 Phyllis Fleming Physics Physics 108
 Answers - Potential Energy, Potential, and Capacitors
 1 The distances from Q1 and Q2 to P are r = (y2 + a2)1/2.  Since V is a scalar quantity, the potentials add algebraically. V = k(Q1/(y2 + a2)1/2 + Q2/(y2 + a2)1/2) = k(Q1 + Q2)/(y2 + a2)1/2. Since a is a constant, V is a function only of y. For y2 >> a2,  V = kQ/y, where Q = Q1 + Q2 which is the same as the potential due to a point charge Q located at the origin.
 2 The potential V is a scalar quantity. In this case you may use a minus sign to calculate the potential. For a point charge Q, the potential at a point P a distance r from P is V = (9 x 109 N-m2/C2)Q/r. For q1,  V1 = (9 x 109 N-m2/C2)(9 x 10-9 C)/3 m = 27 N-m/C = 27 J/C = 27 V. For q2,  V2 = (9 x 109 N-m2/C2)(-1 x 10-9 C)/1 m = -9 V. V at P = (27 - 9) V = 18 V. The potential at P is really the potential difference between point P and a very large distance from P. At an infinite distance, the potential due to q1 and q2 is zero (r = a very large number),  V∞ = 0. The work done to bring a charge q3 from a very great distance to P is the increase of the potential energy of the system. The potential difference VP∞ = the potential energy difference divided by q3 = (UP - U∞)/q3. Work done = q3VP∞ = q(VP - V∞)= ( 3 x 10-9 C)(18 J/C - 0) = 54 nJ. For q2 = +1 nC, V1 = (9 x 109 N-m2/C2)(9 x 10-9 C)/3 m = 27 N-m/C = 27 J/C = 27 V. V2 = (9 x 109 N-m2/C2)(1 x 10-9 C)/1 m = 9 V. V at P = (27 + 9).   V = 36 V. Now the work done = q3(VP - V∞) = ( 3 x 10-9 C){(36 J/C) - 0} = 108 nJ.
 3 In general, the potential due to a point charge Q at a distance r is kQ/r. For q1 = q2 = -200 x 10-6 C, the potential at P is: V = 9 x 109 N-m2/C2(-2 x 10-4 C)/√2 m    = -12.73 x 105 N-m/C    = -12.73 x 105 J/C    = -12.73 x 105 V. For q3 = q4 = 100 x 10-6 C, the potential at P = V = 9 x 109 N-m2/C2(10-4 C)/√2 m = 6.36 x 105 V. Due to all the charges, V = 2(-12.73 + 6.36)105 V    = -12.7 x 105 V. Work to bring q5 = 20 x 10-6 C from infinity = q5(VP – V∞) = 20 x 10-6C(-12.7 - 0)105 V = -25.4 J.
 4 The parallel plate capacitor with its charge and the particle with the forces acting on it are shown in the figure above.  For the particle to be at rest, the net force acting on it must equal zero:  Fnet = ma = m(0).  Taking up to be positive, Fe - mg = 0  or Fe = mg, where the electric force on the particle with charge q in an electric field E is Fe = Eq.  Thus, Eq = mg  or E = mg/q = (10-3 kg)(9.8 m/s2)/(10-6 C) = 9.8 x 103 N/C. The direction of the electric field is up.  Remember the direction of the electric field is that in which a positive charge is urged. Vab = Ed = 9.8 x 103 N/C (10-3 m) = 9.8 J/C = 9.8 V.
 5 This is a problem in conservation of energy. The potential of a sphere of charge equals the potential of a point charge = kQ/r. When the alpha particle is a long distance away from the gold nucleus initially i, the potential energy of the system Ui is 0. When the alpha particle touches the gold nucleus, r = 7.0 x 10-15 m = the radius of the gold nucleus.  The potential energy for the "final" position Uf equals: (9 x 109 N-m2/C2)(79)(2) x (1.6 x 10-19 C)2/(7.0 x 10-15 m) = 5.2 x 10-12 J. From conservation of energy, Ui   +   Ki   =        Uf            +   Kf  0   +   Ki   = 5.2 x 10-12 J  +    0
 6 Ui   +        Ki       =   Uf  +   Kf   or   Ui - Uf = q(Vi - Vf) = 1/2 mvf2 - 1/2 mvi2 1.6 x 10-19 C(100 J/C) = 1/2(1.67 x 10-27 kg)vf2 - 0 vf = (2 x 1.6 x 10-17 J/1.67 x 10-27 kg)1/2 = 1.38 x 105 m/s

7.    300h V/m - 0.005h2 V/m2,  where h is the height above the surface.

