Physics 21
Fall, 2011
Equation Sheet
speed of lightin vacuo c 3.00× 108m/s Gravitational constant G 6.67× 10−11N m2/kg2 Avogadro’s Number NA 6.02× 1023mol−1 Boltzmann’s constant kB 1.38× 10−23J/K charge on electron e 1.60× 10−19C
free space permittivity 0 8.85× 10−12C2/(N m2) free space permeability µ0 4π× 10−7 T m/A gravitational acceleration g 9.807 m/s2
Planck’s constant h 6.626× 10−34J s Planck’s constant/(2π) ¯h = h/2π 1.055× 10−34J s electron rest mass me 9.11× 10−31kg proton rest mass mp 1.6726× 10−27kg neutron rest mass mn 1.6749× 10−27kg atomic mass unit u 1.6605× 10−27kg
1/(4π0) k 8.99× 109N m2/C2 F2 on 1=4π01 q1q2(r1− r2) |r1− r2|3 F = qE dE (at r) = 1 4π0 dQ (r − r) |r − r|3 E = −∇V =− ˆi∂V ∂x + ˆj ∂V ∂y + ˆk ∂V ∂z Vf− Vi=−f i E · dl V = 1 4π0 Q r; dV = 1 4π0 dQ |r − r| uelec= 120E2, umag= 1 2µ0B 2 Work =F · dl Eline= 1 2π0 λ r; Eplane= σ 20 E = Q 0A = σ 0 for plate capacitor Q = CV ; C = 0K A d= A d Ucap= 12CV2=12 Q2 C Uind= 12LI2 V = IR R = ρL A P = IV P = I2R R = mv⊥/(qB) circ. orbit A × B = ˆi ˆj kˆ Ax Ay Az Bx By Bz ξi=Ci(series) or Ri(parallel): 1 ξeffective = 1 ξ1 + 1 ξ2 ξi=Ci(parallel) or Ri(series): ξeffective= ξ1+ ξ2 XR= R, XL= ωL, XC = 1 ωC RC time constant = RC LR time constant = L/R Q(t) for RLC circuit Q0exp(−Rt/2L) cos ωt ω2= 1 LC − R2 4L2 F=qv × B; dF=Idl × B dB = µ0 4π Idl × (r − r) |r − r|3 long wire: B = µ0I 2πR center loop: B = µ0I/2R I =dQ dt I =−neAvd Vs Vp = Ns Np , Is Ip = Np Ns χm= µ µ0 − 1 τ = µ × B µ = IA solenoid B = µ0nI solenoid L = µ0N2A/l E · dA = Q 0 B · dA = 0 E · dl = −d dt B · dA B · dl = µ0I + µ00 d dt E · dA
sin(a± b) = sin a cos b ± cos a sin b sin(θ±π2) = sin θ cosπ2 ± cos θ sinπ2
=± cos θ
cos(a± b) = cos a cos b ∓ sin a sin b sin a + sin b = 2 cos
a− b 2 sin a + b 2 C = 2πr circumference of circle C = πd circumference of circle A = πr2 area of circle
A = 4πr2 surface area of sphere V =4 3πr 3 volume of sphere ax2+ bx + c = 0⇒ x = −b ± √ b2− 4ac 2a du √ a2+ u2 = ln u + a2+ u2 u du √ a2+ u2 = a2+ u2 du a2+ u2 = 1 atan −1u a u du a2+ u2 = 1 2ln a2+ u2 du (a2+ u2)3/2 = u a2√a2+ u2 u du (a2+ u2)3/2 =− 1 √ a2+ u2 eaudu = 1 ae au ln u du = u ln u− u undu = 1 n + 1u n+1 du a + bu = 1 bln(a + bu) du u = ln u 2π 0 cos2θ dθ = 2π 0 sin2θ dθ = π v = T /ρ (T =tension) v = (347.4 m/s) T /300 v = λf = ω/k ω = 2πf k = 2π/λ T = 1/f (T =period) P = 1 2ρA 2 ω2v ∂2D ∂x2 = 1 v2 ∂2D ∂t2 S = 1 µ0 (E × B) S =1 20cE 2 0= 12 c µ0 B20 =E0B0 2µ0 = ErmsBrms µ0 c = 1/√0µ0 E × B ∝ v (plane wave) ∆x∆p >∼¯h (¯h = h/2π) λ = h/p (de Broglie) −¯h2 2M ∂2ψ ∂x2 = i¯h ∂ψ ∂t KE = p2/(2M ) p = ¯hk E = ¯hω = hf eiθ = cos θ + i sin θ
Physics 21
Fall, 2011
Solution to HW-2
21-13 Three point charges are arranged on a line. Charge
q3 = +5.00 nC and is at the origin. Charge q2 =−3.00 nC and is at x2= 4.50 cm. Charge q1 is at x1= 1.00 cm. What is q1(magnitude and sign) if the net force on q3is zero?
x3=0 x1 x2
q3 q1 q2
We can work this problem without using vectors by think-ing it through. Since q2and q3have opposite sign, the force on q3 exerted by q2 is attractive (towards the right). If the total force on q3 is to be zero, the force exerted by q1 must be repulsive (toward the left). Thus q1and q3must have the same sign, and q1 must be positive.
We can find the magnitude of q1 by equating the magni-tude of the forces on q3 exerted by q1 and q2:
1 4π0 |q1q3| x21 = 1 4π0 |q2q3| x22
Cancelling like terms on both sides of the equation, we find
|q1| = x1 x2 2 |q2| = 1.0 cm 4.5 cm 2 (3 nC) = 0.148 nC.
We already concluded that q1 was positive.
A more general way to solve this problem is to use the vector expressions for the Coulomb force. We want
0 = F1 on 3+ F2 on 3, where 0 = 1 4π0 q1q3(r3− r1) |r3− r1|3 + 1 4π0 q2q3(r3− r1) |r3− r2|3 .
and we can evaluate the forces using the locations of the charges. Because q1 is at the origin, r3= 0. Also, r1 = x1ˆi and r2= x2ˆi. Substituting for the vectors gives
0 = 1 4π0 q1q3(0− x1)ˆi |0 − x1|3 + q2q3(0− x2)ˆi |x2|3 = −q3 4π0 q1x1 |x1|2 + q2x2 |x2|2 ˆi
Note that the terms in the denominators are lengths and must be positive. The quantity in brackets must be zero, so we obtain q1=−x2 x1 x1 x2 3q2=−4.5 1.0 1.04.52(−3.0 nC) = 0.148 nC This formula agrees with the previous result. Because we were careful with the signs, the formula gives the correct answer for any combination of signs of the three charges and for the two vector components x1and x2. (We took x3= 0.)
21-15 Three point charges are located on the positive x axis of a coordinate system. Charge q1 = 1.0 nC is 2.0 cm from the origin, charge q2 = −4.0 nC is 4.0 cm from the origin and charge q3= 6.0 nC is located at the origin. What is the net force (magnitude and direction) on charge q1 = 1.0 nC exerted by the other two charges?
This problem is very similar to 21-13, and the same dia-gram applies. Here we need the sum F of F2 on 1and F3 on 1, which is F = 1 4π0 q1q2(r1− r2) |r1− r2|3 + q1q3(r1− r3) |r1− r3|3 ,
where, as before, r3= 0, r1= x1ˆi, and r2= x2ˆi. Then F = 1 4π0q1 q2(x1− x2)ˆi |x1− x2|3 + q3(x1− x3)ˆi |x1− x3|3 = 1 4π0q1 q2(−0.02 m) (0.02 m)3 + q3(0.02 m) (0.02 m)3 ˆi Substituting the other numbers leads to
F = (9 × 109)(1 nC) −4 nC(−.02 m) (.02 m)3 + 6 nC(.02 m) (.02 m)3 ˆi = 2.25 × 10−4Nˆi
21-11 In an experiment in space, one proton is held fixed and another proton is released from rest a distance d away. What is the initial acceleration of the proton after it is re-leased?
From Physics 11 you know that F = ma, or a = F/m. So just find the electrostatic force on one proton due to the other proton, and then divide by the mass. We’ll drop the vector notation and just find the magnitude:
a = 1 4π0 e2 mpd2 =9× 109N m2/C2 1.602 × 10−19C 2 (1.67 × 10−26kg) d2
When you substitute for d, don’t forget to convert to meters. For d = 3 mm = 0.003 m, the result is
a = 1.54 × 104m/s2.
