633 nm is incident on a narrow slit. The angle between the first diffraction minimum on one side of the central maximum and the first minimum on the other side is 1.20. What is the width of the sl

Duke - CHEM - 21L

Chemistry Text Notes Chapter 5Boyle's Law: V = k/po oInverse Relationship Gas that obeys Boyle's Law is called an ideal gas.Charles's Law: V = bToDirect RelationshipAvogadro's Law: V = anoGas at constant temperature and pressure,

University of Texas - M - 91795

Exam: 1 - reviewPage: 1 of 10Review for Midterm Exam 1General stuff : The exam will cover the following sections in the book: 1.1, 1.2, 1.3, 1.4, 1.5, 1.7 and 1.8. : The questions are going to be similar to hw problems and to problems done in cl

University of Texas - M - 91795

Exam: Midterm Exam 1Course: M340L University of Texas at AustinPage: 1 of 5Midterm Exam 1with Solutions.Note: the order of the problems may be different than the one in your exam booklet. Detailed solutions are provided on the pages followi

Duke - BME - 100

BME 100L: topic 1 First law of thermodynamics: the conservation of energyLingchong YouCentral questionsConsider a reaction: C (s, graphite) C (s, diamond) Thermodynamics: Is this process possible? If possible, what conditions should you use?

Duke - EGR - 75

EGR75L: Tension Laboratory Report GuidelinesPlease include the following sections in your lab report. It is completely acceptable and expected that you reference the lab manual in your report. Please cover all of the topics below, but be concise in

Duke - EGR - 75

EGR75L: Tensile Test for Tinius Olsen H50KS Load FrameJoseph Nadeau Civil & Environmental Engineering Duke University September 25, 2006During the course of this laboratory a material specimen (see Fig. 1) is going to be gradually loaded in tension

Duke - EGR - 75

EGR75L: Torsion Laboratory Report GuidelinesPlease include the following sections in your lab report. It is completely acceptable and expected that you reference the lab manual in your report. Please cover all of the topics below, but be concise in

Duke - EGR - 75

EGR75L: Torsion TestJoseph Nadeau Civil & Environmental Engineering Duke University October 18, 2006During the course of this laboratory a material specimen (see Fig. 1) is going to be gradually loaded in torsionuntil it failswhile collecting torq

Duke - BME - 100

BME 100L. Modeling Cellular and Molecular Systems Spring 2009 Tu Th 8:30 to 9:45AM in Hudson 125 This course is an introduction to mathematical modeling of kinetic processes in cellular and molecular systems. It also discusses applications of these m

Duke - BME - 100

Chemical EquilibriaBasic concepts Chemical potential Standard states Biochemist standard state Equilibrium constant Temperature dependence Calculation of system compositionSo far, we have dealt with pure substances, where G = G(T, P, n) Int

Duke - BME - 100

BME 100L. Modeling Cellular and Molecular SystemsDissociation of Double Stranded PolynucleotidesObjective: Inosinic acid (I) is a derivative of adenosine and pairs with cytidine (C), adenosine (A), and uridine (U). In this lab, we will determine

Duke - BME - 153

Duke - BME - 100

Simple kinetic analysis Enzyme kineticsSimple kinetics analysis (continued) Quasi steady state assumption (QSSA) Quasi equilibrium assumption Activation energy Transition state theory Enzyme kineticsIn chemical reactions, Steady state of a sub

Duke - BME - 100

Name:_ Useful Equations Partial derivatives: For function Z = Z(x, y), Z Z dZ = dy , dx + x y y x9 Z Z = y x y x x y x y First Law of Thermodynamics dE = dq + dw closed system H = E + PV dw = -PexdV Ho=(

Duke - BME - 100

Name_ Important Equations for Kinetics First Order Reactions-dCA = k 1CA dtCA = CAoexp(-k1t)t 1/ 2 =- ln0.5 k1dCA 1 1 1 2 = k 2C A t1 / 2 = - = k2 t k2 C A0 dt C A CA o dC Reaction between two different molecules - A = k 2 C AC B dt C If

Duke - BME - 100

Lab Report Guide BME 100L GUIDE TO WRITING LAB REPORTSGeneral Information Include ALL of the following: Abstract, Introduction, Materials and Methods, Results, Discussion, and References. Must be written individually Write in the third perso