 8 dA = 2 πr dr dq = σ dA = 2πσrdr dV = kdq/r = 2πσkrdr/(z2 + r2)1/2  9 VA = 9.0 x 109 N-m2/C2[80/0.10 - 60/0.10]10-9C/m =1.8 x 103 V VB = 9.0 x 109 N-m2/C2[80/0.16 - 60/0.12]C/m = 0 WB A = q(VA - VB) = (10 x 10-6 C)(1.8 x 103 V) = 0.018 J.

10. In the field of these charged wires, draw the circular equipotential traces passing through C and D. The electric field due to the wire through A or B = λ/2πεor, where r is the distance from the wire to the field points. The direction of the field is radially away from the wire.

First consider the wire passing though A: In the field of both wires, the potential difference is twice as large,  or

λ/2πεo(2) = λ/πε

 11 When capacitors are wired in parallel, the equivalent capacitance C = C1 + C2 + C3 But C1 = C2 = C3 = εoA/d so  C = 3 εoA/d = εoA/(d/3) The plate spacing for the single capacitor must be d/3. When capacitors are wired in series, the reciprocal of the equivalent capacitance 1/C = 1/C1 + 1/C2 + 1/C3 = 3/C1   and C = C1/3 = ( εoA/d)/3 = εoA/3d so the plate spacing for the single capacitor must be 3d.
 12 For any capacitance, including C1, C1 = Q/(Vab)i, where Q is the charge on the capacitor when (Vab)i  is the potential difference across it. When the two capacitors are wired in parallel, the potential difference across each is the same = (Vab)f = q1/C1 = q2/C2, where q1 and q2 are the charges on C1 and C2 with potential difference (Vab)f. Also from conservation of charge q1 + q2 = Q,   or q2 = Q - q1 = C1 (Vab)i - C1 (Vab)f = C1{(Vab)i - (Vab)f}. But q2 also equals C2 (Vab)f. Thus, C1{(Vab)i -  (Vab)f} = C2(Vab)f,   or        since C1 = 4πεoR1          and C2 = 4πεoR2, 4πεoR1 {(Vab)i - (Vab)f} = 4πεoR2(Vab)f,   so R2 = R1{(Vab)i - (Vab)f}/(Vab)f.
 13 The capacitance of a parallel plate capacitor in a vacuum  C = εoA/d, where A is the area of the plates and d is the distance between them. When the plate separation is doubled: The capacitance C is halved since the capacitance is inversely proportional to the distance d. C = Q/Vab  or  Q = CVab.  When C is halved and Vab remains the same (the battery is still across the capacitor), the charge Q is halved. E = σ/ εo.  When Q is halved, the charge per unit area σ is halved and E must be halved.  Also E = Vab/d  so when d is doubled E is halved. When a dielectric with κ= 2 is inserted: The capacitance = κεoA/d  so the capacitance is doubled. Q = CVab.  When the capacitance doubles, with constant Vab,  the charge doubles. E = σ/κεo.  When Q doubles, σ doubles, but σ/κ remains the same because κ = 2 and E remains the same.  Also E = Vab/d  so E is the same because Vab and d remain the same.
 14 The electric field in the air gaps is - Eoj = - σ/ εo j= - Q/A εo j. The electric field in the dielectric is - Ej = - Q/Aκεo j. Vab = - a∫b E . dyj = (Q/A εo)(d1 + t/κ + d2). C = Q/Vab = A εo/(d1 + t/κ + d2). Alternatively, we may view the capacitor as three capacitors in series, shown in the lower portion of the figure above. For series 1/Ceq = 1/C1 + 1/C2 + 1/C3 = d1/A εo + t/Aκεo + d2/A εo.                    Ceq = A εo/(d1 + t/κ + d2).
 15 We find the equivalent capacitance by starting with the parallel combination between D and B.  Capacitors in parallel add. CDB = C3 + C4 = (3.0 + 1.0)µF = 4.0 µF   (Fig. 4b ). CDB and C2 are in series. The reciprocal of the equivalent capacitance equals the sum of the reciprocals of the capacitors in series. 1/CA’B’ = 1/C2 + 1/CDB = 1/12 µF + 1/4.0 µF = (1 + 3)/12 µF  or CA’B’ = 3.0 µF  (Fig. 5c).   