21-46 Two particles having charges q1 = 0.600 nC and
q2= 5.00 nC are separated by a distance of d = 1.60 m. At what point along the line connecting the two charges is the total electric field due to the two charges equal to zero?
0
x
d
q
1q
2Since both charges are positive, it’s easy to keep track of the direction of the electric field. The field at x from q1 points to the right, and the one from q2 points to the left. These two fields must be equal in magnitude for their vector sum to be zero. Therefore
1 4π0 q1 x2 = 1 4π0 q2 (x − d)2
Cancelling the common factor of 1/(4π0), we can rewrite the above equation as
(d − x)2 x2 = q2 q1 ⇒ d − x x = q2 q1 Solving for x, we find
x = d
1 + q2/q1
Substituting the specific numbers given above leads to
x = 0.412 m.
Note that instead of taking the square root and solving a linear equation for x, one could also set up a quadratic equation. One must identify the correct root of the quadratic equation, but the result is the same.
YF 21-50 mod A point charge q1 =−4.00 nC is at the point x = 0.60 m, y = 0.80 m, and a second point charge
q2 = +6.00 nC is at the point x = 0.60 m, y = 0. (a,b) Calculate the x and y components of the net electric field at the origin due to these two point charges. (c,d) Calculate the x and y components of the net electric field at the point
x = 0.90 m, y = 0.40 m due to these two point charges.
Use the vector expression given in class for the field E at r due to a charge Q at point r. Apply this formula to get the field at r due to Q1; apply it again to get the field at r due to Q2, and then add the results (superposition).
E (at r) = 1 4π0
Q (r − r)
|r − r|3
Remember, r = field point; r= charge point. y x (0.6, 0.8) (0.6, 0.0) Q1= -4.0 nC Q2= +6.0 nC (0.9, 0.4)
(a,b) Find E at origin, r = 0ˆi+0ˆj. For Q1, r= 0.6ˆi+0.8ˆj, so r − r=−0.6ˆi− 0.8ˆj and |r − r| = 1.0 m. For Q2, r= 0.6ˆi, so r − r=−0.6ˆi and |r − r| = 0.6 m. E = 1 4π0 −4 nC(−.6ˆi − .8ˆj)m (1.0 m)3 + 6 nC(−.6ˆi)m (.6 m)3 = 1 4π0 ˆi 2.4 13 − 3.6 (.6)3 + ˆj 3.2 13 nC m2 = 1 4π0 −14.3ˆi + 3.2ˆj nC m2 = −128.3ˆi + 28.77ˆjN/C
(c,d) Find E at point r = 0.9ˆi+0.4ˆj. For Q1, r= 0.6ˆi+0.8ˆj, so r−r= 0.3ˆi−0.4ˆj and |r − r| = 0.5 m. For Q2, r= 0.6ˆi, so r − r= 0.3ˆi + 0.4ˆj, and |r − r| = 0.5 m. E = 1 4π0 −4 nC(.3ˆi − .4ˆj)m (0.5 m)3 + 6 nC(.3ˆi + .4ˆj)m (0.5 m)3 = 1 4π0 ˆi(−1.2 + 1.8) + ˆj(1.6 + 2.4) 0.125 nC m2 = 1 4π0 4.8ˆi + 32ˆj nC m2 = 43.2ˆi + 287.7ˆj N/C
Physics 21
Fall, 2011
Solution to HW-3
21-96 Positive charge Q is uniformly distributed around a semicircle of radius a. Find the electric field (magnitude and direction) at the center of curvature P .
x y
dθ dQ = λds
= λadθ
We start with the equation from the equation sheet that gives the field at the field point r in terms of a charge dQ at the charge point r′:
dE(r) = 1 4πǫ0
dQ (r − r′) |r − r′|3
The field point r (where we want to know E) is at the origin, so r = 0, and the charge point (where dQ is located) is
r′= a cos θˆi + a sin θˆj,
where θ is the angle above the x-axis, so
r− r′ = −a cos θˆi − a sin θˆj, and |r − r′| = a. Since we want to add up all the contribution from all dQ, we must relate dQ to dθ so that we can integrate over the angle θ spanned by the semicircle. The arc length ds swept out by an angle dθ is adθ, so the charge dQ on ds is
dQ = λds = λa dθ,
where λ is the linear charge density (charge per unit length):
λ = Q πa.
With all these substitutions, the original equation becomes
dE = 1 4πǫ0
(πaQadθ)(−a cos θˆi − a sin θˆj) a3
= −Q
4π2ǫ0a2(cos θˆi + sin θˆj)dθ.
We can find the total field at the origin by integrating:
E= Z dE = Z π 0 −Q
4π2ǫ0a2(cos θˆi + sin θˆj) dθ.
We will look at each component of the integral over θ sepa-rately. We see that
Z π 0 cos θ dθ = 0 and Z π 0 sin θ dθ = 2, so we get E= − Q 4π2ǫ0a2(0ˆi + 2ˆj) = − Q 2π2ǫ0a2ˆj.
From this result we see the E has a magnitude of Q
2π2ǫ0a2
and points in the −y direction or downward. The x compo-nent of the field is zero, as one would expect from symmetry.
21-105 Three charges are placed as shown in the figure. The magnitude of q1is 2.00 µC, but its sign and the value of the charge q2are not known. Charge q3is +4.00 µC, and the net force F on q3 is entirely in the negative x-direction. a) Calculate the magnitude of q2. b) Determine the magnitude of the net force F on q3.
(a) We first determine the sign of the charge q1. We can do this by thinking about which direction the force will be in for the different combinations of signs for charges q1 and q2. Since there is no y-component of the force on q3 we know that q1 and q2 must have opposite signs. Since the force is directed in the negative x-direction we can infer that q1must be negative and q2 must be positive.
To determine q2we calculate the the total force F = F1 on 3+ F2 on 3on q3. Expressions for F1 on 3 and F2 on 3follow from the general expression for F2 on 1on the equation sheet: F1 on 3= 1 4πǫ0 q3q1(r3− r1) |r3− r1|3 , F2 on 3 1 4πǫ0 q3q2(r3− r2) |r3− r2|3 . We need the position vectors of each charge. Let q1 be at the origin. Then the position vectors r1 and r2of q1 and q2 are trivial. For r3, we notice that cos θ = 4/5 = x/4 where x is the x-component of the position vector of q3. Along with this and Pythagorean’s theorem we can find both x and y-components of r3. The result is
r1= 0, r2= 5 cmˆi, r3= 3.2 cmˆi + 2.4 cmˆj Knowing these position vectors we can write F1 on 3 and F2 on 3, in terms of the unknown charge q2:
F1 on 3= −36 Nˆi − 27 Nˆj
F2 on 3= (−24 × 106N/C) q2ˆi + (32 × 106N/C) q2ˆj Since the net force on q3 is in the negative x-direction, the sum of the y-components must be 0. From the sum of the y-components we find that the charge q2= +0.844 µC. The charge is positive, as expected by the reasoning above. (b) Using q2, we can now determine the net force on q3 by adding the components together. We already know that there is no net y-component, so the net force has only an x-component, which is the sum of the x components of F1 on 3 and F2 on 3. The magnitude of the total force is the absolute value of its x component, |F| = 56.26 N.
21-87 A proton with the mass m is projected into a uni-form electric field that points vertically upward and has mag-nitude E. The initial velocity of the proton has a magmag-nitude v0 and is directed at an angle α below the horizontal. (a) Find the maximum distance hmax that the proton descends vertically below its initial elevation. You can ignore gravita-tional forces. (b) After what horizontal distance d does the proton return to its original elevation? (c) Find the numer-ical value of hmaxif E = 520 N/C, v0= 5.00 × 104m/s, and α = 35.0◦. (d) Find the numerical value of d if E = 520 N/C, v0= 5.00 × 104m/s, and α = 35.0◦.
Because the proton is a charged particle (charge e), when it enters a region of uniform electric field, it experiences a constant force according to the relationship F = eE. In previous physics classes, you have studied how a constant force influences the motion of an object. In particular, recall that Newton’s Second Law tells us how force and acceleration are related through the equation F = ma. Combining these two equations allows us to calculate the acceleration:
a=eE m.
Since the electric field is uniform, the acceleration will be constant, and the proton will follow a parabolic trajectory, as shown in the figure.