Duke - BME - 100

Homework 1Jan. 25, 20071.a.At constant pressure, qp = H and H does not depend on paths. qp = H(20 C) = (100 C) + Cp ( )(100 - 20) + Cp (g)(20 - 100) = (100 C) + (Cp ( ) - Cp (g)(100 - 20) = 2257 + (4.18 - 1.874) 80 = 2441.48 kJb.Thiker = qp

Duke - BME - 100

Homework 2Feb. 1, 20072.a.efficiency = -w = 0.75 q1 q1 = 133.3 kJAlso, efficiency = 1 + q3 = 0.75 q1 q3 = -33.3 kJThe negative sign of w means the system does work on surroundings. The positive sign of q1 means heat enters into the system f

Duke - BME - 100

Homework 3Feb. 8, 20071.a.By using the dilute solution standard states, K= [G-1-P][ADP] [G][ATP] 10-4 = 770 -3 10 = 77.0where G-1-P and G refer to glycerol-1-phosphate and glycerol, respectively. G0 = -RT ln K = -8.314 298 ln 77 = -10.76 k

Duke - BME - 100

Homework 4Feb. 15, 200710.1 CO(g) + O2 (g) CO2 (g) 2 = 0.2 atm, PCO2 = 3 10-4 atmPO2 From Table A.5,G0 = G0 (CO2 ) - (G0 (CO) + G0 (O2 ) 298 298 298 f = -394.359 - (-137.168 + 0) = -257.191 kJ mol-1 -G0 -257.191 298 = = 103.808 RT 0.008314

Duke - BME - 100

7.a.d[B] = k1 [A] - k2 [B] - k3 [B][C] dt Assuming steady state of B,d[B] dt= 0 thenk1 [A] - k2 [B] - k3 [B][C] = 0 [B] = d[D] dt k1 [A] k2 + k3 [C]= k3 [B][C] = k1 k3 [A][C] k2 + k3 [C]b.By equilibrium assumption, [AB] [A][B]K = d[D] d

Duke - BME - 100

Homework 6Mar. 821.a.d[P ] = k[A][B] = 1 105 0.1 0.1 = 103 M/s dtb.d[P ] = k[A][B] = 1 105 1 10-4 1 10-6 = 10-5 M/s dtc.For [A]0 = [B]0 = 0.1 M, - d[A] = k[A]2 . Then, dt0.1-0.05-[A]0d[A] dtt= k0dt1 0.05 | = kt [A]

Duke - BME - 100

Ch8-5a.Michaelis-Menten equation shows that d[P ] k2 [E]0 [S] =- dt KM + [S] When [S] = 0.1 M and [E]0 = 1.0 10-5 M at 280 K, d[P ] 100 1 10-5 0.1 = = -1 10-3 Ms-1 . dt 1 10-4 + 0.1b.Ea = k2 RT T ln (see Example 7.5 in textbook) T -T k2 8.

Duke - BME - 100

Introduction to reaction kineticsLingchong YouThermodynamics is the study and understanding of whether a process can occur, regardless how long it takes. The 2nd law dSisolated > 0 (or dSsys + dSsurrondings >0) for spontaneous processes At cons

Duke - BME - 100

BME 100LModeling cellular and molecular systems IntroductionInstructor: Lingchong YouTAs: Stephen Payne & Pavel YarmolenkoWhat is a model?Webster definition: Main Entry: 1model Pronunciation: 'm-d&l Function: noun Etymology: Middle French mod

Duke - BME - 100

POPULATION CONTROL GENE CIRCUIT: WEEK ONE Safety: The Escherichia coli bacteria used in this lab have been modified to be non-virulent, but may induce an acute immune response in a very small percentage of individuals (this may occur whenever one is

Duke - BME - 100

POPULATION CONTROL GENE CIRCUIT: WEEK TWO Safety: The Escherichia coli bacteria used in this lab have been modified to be non-virulent, but may induce an acute immune response in a very small percentage of individuals (this may occur whenever one is

Duke - BME - 100

Review topics for BME100 Exam 2. Lingchong You General comments: Covers chapters 7, 8 (supplemented with notes) [Homework sets 5, 6, 7] The exam will be close book and close note. But major equations will be provided for your reference you probabl

Duke - BME - 100

Review topics for BME100 Exam 2. Lingchong You General comments: Covers chapters 7, 8 (supplemented with notes) [Homework sets 5, 6, 7] The exam will be close book and close note. But major equations will be provided for your reference you probabl

Duke - BME - 100

Survival in changing environmentsInput Temperature Light Nutrient pH Cell density .cellular networkOutput Gene expression Enzymatic activity Motility .Example: Signaling in animals 1) (a-c) cell-to-cell signaling by extracellular chemical