Then C1 and CA’B’ are in parallel. Cequivalent = (4.0 + 3.0) µF = 7.0 µF. Now to find the charges and potential differences we work backwards. For Fig. 4d, VAB = 100V = Qequivalent/Cequivalent = Qequivalent/7.0 x 10-6 C/V. Qequivalent = 7.0 x 10-4 C. For Fig. 4c, VAB = 100V = Q1/C1 = Q1/4.0 x 10-6 C/V. Q1 = 4.0 x 10-4 C. QA’B’ = 100 V(CA’B’) = 100 V (3.0 x 10-6 C/V) = 3.0 x 10-4 C. Notice Q1 + QA’B’ = 7.0 x 10-4 C. For Fig. 4b, C2 and CDB are in series and have the same charge. Q2 = QDB = QA’B’ = 3.0 x 10-4 C. VAD = Q2/C2 = 3.0 x 10-4 C/12 x 10-6 C/V = 25 V. VDB = QDB/CDB = 3.0 x 10-4 C/4 x 10-6 C/V = 75 V. Notice VAD + VDB = 100 V. Q3 = C3 VDB = 3.0 x 10-6 C/V (75 V) = 2.25 x 10-4 C. Q4 = C4 VDB = 1.0 x 10-6 C/V (75 V) = 0.75 x 10-4 C. Notice that Q3 + Q4 = 3.0 x 10-4 C = Q2. Energy stored in a capacitor = U = 1/2 QV = 1/2 Q2/C. U1 = 1/2 Q12/C1 = 1/2(4.0 x 10-4 C)2/(4.0 x 10-6 C/V) = 2.0 x 10-2 J. U2 = 1/2 Q22/C2 = 1/2(3.0 x 10-4 C)2/(12.0 x 10-6 C/V) = 0.375 x 10-2 J. U3 = 1/2 Q32/C3 = 1/2(2.25 x 10-4 C)2/(3.0 x 10-6 C/V) = 0.843 x 10-2 J. U4 = 1/2 Q42/C4 = 1/2(0.75 x 10-4 C)2/(1.0 x 10-6 C/V) = 0.281 x 10-2 J. U1 + U2 + U3 + U4 = 3.5 x 10-2 J. Uequivalent = 1/2 (Qequivalent)2/Cequivalent                    = 1/2(7 x 10-4 C)2/(7 x 10-6 C/V) = 3.5 x 10-2 J.
 16 For a uniform spherical distribution of charge, we find from Gauss' law that E = kQr/R3 for r ≤ R  and E = kQ/r2 for r ≥ R. In both cases, the field is radially outward.  A plot of  E vs. r  and  V vs. r  are shown in the graphs above. For r ≤ R, the electric field is directly proportional to r. For r ≥ R, E is inversely proportional to the square of r. At r = R, the expression for E for r ≤ R equals the expression for E for r ≥ R. The potential equals zero when r approaches a very large number. The potential at any point is the area under the E vs. r curve from infinity to that point. The electric field at any point equals the negative slope of V vs. r at that point. This occurs, of course, because of the definition of the potential function.
 17 From Gauss' theorem, we found that the electric field a distance r from the axis of a cylinder equals 2k λ/r, where λis the charge per unit length. For a cylinder of total charge Q and length L,   λ= Q/L and E = 2kQ/rL. In this problem, we have coaxial cylinders of radius a and radius b with b > a.  Using Gauss's theorem, the electric field between the cylinders still equals 2kQ/rL because a Gaussian surface there does not enclose the charge on the outer cylinder. By definition, C = Q/Vab = Q/[(2kQ/L) ln b/a] = L/[2k (ln b/a)].
 18 To move the plates a distance ds, takes dW = F ds, where F is the force we are seeking. The energy stored in the electric field is U = (1/2)Q2/C. For a parallel plate capacitor of area A and separation s, C = εoA/s  and  U = Q2s/2 εoA, where Q,   εo,  and  A  are constants. The work dW would appear as an increase dU in the potential energy: dU = (Q2/2 εoA)ds Equating dW and dU, we find F ds = (Q2/2 εoA)ds   and F = (Q2/2 εoA).
 19 q = (charge/volume)(volume) = ρ(4 πr3/3) q’ = ρ(4 πr2 dr) dU = kqq’/r = kρ2(4 π)2r5 dr/3r U = kρ2(4 π)2/3 o∫R r4 dr = kρ2(4 π)2 R5/15 = 3k(4πR3ρ/3)2/5R = 3kQ2/5R, where the total charge of the sphere of radius R is Q = (4 πR3ρ/3).
 20 3kQ2/5ro = mc2. ro = 3kQ2/5mc2    = 3(9.0 x 109 N-m2/C2)(1.6 x 10-19 C)2/5(9.1 x 10-31 kg)(3.0 x 108 m/s)2    = 1.7 x 10-15 m.
 21 From conservation of angular momentum,                    LA = LB rAmvA sin 90o = rBmvB sin 90o vB/vA = rA/rB = 6 x 10-5/3 x 10-5 = 2 From conservation of energy,    UA      +     KA        =    UB      +      KB -kqq’/rA + 1/2 mvA2  = -kqq’/rB + 1/2 mvB2 -kqq’/rA + kqq’/rB = 1/2 mvB2 - 1/2 mvA2 kqq’(1/rB - 1/rA) = m/2(vB2 - vA2) = m/2(4vA2 - vA2) = 3mvA2/2 9 x 109 N-m2/C2(10-6 C)(10-6 C)(105/m)[1/3 -1/6] = 3(10-10 kg)vA2/2 vA = 106 m/s vB = 2vA = 2 x 106 m/s