The acceleration of the proton will be in the same direction as the electric field. Thus, there is zero acceleration along the x direction. This leads to the same type of problem as that studied in projectile motion near the surface of the earth. The only difference is that the acceleration is upward instead of downward. The equations we need to use are:
x direction y direction x = x0+ vx0t y = y0+ vy 0t +1 2at 2 vx= vx0= constant vy= vy 0+ at v2 y= vy02 + 2a∆y
Before we go any further, we must resolve the intial velocity vector into x and y components:
v0= vx0ˆi + vy0ˆj, where
vx0= v0cos α and vy0 = −v0sin α.
(a) When the proton reaches its maximum “height”, the y component of the velocity reaches 0 (see v1 in the figure). Substituting into v2
y = v2y0+ 2a∆y yields 0 = v02sin 2α + 2ay1, where y1is the position at maximum “height”. By the choice of coordinate system, this will be a negative position.
However, “height” is a scalar quantity, so we can solve for y1 and take the absolute value. We get an answer of
hmax=mv 2 0sin
2α 2eE .
(b) In order to calculate x2, the position when the proton comes back up to its original height, we need to find the time it takes to get there, t2. Because of the symmetry of the motion, t2 is just twice the time t1 needed to reach the “peak”. Using vy = vy0+ at, we substitute in vy = 0 at t1, use our expression for vy0, and solve for t1. Doubling this result gives us
t2= 2v0sin α
a .
Substitute this expression into x = x0+ vx0t, use our ex-pression for a, and set x0= 0 to get
x2= 2mv 2
0sin α cos α
eE .
This expression is also the answer for d, the distance travelled in the x direction, since d = x2− x0= x2− 0.
(c) and (d) Substituting in the given values, along with the fundamental constants e = 1.602 × 10−19C and m = 1.673 × 10−27kg (see the equation sheet), yields
hmax= 8.26 × 10−3m and d = 4.72 × 10−2m.
21-99 Two 1.20 m nonconducting wires meet at a right angle. One segment carries 2.00 µC of charge distributed uniformly along its length, and the other carries −2.00 µC distributed uniformly along it, as shown in the figure. Find (a) the magnitude and (b) the direction of the electric field these wires produce at point P , which is 60.0 cm from each wire. If an electron is released at P , what is (c) the magni-tude and (d) the direction of the net force that these wires exert on it?
(a) In class we worked out the vector electric field due to a finite line of charge at a field point located directly outward from the midpoint of the line. The magnitude of the field is
E = 1 2πǫ0 λ d a √ d2+ a2,
where d is the distance of the point from the line of charge, and a is half the length of the line. In this exercise, the two wires have the same length, the same magnitude of charge,
and are the same distance from the point P . Thus, we can adapt the formula above to the charge and spatial orientation of each wire by putting in the correct direction for each field. Recalling that λ = Q/L and a = L/2, we find
E = 1 4πǫ0 Q d 1 pd2+ (L/2)2 = (9 × 109Nm2/C2)2 × 10−6C 0.6m 1 p(0.6 m)2+ (0.6 m)2 = 35355N/C.
The field E− due to the negatively charged wire is directed to the left, while the field E+ due to the positively charged wire is straight down. We can use superposition to find the net electric field
Enet= E−+ E+ = −35355 N/Cˆi − 35355 N/Cˆj Enet=qE2 −+ E 2 += 5.00 × 10 4N/C
(b) Since the field vectors are perpendicular and of equal magnitude, the net electric field will be 45◦ from either vec-tor, and 135◦ counterclockwise from the +y axis.
(c) The force acting on a charge q is F = qE, so at P the magnitude of the force on the electron is
Fnet= eEnet= (1.60 × 10−19C)(5.00 × 104N/C) = 8.00 × 10−15N.
(d) Since the electron has a negative charge, the force will be directed opposite (180◦) from the electric field. Counter-clockwise from the +y axis, θ = 315◦.
21-104 A thin disk with a circular hole at its center, called an annulus, has inner radius R1 and outer radius R2. The disk has a uniform positive surface charge den-sity σ on its surface. (a) Determine the total electric charge on the annulus. (b) The annulus lies in the yz -plane, with its center at the origin. For an arbitrary point on the x -axis (the axis of the annulus), find the magnitude of the electric field E. Consider points above the annu-lus in the figure. (c) Find the direction of the electric field E. Consider points above the annulus in the figure.
(a) Given a uniform positive surface charge density σ, the total electric charge on any surface is σA, where A is the
surface area. For an annulus, A = πR2 2− πR
2
1. Mastering Physics is expecting a symbolic expression, so simply enter
σπ¡R2 2− R21
¢
(b) This problem is similar to Example 21.12 in the textbook. Our target is the electric field along a symmetry axis of a continuous charge distribution. We can represent the charge distribution as a collection of concentric rings of charge dQ. From lecture, we know the field of a single ring on its axis of symmetry, so all we have to do is add the contributions of the rings. For a single ring,
Ering= ˆi 1 4πǫ0
Qx (x2+ a2)3/2
where Q is the total charge on the ring, x is the distance along the axis from the ring to the point at which we are finding the field, and a is the radius of the ring. We will now assume an infinitesimally thin ring, with charge dQ on it, of radius r′, and thickness dr′. It will have a contribution to the electric field
dE = ˆi 1 4πǫ0
dQ x (x2+ r′2)3/2
Before we can integrate this, we need to come up with an expression that tells us how much charge is on our infinites-imally thin ring. We will approximate the area of the ring as circumference × thickness (this approximation works be-cause the ring is infinitesimally thin). So, dQ = σdA = σ2πr′dr′. We now have
dE = ˆiσx 2ǫ0
r′dr′ (x2+ r′2)3/2
To get the net electric field, we integrate this expression from r′= R1to r′= R2. E= ˆiσx 2ǫ0 Z R2 R1 r′dr′ (x2+ r′2)3/2 = ˆi σx 2ǫ0 · −√ 1 x2+ r′2 ¸R2 R1 = ˆi σ 2ǫ0 1 q 1 + (R1/x)2 −q 1 1 + (R2/x)2
Note that the integral used here is on the table of integrals on the equation sheet. Mastering Physics is only asking for the magnitude of the electric field here, so enter
σ 2ǫ0 1 q 1 + (R1/x)2 −q 1 1 + (R2/x)2
(c) Here we select the direction of the electric field that was found in part (b), the positive x -direction.
Physics 21
Fall, 2011
Solution to HW-4
22-7 The electric field due to an infinite line of charge is perpendicular to the line and has magnitude E = λ/2π0r.
Consider an imaginary cylinder with a radius of r = 0.200 m and length l = 0.465 m that has an infinite line of positive charge running along its axis. The charge per unit length on the line is λ = 7.15µC/ m. (a) What is the electric flux through the cylinder due to this infinite line of charge? (b) What is the flux through the cylinder if its radius is increased to r = 0.515 m? c) What is the flux through the cylinder if its length is increased to l = 0.760 m?
(a) Because the electric field lines are perpendicular to the line of charge, they will also be perpendicular to the surface of the curved or “barrel” part of the cylinder, and parallel to the ends of the cylinder. Because the line of charge is concentric with the cylinder, the field strength will also be constant along the barrel part of the cylinder. Then the electric flux can just be calculated by:
Φ = EAbarrel= λ
2π0r2πrl = 375, 700 Nm
2/C
Note that the r’s cancel; the flux is independent of the ra-dius of the cylinder. We can also use Gauss’ law to do the calculation and obtain the same expression.
Φ = Qenc
0 = λl
0 = 375, 700 Nm 2/C
(b) We already found in (a) that the flux is independent of
r; Gauss’ law also tells us that the flux through the cylinder
only depends on the charge enclosed. Because increasing the radius does not change how much charge is enclosed, the answer for part (b) should be the same as the answer for part (a).
Φ = 375, 700 Nm2/C
(c) If the length of the cylinder is increased, the new flux must be calculated because now more charge on the line is enclosed. Φ = Qenclosed 0 = λl 0 = 614, 000 Nm 2/C
22-10 A point charge q1= 4.00 nC is located on the x-axis at x = 2.00 m, and a second point charge q2 = −6.00 nC is on the y-axis at y = 1.00 m. What is the total electric flux due to these two point charges through a spherical sur-face centered at the origin and with radius (a) 0.500 m, (b) 1.50 m, (c) 2.50 m? q1 q2 Sa Sb Sc
The problem asks for total electric flux through a surface. Gauss’s Law helps here. The Equation Sheet gives Gauss’s Law in the form:
E · dA = Q
0,
where the left hand side is the total electric flux ΦEthrough a surface S, and Q is the charge enclosed by S.