Duke - BME - 100

Common state variablesP, V, T, n, Cp, Cv, E, H = E+PV, S (dS =dqrev/T), G = H-TS, A = E-TS, i = (G/ni)T,P,njsurroundings systemPath-dependent variables: dw = -Pex dV; dq = C dT Reversible vs irreversible processes. Equations of state: 1st law:

Duke - BME - 100

The second law of thermodynamicsLingchong YouFrom 1st law to 2nd law E = q + w If I can somehow reduce the energy of a system by E, I can then extract work. I can also do this continuously through a cyclic process (for each cycle E = 0) For one

Duke - BME - 153

'XNH8QLYHUVLW\(GPXQG73UDWW-U6FKRRORI(QJLQHHULQJComplex NumbersMichael R. Gustafson II version 2.4 last reviewed: February 1, 20091IntroductionIn the field of Mathematics, people had to come up with some satisfactory way to deal with the pro

Duke - BME - 153

This exam is cumulative. Note, however, that specific questions about digital logic, diodes, and transformers will not appear on this final. The focus will be on the material presented in lectures 1-24. 1. Circuit elements 1. Know the voltage/current

Duke - BME - 153

BME 153L - Spring 2009Homework 1: Circuits and Circuit VariablesIntroductionThe problems for this week focus on basic circuit topology and variables.Homework Format GuidelinesPlease follow the guidelines below for homework solution presentatio

Duke - BME - 153

BME 153L - Spring 2009Homework 2: More Circuits and Circuit VariablesIntroductionThe problems for this week focus on circuit solution methods and measurement devices.Homework Format GuidelinesPlease follow the guidelines below for homework sol

Duke - BME - 153

BME 153L - Spring 2009Homework 3: Circuit Solution Techniques IIntroductionThe problems for this week focus on circuit solution methods - specifically the Node Voltage Method and the Mesh Current Method.AssignmentRemember, each Part needs to b

Duke - BME - 153

BME 153L - Spring 2009Homework 4: Circuit Solution Techniques IIIntroductionThe problems for this week focus on circuit solution methods - specifically superposition and equivalent circuits. There are also applications - specifically Wheatstone b

Duke - BME - 153

BME 153L - Spring 2009Homework 5: Maple Introduction5.1 IntroductionThis homework focuses using Maple to find both the symbolic and the numeric solutions to the linear algebra equations involved with solving electric circuits. Specifically, this

Duke - BME - 153

BME 153L - Spring 2009Homework 6: Reactive Elements and Complex NumbersIntroductionThe problems for this assignment focus on capacitors, inductors, complex numbers, and phasors.AssignmentRemember, each Part needs to be turned in separately, so

Duke - BME - 153

BME 153L - Spring 2009Homework 7: Phasors and AC Circuit AnalysisIntroductionThe problems for this assignment focus on using phasors to solve circuits.AssignmentRemember, each Part needs to be turned in separately, so when you turn in this par

Duke - BME - 153

BME 153L - Spring 2009Homework 8: Filters; Transient AnalysisIntroductionThe problems for this assignment focus on investigating the responses of reactive circuits, including the frequency response of filters and the transient response of RC and

Duke - BME - 153

BME 153L - Spring 2009Homework 9: More Transient Analysis; Introduction to AmplifiersIntroductionThe problems for this assignment focus on further investigating the responses of reactive circuits, including the transient response of RLC circuits.

Duke - BME - 153

BME 153L - Spring 2009Homework 10: Amplifiers; Active FiltersIntroductionThe problems for this assignment focus on operational amplifiers.AssignmentRemember, each Part needs to be turned in separately, so when you turn in this particular assig

Duke - BME - 153

BME 153L - Spring 2009Homework 11: Big Box of RandomIntroductionThe problems for this assignment focus on logic, transformers, and diodes.AssignmentRemember, each Part needs to be turned in separately, so when you turn in this particular assig

Duke - BME - 153

BME 153L.1 Spring 2009Test 1 Coverage(1) Basic electrical entities - be able to fill in the following chart: Name charge current work voltage power resistance conductance Variable q i w v p R G Units Coulombs (C) Amperes (A) Joules (J) Volts (V)

Duke - BME - 153

While the test is, by nature, cumulative, there will be certain aspects of Biomedical Electronic Measurements which form the core of this test. Specifically, topics from lectures 8-17. More specifically, topics including, but not limited to, 1. React