(a) Surface Sa, whose radius is 0.500 m, encloses no charge. The total electric flux is zero (in any units).
(b) Surface Sb encloses q2 but not q1, so Q = q2:
ΦE= q2
0
= −6.00 × 10
−9C
8.85 × 10−12C2Nm−2 =−678 NC−1m2. Note the units: NC−1 is electric field and m2is area. (c) Surface Sc encloses both charges, so Q = q1+ q2:
ΦE= q1+ q2 0 =4.00 × 10 −9− 6.00 × 10−9 8.85 × 10−12 =−226 NC−1m2. September 12, 2011
22-20 The electric field 0.400m from a very long uniform line of charge is 840 N/C. How much charge is contatined in a 2.00-cm section of the line?
The equation for the electric field of an infinite wire is
E = 1
2π0
λ r,
where r is the perpendicular distance to the wire, and λ is charge per unit length. Solving for λ gives
λ = 2π0rE
= 2π × 8.8541 × 10−12× 0.4 × 840 = 1.86 × 10−8 C m If we muliply λ (in C/m) by the length (in m) we get charge:
λ × 0.02 = 3.74 × 10−10C
22-26 A conductor with an inner cavity, like that shown below, carries a total charge of Qcond= +4.80 nC. An addi-tional charge within the cavity, insulated from the conductor, is q = −6.20 nC. (a) How much charge is on the inner sur-face of the conductor? (b) How much charge is on the outer surface of the conductor?
(a) The Gaussian surface G shown includes the inner surface of the conductor. The electric field must be zero inside the conductor, makingGE · dA = 0, and thus the total electric charge Qencl enclosed by G must be zero. The charges con-tained within the surface are the charge in the cavity q and the charge on the conductor’s inner surface qinner, so
Qencl= 0 = qinner+ q ⇒ qinner=−q = 6.20 nC. (b) The total charge on the conductor Qcond is the sum of the charge on the inner surface qinner and the charge on the outer surface qouter. Hence
Qcond= 4.8 nC = qinner+ qouter
qouter= 4.8 nC − qinner= 4.8 nC − 6.20 nC = −1.4 nC
22-24 A point charge of−2.11 µC is located in the center of a spherical cavity of radius rcav = 6.50 cm inside an in-sulating spherical charged solid. The charge density in the solid is ρ = 7.36 × 10−4 C/m3. Calculate (a) the magnitude and (b) the direction of the electric field inside the solid at a distance r = 9.48 cm from the center of the cavity.
(a) To find the electric field we can use Gauss’ Law,
SE · dA =
Qencl
0 ,
where we choose the Gaussian surface S to be a sphere of radius r = 9.48 cm centered at the point charge. At every point on S the vector dA and E are perpendicular to S and are therefore parallel (or antiparallel – we’ll provisionally assume the former). By symmetry, the electric field has the same magnitude E everywhere on the surface, so we can evaluate the surface integral in terms of the unknown E,
SE · dA = SE dA = 4πr 2E = Qencl 0 .
All that remains is to determine the charge enclosed by the Gaussian surface, which consists of the point charge qpoint and the charge Qinsul on the insulator, which we can calcu-late as ρV . (Note that the charge on an insulator can be evenly distributed throughout its volume with density ρ.) The volume V of the insulator inside S is the volume of S less the volume of the cavity,
V = 4 3πr 3−4 3πr 3 cav= 4 3π(r 3− r3 cav) = 4 3π (.0948 m)3− (.0650 m)3= .00242 m3.
Thus the total charge enclosed is given by
Qencl= qpoint+ Qinsul= qpoint+ ρVencl
= − 2.11×10−6 C+(.000736 C/m3)(.00242 m3) = − 3.3 × 10−7 C.
Now we can use the relation between E and Qencl:
E = Qencl 4πr2 0 = −3.3 × 10 −7 C 4π(9.48 × 10−2 m)2(8.854 × 10−12 C2/(Nm2)) = − 3.3 × 10−5 N/C.
(b) Since the magnitude E must be positive, the minus sign we obtained for E above tells us that E and dA must be antiparallel, that is,E points into the surface.
Physics 21
Fall, 2011
Solution to HW-5
22-4 A cube has sides of length L = 0.330m. It is placed with one corner at the origin as shown in the figure. The electric field is not uniform but is given by
E= [−4.41 N/(C m)] xˆi + [3.29 N/(C m] z ˆk.
(a) Find the electric flux through each of the six cube faces S1, S2, S3, S4, S5, S6. (b) Find the total electric charge inside the cube.
(a) The electric flux is defined as Φ = R E · dA. Because dA is always perpendicular to the surface, we only need the component of the field that is normal to the surface of in-terest. (The dot product of dA with the parallel component of the field will be zero.) For each of the six surfaces, it will work out that the component normal to the surface is con-stant, so the surface integral will just be the constant value of the normal component times the surface area L2.
For surface S1, there is no component of the electric field normal to the surface (in the y direction ˆj), so the electric flux Φ1= 0.
For surface S2, the component of the electric field normal to the surface is in the +z direction, so take only the ˆk component of E. The integral over the surface is
Z L 0
Z L 0
3.29z dx dy = 3.29zL2.
To evaluate the electric field at the surface we set z = L, and then the flux is
Φ2= 3.29L3= 0.118 N m2/C
Surface S3 is similar to S1; there is no electric field compo-nent in the normal direction ˆj, so the flux Φ3= 0.
For surface S4, the normal direction is in the −z direction, so we need to take the negative of the ˆk component. The flux is that component integrated over the surface.
Φ4= − Z L 0 Z L 0 3.29z dx dy = −3.29zL2
But evaluating the electric field at z = 0 results in Φ4= 0.
For surface S5, the normal direction is in the +x direction, so we integrate the ˆi component over the surface.
Φ5= − Z L 0 Z L 0 4.41x dy dz = −4.41xL2
Evaluating this at x = L, gives the flux: Φ5= −4.41L3= 0.158 N m2/C.
Surface S6 is similar, but the surface is located at x = 0 so the flux evaluates to Φ6= 0.
(b) To find the total electric charge inside the cube, we can apply Gauss’ Law, which tells us that the total flux through this cube is equal to the charge enclosed, divided by ǫ0.
Φ1+ Φ2+ Φ3+ Φ4+ Φ5+ Φ6=Qencl ǫ0 Solving for Qencl and substituting the numbers gives
Qencl= (0 + 0.118 + 0 + 0 − 0.158 + 0)ǫ0 = −3.54 · 10−13C.
22-26 A conductor with an inner cavity, like that shown below, carries a total charge of qtot= +5.80 nC. The charge within the cavity, insulated from the conductor, is qcenter= −7.10 nC. (a) How much charge is on the inner surface of the conductor? (b) How much charge is on the outer surface of the conductor? Gaussian Surface qtot= +5.80 nC qouter qinner qcenter --7.10 nC
(a) The insulated point in the center of the cavity will in-duce an opposite charge on the inner surface of the conduc-tor. The induced charge will have the same magnitude as the center charge, qinner= +7.10 nC. We can prove this by placing a Gaussian surface within the conductor that en-compasses the entire cavity and point charge. Inside the conductor E = 0, and therefor Qencl= qcenter+ qinner= 0. (b) The total charge qtot on the conductor is a constant and does not change, but we must find how it is divided between the inner and the outer surface.
qtot= qinner+ qouter =⇒ qouter= qtot− qinner qouter= 5.80 nC − 7.10 nC = −1.30 nC
23-8 Three equal 1.20µC point charges are placed at the corners of an equilateral triangle whose sides are 0.500 m long. What is the potential energy of the system? (Take as zero the potential energy of the three charges when they are infinitely far apart.)
The total potential energy is obtained by summing up the potential energy of interaction between each pair of charges (superposition). If we label the three charges 1, 2, and 3 then
U = U12+ U13+ U23= 1 4πǫ0( q1q2 r12 + q1q3 r13 + q2q3 r23 ) Since all three charges are identical and the separation of any pair is the same, then
U = 1 4πǫ0 3q2 r = (9 × 10 9)3(1.2 × 10−6µC)2 0.500m = 0.078 J
23-5 A small metal sphere, carrying a net charge of q1= −2.90 µC, is held in a stationary position by insulating sup-ports. A second small metal sphere, with a net charge of q2 = −7.80 µC and mass 1.80 g, is projected toward q1. When the two spheres are ri= 0.800 m apart, q2 is moving toward q1 with speed vi = 22.0 m/s. Assume that the two spheres can be treated as point charges. You can ignore the force of gravity. (a) What is the speed of q2when the spheres are rf = 0.430 m apart? (b) How close does q2get to q1?
For this problem we will use the conservation of energy. We first notice that the two charges are both negative, so the Coulomb force is repulsive. Particle 2 must therefore start with positive kinetic energy KE, and as it approaches particle 1, KE will diminish and the potential energy U will increase by the same amount. We know
KE = 12mv2 and U = 1 4πǫ0
q1q2 |r1− r2|,
where |r1− r2| is the distance between the charges. We can evaluate the total energy Etot by substituting the given ini-tial values ri and vi:
Etot=12mv2i + 1 4πǫ0
q1q2
ri = 0.689841 J.
Since energy is conserved, Etot has the same value at any other value of rf: Etot= 12mv2f+ 1 4πǫ0 q1q2 rf . By equating the two expressions for Etot, we see
1 2mv 2 f+ 1 4πǫ0 q1q2 rf = 0.689841 J. (1)
For part (a) we are given rf, and we need to solve Eq. (1) for vf. The result is
vf = 15.5 m/s.
For part (b), we know that when q2 is closest to q1, the KE reaches its minimum value of zero, so we must set vf= 0 in Eq. (1) and solve for rf. The result is
rf = 0.295 m.
23-79 Electric charge is distributed uniformly along a thin rod of length a, with total charge Q. Take the potential to be zero at infinity. (a) Find the potential at the point P , a distance x to the right of the rod. (b) Find the potential at the point R, a distance y above the right-hand end of the rod. (c) In part a, what does your result reduce to as x becomes much larger than a? (d) In part b, what does your result reduce to as y becomes much larger than a?
We will express the potential dV at points P and R from charge dQ at the charge point r′= x′ˆi, and then we integrate over the length of the charged rod (from x′= −a to x′ = 0). The general formula is
dV = 1 4πǫ0
dQ |r − r′|.
The charge dQ can be related to the length dx′ using the linear charge density:
dQ = λdx′= Q adx
′.
(a) For point P , r = xˆi and r − r′= (x − x′)ˆi. Hence dV = 1 4πǫ0 Q a dx′ (x − x′) V = Z 0 −a dV = 1 4πǫ0 Q a Z 0 −a dx′ x − x′ = 1 4πǫ0 Q a ln x + a x (b) For point R, r = y ˆj and r − r′ = −x′ˆi + y ˆj. Hence
dV = 1 4πǫ0 Q a dx′ p(−x′)2+ y2 V = 1 4πǫ0 Q a Z 0 −a dx′ p(x′)2+ y2 = Q 4πǫ0aln y pa2+ y2− a Another expression for the argument of the log follows from
y
pa2+ y2− a =
pa2+ y2+ a y
(c,d) For x or y large compared to a, we can consider the whole charge Q on the line to be a point charge at the ori-gin. The potential is then 1/(4πǫ) times Q/x for c or times Q/y for d. Mathematically the result follows if we use the approximation ln(1 + δ) ≈ δ for δ ≪ 1. The key steps are
lnx + a x = ln(1 + a x) ≈ a x lnpa 2+ y2+ a y ≈ ln py2+ a y ≈ ln µ 1 + a y ¶ ≈ a y.
Physics 21
Fall, 2011
Solution to HW-6
23-9 A point charge q1 = 4.10 nC is placed at the origin,
and a second point charge q2 = −2.9 nC is placed on the x-axis at x = +21.0 cm. A third point charge q3 = 2.1 nC is to be placed on the x-axis between q1 and q2. (Take as zero the potential energy of the three charges when they are infinitely far apart.) a) What is the potential energy of the system of the three charges if q3 is placed at x = +11.0 cm? b) Where should q3 be placed to make the potential energy of the system equal to zero?
(a) The general formula for the potential energy between any two point charges labeled 1 and 2 is
U12= 1 4πǫ0
q1q2 |r1− r2|
,
where |r1− r2| is the distance between the two charges. To find the total potential energy of several charges, we must find the potential energy due to each pair of point charges. So the total potential energy here is
Utotal= U12+ U13+ U23 = 1 4πǫ0 µ q1q2 |0 − .21 m|+ q1q3 |0 − .11 m|+ q2q3 |.21 m − .11 m| ¶ = −3.53 × 10−7J
(b) We can say that q3 is placed at some point x between the other two charges and determine which value of x gives the system a total potential energy of zero. Incorporating this into the expression for the total potential energy gives:
Utotal= 1 4πǫ0 µ q1q2 .21 m+ q1q3 x + q2q3 (.21 m − x) ¶ = 0
We can save ourselves some trouble by eliminating several overall factors that won’t affect the solution x. Since all the terms have exactly the same units, we can drop the conver-sion factors to SI units and write distances in cm and charges in nC. The result is (4.1)(−2.9) 21 + (4.1)(2.1) x + (−2.9)(2.1) (21 − x) = 0
Multiply through by x(21 − x) to get a quadratic equation: 0.5662x2
− 26.59x + 180.81 = 0
Using the quadratic formula we find x = 38.72 or 8.25. Re-member that these values are in centimeters. Since the first solution is not between the two other point charges, we can ignore it. So q3 must be placed in between the two charges at x = 8.25 cm to give the system a total potential energy of zero.
23-31 (a) An electron is to be accelerated from a velocity
of 2.50 × 106
m/s to a velocity of 8.00 × 106
m/s. Through what potential difference must the electron pass to accom-plish this? (b) Through what potential difference must the electron pass if it is to be slowed from 8.00 × 106m/s to a halt?
This problem is a conservation of energy problem. The only forms of energy involved are electrical potential energy (U ) and kinetic energy (K). Note that “potential difference” means the “change in voltage” (V2−V1) from the initial value V1 to the final value V2, and that the relationship U = qV is not on your equation sheet.
(a) K1+ U1 = K2+ U2 1 2mv 2 1+ qV1= 1 2mv 2 2+ qV2 q(V2− V1) = −12m(v22− v 2 1)
The final equation shows that the change in potential energy is the negative of the change in kinetic energy, as expected. Dividing by the charge gives
V2− V1= − m 2q(v 2 2− v 2 1).
Using the velocities given and remembering that q = −e for an electron gives us the answer
V2− V1= 164 V.
Think carefully about the sign. A positively charged particle would slow down when the potential goes up (like a mass going uphill), but an electron behaves in the opposite way. (b) With the same equation we found for part (a), we can recalculate for the new velocities that
V2− V1= −182 V.
In this case, the particle slows down, so a positively charged particle would go through a positive change in potential, but again the electron does the opposite.
Note that on Mastering Physics, the question asks for the value of V1− V2, not V2− V1. So, the numerical answers you should give are −164 V and 182 V.
23-36 A very long insulating cylinder of charge of radius
3.00 cm carries a uniform linear density of 15.0 nC/m. If you put one probe of a voltmeter at the surface, how far from the surface must the other probe be placed so that the voltmeter reads 175 V?
The symmetry of such a uniformly charged cylinder is the same as the symmetry of a uniform line of charge. In fact, outside of the cylinder, the electric field is the same as though all the charges were concentrated on a line along its axis. (We can see this by applying Gauss’s law to a Gaussian surface like that used for the uniform line of charge.) This electric field is listed on the equation sheet as Eline= 1
2πǫ0 λ
r where λ is the linear charge density and r is the distance away from the line. The electric field lines are directed radially outward from the surface of the cylinder.
Now that we know what the electric field looks like, we can calculate the potential difference (the quantity measured by the voltmeter) by evaluating the line integral of the field between the probes of the voltmeter. We will evaluate
Vf− Vi= − Z f
i E· dl
along a path radially outward, beginning at the surface of the cylinder. This will make E and dl parallel, which simplifies the dot product in the integral. Let’s call the radius of the cylinder r0 and call the distance away from the cylinder’s surface that we are looking for d.
∆V = − Z f i Edl = − Z r=r0+d r=r0 1 2πǫ0 λ rdr = − λ 2πǫ0 Z r=r0+d r=r0 dr r = − λ 2πǫ0[ln (r0+ d) − ln r0] = −2πǫ0λ ln µ 1 + d r0 ¶
Be careful with the signs. The argument of the logarithm is greater than one, so the logarithm is positive. Thus ∆V is negative. That makes sense; the potential decreases as one goes further away from the positively charged cylinder. Now, we solve for d and evaluate.
−2πǫ0∆V λ = ln µ 1 + d r0 ¶ expµ −2πǫ0∆V λ ¶ = 1 + d r0 d = r0 · expµ −2πǫ0∆V λ ¶ − 1 ¸ = 2.74 cm Again, we used ∆V = −175 V.
23-47 In a certain region, the electric potential is
V (x, y, z) = Axy −Bx2+Cy, where A, B, and C are positive constants. Find Ex, Ey, and Ez. Where is the electric field zero (multiple choice)?
Ex= −∂V∂x = −Ay + 2Bx Ey = −∂V
∂y = −Ax − C Ez = −∂V∂z = 0
From the second equation, Ey = 0 when x = −C/A. Then (from the first equation) Ex= 0 when
−Ay + 2B(−C/A) = 0 ⇒ y = −2BC/A2. That analysis gives x and y; any value of z is OK.
23-66 A disk with radius R has a uniform charge density
σ. (a) By regarding the disk as a series of thin concentric rings, calculate the electric potential V at a point on the disk’s axis a distance x from the center of the disk. Assume that the potential is zero at infinity. (Hint: Use the result that the potential at a point on the ring axis at a distance x from the center of the ring is
V = 1 4πǫ0
Q √
x2+ a2
where Q is the charge of the ring.) (b) Find Ex= −∂V/∂x.
(a) Using the hint, we can modify the result for the finite ring to determine the potential dV due to a thin ring of radius r and charge dQ. We have
dV = 1 4πǫ0
dQ √
x2+ r2.
Because electric potential is a scalar quantity, we can inte-grate dV without keeping track of vectors. First, we must determine dQ in terms of known quantities. It is equal to the surface charge density times the surface area dA of the thin ring,
dQ = σdA = σ2πrdr,
where dA is the product of the length (circumference) 2πr and width dr of the ring of radius r.
Now it is possible to do the integration
V = Z dV = 1 4πǫ0 Z R 0 2πσr dr √ x2+ r2 = σ 2ǫ0 Z R 0 r dr √ x2+ r2 The integral is on the equation sheet:
Z udu
√
a2+ u2 = p
a2+ u2.
Evaluating it at the limits results in V = σ
2ǫ0 hp
x2+ R2− xi.
(b) The x component of the electric field is equal to −∂V/∂x:
−∂V ∂x = − σ 2ǫ0 · 1 2¡x 2 + R2¢−12 2x − 1 ¸ = σ 2ǫ0 · 1 − √ x x2+ R2 ¸ .
Notice that as R → ∞, this result reduces to the field of an infinite sheet.
Physics 21
Fall, 2011
Solution to HW-7
24-25 A 6.60 µF, parallel-plate, air capacitor has a plate
separation of 3.00 mm and is charged to a potential difference of 300 V. Calculate the energy density in the region between the plates.
Energy density is calculated using the formula u = 12ǫ0E2. In this problem, we know the voltage on the capacitor, not the electric field strength between the plates. But the volt-age and electric field strength are related. Because the elec-tric field between the plates is uniform and directed straight across the gap between the plates, it is easy to show that V = Ed where d is the separation between the plates. Do-ing the substitutions and evaluatDo-ing for the given values, we get u = 1 2ǫ0 µ V d ¶2 =1 2 µ 8.85 × 10−12 C 2 N · m2 ¶ µ 300 V .003 m ¶2 = 4.43 × 10−2J/m3.
24-38 A parallel-plate capacitor has capacitance C0 =
5.00 pF when there is air between the plates. The separation between the plates is 1.50 mm. (a) what is the maximum magnitude of charge Q that can be placed on each plate if the electric field in the region between the plates is not to exceed 3.00 × 104 V/m? (b) A dielectric with K = 2.70 is inserted between the plates of the capacitor, completely filling the volume between the plates. Now what is the max-imum magnitude of charge on each plate if the electric field between the plates is not to exceed 3.00 × 104 V/m? (a) Without the dielectric, Q = C0V and V = Ed. There-fore, the charge for the given values of C0, E, and d are
Q = C0Ed =¡5.00×10−12F¢ µ 3.00×104V m ¶ (.0015 m) = 2.25 × 10−10C = 225 pC
(b) With the dielectric, C increases to KC0, and V = Ed still holds. The charge is K times the previous value:
Q = KC0Ed = 2.7 ס2.25 × 10−10C¢ = 6.08 × 10−10C = 608 pC
24-59 In the figure, C1 = C5 = 8.3 µF and C2 = C3 =
C4 = 4.8 µF. The applied potential is Vab = 230 V. (a) What is the equivalent capacitance of the network between points a and b? (b) Calculate the charge on each capacitor and the potential difference across each capacitor.
(a) The reciprocal of the equivalent capacitance for capaci-tors in series is the sum of the reciprocals of each capacitor. The equivalent capacitance for capacitors in parallel is the sum of each capacitor. Using these rules we can break down the circuit by combining each capacitor until we are left with a single equivalent capacitance. C3 and C4 are in series so we can find the equivalent capacitance:
1 C3,4 = 1 C3 + 1 C4 =⇒ C3,4= C3C4 C3+ C4.
Substituting, we find C3,4 = 2.4 µF. Next we combine this with C2. The capacitors are in parallel so we find:
C2,3,4= C2+ C3,4= 7.2 µF.
Finally, we combine C1, C2,3,4, and C5 which are in series. 1 Ceq = 1 C1 + 1 C2,3,4+ 1 C5. We find Ceq= 2.63 µF.
(b) We know that for capacitors in series, the charge on the equivalent capacitor is the same as the charge on each individual capacitor. Also, for capacitors in parallel, the po-tential difference across the equivalent capacitor is the same as the potential difference across each individual capacitor. Keeping these rules in mind, we will determine the charge and potential difference across each capacitor by rebuilding circuit in the opposite way that we broke it down in part (a), calculating the charge and potential at each point. We can determine the total charge on Ceq since we know the potential across a and b.
Qtot= CeqVab= 6.1 × 10−4C = 610 µC.
Ceqis made up of C1, C2,3,4, and C5, which are in series, so this charge is the charge on each of these three capacitors, so we know Q1 = Q5 = 610 µC. We can find the potential difference by applying V = Q/C to find V1= V5= 73 V.
We can determine the potential across C2,3,4 the same way to find V2,3,4 = 84 V. Again, C2,3,4 is made up of C2 and C3,4, which are in parallel, so this is the potential difference across both capacitors. So we know V2 = V3,4 = 84 V, and applying Q = CV gives Q2= 400 µC and Q3,4= 200 µC.
C3,4 is made up of C3 and C4 in series so we know Q3,4 = Q3 = Q4 = 200 µC. By applying V = Q/C we find V3 = V4= 42 V.
Capacitor Charge Potential Difference
1 610 µC 73 V 2 400 µC 84 V 3 200 µC 42 V 4 200 µC 42 V 5 610 µC 73 V September 21, 2011
24-61 Three capacitors having capacitances of
8.0 µF, 8.3 µF, and 4.1 µF are connected in series across a 40 V potential difference. (a) What is the charge on the 4.1 µF capacitor? (b) What is the total energy stored in all three capacitors? (c) The capacitors are disconnected from the potential difference without allowing them to discharge. They are then reconnected in parallel with each other, with the positively charged plates connected together. What is the voltage across each capacitor in the parallel combination? (d) What is the total energy now stored in the capacitors?
(a) In a series connection the charges on the plates is as shown. We can determine the effective capacitance of the three capacitors and use this to determine the charge Q. The effective capacitance is found using
1 Ceff = 1 C1+ 1 C2 + 1 C3 = 1 8.0 µF + 1 8.3 µF + 1 4.1 µF Ceff = 2.04 µF Then Q = CeffV = (2.04 µF) × (40 V) = 8.2 × 10−5C. (b) The total energy stored in all the capacitors can be found by adding up the energies on each capacitor. Use the form of the equation that involves the charge and the capacitance:
Ucap=1 2 Q2 C1 + 1 2 Q2 C2 + 1 2 Q2 C3 = 1 2Q 2 µ 1 C1 + 1 C2 + 1 C3 ¶ =1 2 Q2 Ceff = 1 2 ¡8.2 × 10−5C¢2 2.04 × 10−6F = 1.6 mJ Note one gets the same answer using the form Ucap= 1
2CV 2, with Ceff and V = 40 V.
(c) When the capacitors are disconnected from the serial cir-cuit, the charge on each one remains the same (+Q). When they are reconnected in parallel, the total charge (3Q) is free to move around on the three positive plates shown in the dia-gram. After the charges equilibrate, the potential difference Va−Vbmust be the same across every one of the capacitors.
The parallel combination is equivalent to a single effective capacitance Ceff = C1+ C2+ C2with the same total charge 3Q. The voltage V = Va−Vb is then given by
V = Qtot Ceff = 3Q C1+ C2+ C3 = 3¡8.2 × 10−5C¢ (8.0 + 8.3 + 4.1) µF= 12 V One could determine the charges on each capacitor with this result, using Qi= CiV .
(d) As in part (b), the total energy stored in all the capaci-tors can be found by adding up the energies on each capac-itor. This time use the form of the equation that involves the voltage and the capacitance:
Ucap=1 2C1V 2+1 2C2V 2+1 2C3V 2= 1 2(C1+C2+C3) V 2 =1 2CeffV 2=1 2(2.04 × 10−5F)(12 V) 2= 1.5 mJ.
24-65 A parallel-plate capacitor with only air between the
plates is charged by connecting it to a battery. The capac-itor is then disconnected from the battery, without any of the charge leaving the plates. (a) A voltmeter reads 41.0 V when placed across the capacitor. When a dielectric is in-serted between the plates, completely filling the space, the voltmeter reads 11.5 V. What is the dielectric constant of this material? (b) What will the voltmeter read if the di-electric is now pulled partway out so it fills only one third of the space between the plates?
(a) With no dielectric, Q = C0V0, where V0 has been mea-sured. With the dielectric, the charge does not change, but a new V is measured, so Q = KC0V . Equating both expres-sions for Q,
C0V0= KC0V ⇒ K =V0 V =
41
11.5 = 3.57 V (b) This situation is essentialy two capacitors in parallel:
(2/3) C0 (1/3) KC0
Ceff =2 3C0+
1 3KC0 Therefore the measured voltage will be
V = Q Ceff = C0V0 2 3C0+ 1 3KC0 = 2 V0 3+ 1 3K = 3V0 K + 2 = 22.1 V
Physics 21
Fall, 2011
Solution to HW-8
25-44 If a “75 W” bulb (75 W are dissipated when
con-nected across 120V) is concon-nected across a 220 V potential difference (as is used in Europe), how much power does it dissipate?
(a) The power dissipation P and potential difference V across the bulb are variables, but the resistance R across the light bulb is a physical constant whether you are in America or Europe. From the equation sheet we know that P = IV and V = IR. We can combine these formulas to eliminate the current I, which we don’t know, and get a relation be-tween power, voltage, and resistance:
P = IV = V RV = V2 R ⇒R = V2 P .
Using the US values P = 75 W and V = 120 V, we find
R = (120 V) 2
75 W = 192 Ω.
We can then use this resistance to solve for the power dis-sipated when the bulb is connected to the new potential difference: P = V 2 R = (220 V)2 192 Ω = 252 W.
25-46 A battery-powered global positioning system (GPS)
receiver operating on a voltage of 9.1 V draws a current of 0.20 A. a) How much electrical energy does it consume during a time of 2.0 h?
(a) We can calculate the energy use over a period of time if we know the rate that the device uses electrical energy (power). The power can be determined from:
P = IV = 9.1 V × 0.2 A = 1.82 J/s = 1.82 W The total energy used over a time of two hours is:
Etot = P t = 1.82 J/s × 7200 s = 1.31 × 104J
25-70 A person with body resistance between his hands
of 10 kΩ accidentally grasps the terminals of a 14 kV power supply. (a) If the internal resistance of the power supply is 2000 Ω, what is the current through the person’s body? (b) What is the power dissipated in his body? (c) If the power supply is to be made safe by increasing its internal resistance, what should the internal resistance be for the maximum current in the above situation to be 1.00 mA or less?
Rint V Rperson
Power supply
(a) We can model the power supply as a battery with V = 14 kV in series with an internal resistor of Rint. When the person holds the terminals of the power supply (don’t do this at home, kids), he completes the circuit shown in the diagram. The total resistance of resistors in series is the sum
Rtot= Rint+ Rperson,
and the current going through them will be the same as the current going through the equivalent Rtot. This current is
I = V Rtot = V Rint+ Rperson = 14 kV 2 kΩ + 10 kΩ = 1.167 A. (b) The power dissipated in a device with current running through it is given by:
P = IV.
We already found the current going through the person in part (a). However the potential V here is not the same potential as the battery. It is the potential across just the person’s body, which we can find using V = IRperson:
P = IV = I2Rperson = (1.167 A)210 kΩ = 13.6 kW. (c) To make the battery safe we need to increase the internal resistance to Rsafe
int . From the second equation in part (a) we can see that
Isafe= V
Rsafe
int + Rperson .
Solving this equation for Rsafe
int , we have Rsafe int = V 0.001 A−Rperson= 14 kV 0.001 A−10 kΩ = 13.99 MΩ. September 23, 2011
26-8 Three resistors having resistances of R1 = 1.60 Ω,
R2 = 2.90 Ω, and R3 = 4.60 Ω are connected in parallel to a 26.0 V battery that has negligible internal resistance. (a) Find the equivalent resistance of the combination. (b,c,d) Find the current in through each resistor. (e) Find the total current through the battery. (f,g,h) Find the voltage across each resistor. (i,j,k) Find the power dissipated through each resistor. (l) Which resistor dissipates the most power: the one with the greatest resistance or the least resistance?
(a) The equivalent resistance for resistors in parallel is given by the sum of their reciprocals,
1 Req = 1 R1 + 1 R2 + 1 R3. Thus, Req= 0.842 Ω.
(b,c,d) The voltage is the same across parallel resistors, therefore V = IR ⇒ I = V /R. Thus, I1= V R1 = 16.3 A; I2= V R2 = 8.97 A; I3= V R3 = 5.65 A. (e) The current from the battery is the sum of the currents through each resistor:
Itotal= I1+ I2+ I3= 30.9 A One could also use the equivalent resistance,
Itotal= V /Req= 26 V/(0.842 Ω) = 30.9 A. (f,g,h) The voltage is the same across resistors in parallel.
V1= V2= V3= 26.0 V
(i,j,k) Any formulation of P = V I = I2R = V2/R will work, P1= 423 W, P2= 233 W, P3= 147 W.
(l) P = V2/R. Since the voltage is the same across all the resistors, the power is greatest in the smallest resistor.
OE 41-1 For this problem, you must write and then solve
the loop and node equations needed to find the currents I1, I2, and I3shown in the figure. (a) In the circuit shown in the figure, what is the value of the current I1? Remember that I1may be positive or negative. (b) What is the value of the current I2? Remember that I2 may be positive or negative. (c) What is the value of the current I3? Remember that I3 may be positive or negative.
1Ω 4Ω 2Ω 2Ω I1 I3 I2 1 V 8 V 9 V 1 2
(a) Write the loop and node equations needed to determine the currents I1, I2, and I3in the circuit shown. Indicate clearly the loop used to determine each loop equation.
node: I2= I1+ I3
loop 1: 9 − I1−2I1−4I2= 0 loop 2: 8 − 2I3−1 − 4I2= 0
(b) Determine the currents by explicit solution of the equa-tions. You must show your work.
Rearrange loop 1 and loop 2 equations: loop 1: 3I1+ 4I2= 9
loop 2: 4I2+ 2I3= 7
Use the node equation to eliminate I3= I2−I1:
3I1+ 4I2= 9 (1)
−2I1+ 6I2= 7 (2)
Multiply (1) by 2 and (2) by 3; add and solve for I2. Sub-stitute back for I1then I3. Results are
Physics 21
Fall, 2011
Solution to HW-9
26-27 In the circuit shown in the figure the batteries have
negligible internal resistance and the meters are both ideal-ized. With the switch S open, the voltmeter reads 15 V. (a) Find the emf E of the battery. (b) What will the ammeter read when the switch is closed?
(a) For this part we can ignore the battery with the switch, since the switch is open and there will be no current through it. So let’s draw the circuit that we are concerned with, showing also the currents in each branch and the loops we will use to write Kirchhoff’s equations:
We can find I3 since we were given the measured voltage across the 50 Ω resistor. Using Ohm’s Law,
V = IR =⇒ I3=V R =
15 V
50 Ω = 0.30 A. Now, let’s write out the loop and node equations:
Node : I1= I2+ I3
Left loop : E −(20 Ω)I1−(75 Ω)I2= 0
Right loop : E − (20 Ω)I1−(30 Ω)I3−(50 Ω)I3= 0 Since we already know I3, there are three unknowns in this system, E, I1, and I2. Here are the equations after substi-tuting for I3 and combining some terms:
Node : I1= I2+ 0.3 Left loop : E −20I1−75I2= 0 Right loop : E − 20I1−24 = 0
Now we have three equations for three unknowns. Next we simplify the equations and solve for E. The solution is
E= 36.4 V, I1= 0.62 A, I2= 0.32 A.
(b) The circuit is different for this part and is shown below. Conveniently, to find the new current in the ammeter, we only need to consider the one loop shown.
Set up the loop equation for this new loop: 25 V − (50 Ω)I = 0 =⇒ I = 25 V
50 Ω = 0.50 A
26-28 In the circuit shown in the figure both batteries have
insignificant internal resistance and the idealized ammeter reads 1.60 A in the direction shown. (a) Find the emf E of the battery. (b) Is the polarity shown correct?
(a) To solve this problem, we will use Kirchhoff’s rules, but this time instead of solving for three currents, we will solve for two currents and the emf. We assign currents I1and I2to the remaining branches and loops as shown in the diagram:
The node equation is
I2= I1+ 1.60 A. The left loop equation is
0 = 48.0 Ω(I2) + 12 Ω(1.6 A)− 75.0 V = 48.0 Ω(I2) − 55.8 V,
and the right loop equation is
0 = −15.0 Ω(I1) + E − 48.0 Ω(I2) We can solve the left loop equation directly for I2:
I2= 55.8 V
48.0 Ω = 1.1625 A.
Now substitute this I2into the node equation to solve for I1: I1= I2−1.60 A = −.4375 A
Substituting the currents found above into the right loop equations gives
0 = − 15.0 Ω(−.4375 A) + E − 48.0 Ω(1.1625 A) E= 15.0 Ω(−.4375 A) + 48.0 Ω(1.1625 A) = 49.24 V. (b) The polarity of the battery as shown is correct, because the E we calculated was positive. If the calculated E had been negative, it would imply that the assumed polarity in the drawing was incorrect.
26-40 A 12.8µF capacitor is connected through a 0.890
MΩ resistor to a constant potential difference of 60.0 V. (a) Compute the charge on the capacitor at the following times after the connections are made: 0 s, 5.0 s, 10.0 s, 20.0 s, and 100.0 s. (b) Compute the charging currents at the same instants.
(a) As derived, the formula for the charge on a charging capacitor as a function of time is:
q(t) = Qf³1 − e−t/RC´,
where the final charge Qf = CV = 7.68 × 10−4
C. The time constant τ = RC = 11.392 s. The table below gives q(t) at the times specified.
(b) The relationship between charge and current is i = dq/dt, so we can determine the current as a function of time by differentiating the expression for q(t) above:
i(t) = dq dt =
Qf RCe
−t/RC = I0e−t/RC
where we substituted Qf = CV and noted that the initial current I0that flows is the battery voltage V divided by the resistance R. I0 is 6.74 × 10−5 A.
Here is a table of the charge and current at various times. Note that 100 s is about nine times the time constant, and at that point the capacitor is essentially fully charged, and the current from the battery is essentially zero.
t (s) q(t) (C) i(t) (A) 0 0 6.74 × 10−5 5 2.73 × 10−4 4.35 × 10−5 10 4.49 × 10−4 2.80 × 10−5 20 6.35 × 10−4 1.16 × 10−5 100 7.68 × 10−4 1.00 × 10−8
26-43 An emf source with a magnitude of E = 120 V, a
resistor with a resistance of R = 87.0 Ω, and a capacitor with a capacitance of C = 3.90 µF are connected in series. As the capacitor charges, when the current in the resistor is 0.700 A, what is the magnitude of the charge on each plate of the capacitor?
(a) The simplest way is to apply the loop equation. Let q(t) and i(t) be the instantaneous charge on the capacitor and current in the circuit. Then
E −VR−VC= 0 E −i(t)R −q(t)
C = 0.
Now we can solve for the charge on the capacitor as a func-tion of the current:
q(t) = C[E − i(t)R] = 3.90 µCh120 V − (0.7 A)(87 Ω)i = 230 µC.
26-48 In the circuit shown below, C = 5.90 µF, E = 28.0 V,
and the emf has negligible resistance. Initially the capacitor is uncharged and the switch S is in position 1. The switch is then moved to position 2, so the capacitor begins to charge. (a) What will be the charge on the capacitor a long time after the switch is moved to position 2? (b) After the switch has been moved to position 2 for 3.00 ms, the charge on the capacitor is measured to be 110 µC. What is the value of the resistance R? (c) How long after the switch is moved to position 2 will the charge on the capacitor be equal to 99.0% of the final value found in part (a)?
(a) We know that after a long time the circuit will approach a steady state where the charge on the capacitor will be simply given by
Q = CE.
Substituting C = 5.90 µF and E = 28.0 V we find the charge on the capacitor after a long time will be Q = 165.2 µC. (b) We know that the resistance R is part of the time con-stant in the function q(t). We found q(t) by solving the differential equation obtained from Kirchoff’s loop equation.
q(t) = CE(1 − e−RCt )
Since we know q(t = 3 s) we can rearrange this equation to solve for R.
R = −t C
1 ln(1 −CEq )
We insert our value for q(t = 3 s) = 110 µC and find the resistance R = 464 Ω.
(c) We want to find at what time t will the charge on the capacitor be 99% of its final value which we found in part (a). In other words we want to solve for t when q = 0.99 × CE. We rearrange our equation for q(t) to get:
t = −RC ln(1 − q CE). Substituting our values of R, C, and q we find:
t = −(464 Ω)(5.9 × 10−6
Physics 21
Fall, 2011
Solution to HW-10
27-3 In a 1.35 T magnetic field directed vertically upward, a particle having a charge of magnitude 8.90 µC and initially moving northward at 4.72 km/s is deflected toward the east. (a) What is the sign of the charge of this particle? (b) Find the magnetic force on the particle.
The magnetic force on a charged particle moving in a mag-netic field is given by the equation
F= qv × B
Since, in this case, F, v, and B are mutually perpendicular, the magnitude of F is simply given by F = qvB, with the direction determined by the right hand rule.
(a) We need to apply the right hand rule to see if the di-rection of the force is consistent with a positive charge or a negative charge. Imagine you are seated so that north is in front of you. The other directions are then determined: east is to the right, south is behind you, and west is to the left. So northward velocity means the particle is moving forward. Point the fingers of your right hand straight forward. The magnetic field is upward, so curl the fingers of your right hand upward. In order to do this, your palm must be facing upward. Then, the thumb of your right hand is pointing to the right (eastward). Eastward is the direction the particle is deflected. Thus, the particle must have a positive charge. Here is a diagram: N E S W B v F (b) F = qvB = (8.90 µC)(4.72 km/s)(1.35 T) = 0.0567 N
27-4 A particle with mass m = 1.81×10−3 kg and a charge of q = 1.22 × 10−8 C has, at a given instant, a velocity v = (3.00 × 104
m/s)ˆj. (a) What is the magnitude of the particle’s acceleration produced by a uniform magnetic field B = (1.63 T)ˆi + (0.980 T)ˆj? (b) What is the direction of the particle’s acceleration?
To determine the acceleration of the particle we need to know what force is acting on it. We can assume the only force is due to the magnetic field. The force on a charged particle moving in a magnetic field is given by:
F= q (v × B)
For the cross product, we notice that many components of vand B are zero:
v= vyˆj and B = Bxˆi + Byˆj.
We can evaluate the cross product v × B by using the cross product of each pair of unit vectors:
ˆj ׈i = −ˆk ˆj × ˆj = 0 Then
F= qv × B = qhvyˆj ׳Bxˆi + Byˆj´i= −qvyBxkˆ Substituting the given quantities we get:
F= −(1.22 × 10−8C)(3.00 × 104m/s)(1.63 T) = −5.97 × 10−4N ˆk
Using Newton’s second law, we can find the acceleration of the particle:
a= F/m = −0.330 m/s ˆk.