22. U = -ke2/r    Fnet = ma ke2/r2 = mv2/r  or  mv2 = ke2/r E = U + K    = -ke2/r + 1/2 mv2    = -ke2/r + 1/2 ke2/r    = -ke2/2r L = mvr

 23 U = -ke2/2r At distance 2r,  the electron is momentarily at rest and K = 0. E = U + K = -ke2/r + 0. The total energy is the same for motion in a circle of radius r and for the skinny ellipse approximated by motion back and forth in Fig. 6b. For Fig. 6c,  L = rmv sin r,v = rmv sin 0o = 0. Note: The quantum number "represents" angular momentum. For the orbital model, the total energy would be the same for both, but the quantum number would be different. For more complex atoms, there is penetration of the "inner core"  for low angular momentum and the total energy depends on the quantum number .
 24 Let the charge/area = σ. The area of the washer = πb2 - πa2 = π(b2 – a2). So for a total charge q,   σ = q/ π(b2 – a2). Area of an annulus of radius r and thickness dr is dA = 2 πr dr. The charge dq = σdA = [q/ π(b2 – a2)] 2 πr dr.  Each little point on the annulus is a point charge. The distance of each of these to point P is the same distance. The potential at point P due to a charge dq at distance (r2 + y2)1/2   is dV = k (dq)/(r2 + y2)1/2,  where k = 9 x 109 N-m2/C2  or dV = k([q/ π(b2 – a2)] 2 πr dr)/(r2 + y2)1/2.  For  y = 0,  Ey = 0,  as you might expect because point O is at the center of the distribution.
 25 In general, In all cases,  E is constant over the Gaussian surface and the angle between E and dA is zero. Thus, . For  a < r < b,  the dashed Gaussian surface is inside a conductor so the electric field is zero.  The charge enclosed must be zero  or 0 = qR + qa   and   qR = - qa = -5 µC. For  0 < r < R,  the Gaussian surface is again inside a conductor and E = 0. For  R < r < a,  the charge enclosed is qa. E4 πr2 = qR/ εo.  E = -5 x 10-6C/4πεor2 = k(-5 x 10-6C/r2). The minus sign means the electric field for  R < r < a  is radially inward. For  a < r < b,  as we said before,  E = 0. For  r > b,  the charge enclosed = qR + qa + qb = qb  since qR = - qa. E4 πr2 =qb/ εo.   And E = qb/4πεor2 = kqb/r2. For r >b,  V(r) = kqb/r. For  a < r < b,  V(r) = kqb/b. For  R < r < a,  V(r) = kqb/b - kqa(1/r - 1/a). For  r < a,  V(r) = kqb/b - kqa(1/R - 1/a). Plots of E and V as a function of r are below in Fig. for 25. 26 E(r) = λ/2πεo ln ∞ ≠ 0.  With an infinite wire you can't get away with choosing  V = 0 at infinity. E = -dV/dr = -d{( λ/2πεo)(ln a – ln r)/dr = λ/2πεor!
 27 E = - dV/dx  or  E = - slope of  V vs x For –0.4 m < x < -0.3 m, - Slope = - (-150 – 0)V/{(-0.3 –(-0.4)}m = +1500 V/m. For –0.3 m < x < -0.1 m, - Slope = -{150 – -(150)}V/{(-0.1 –(-0.3)}m = -1500 V/m. For –0.1 m < x < 0.1 m, slope = 0. For 0.1 m < x < 0.3 m, - slope = -(-150 – 150)V/(0.3 –0.1)}m = +1500 V/m. For 0.3 m < x < 0.4 m, - slope = -{0 – (-150)}V/(0.4 –0.3)}m = -1500 V/m. If you are so inclined, you can work backwards. V – 0 = -∫ E dx  or  V = - area under E vs x. For example,  - the area from  x = 0.40 m  to  x = 0.3 m  is 150 V. V(0.3m) = 0 –150 V = - 150 V.   Try it for a few other points. Homepage Sitemap
 Website Designed By: Questions, Comments To: Date Created: Date Last Updated: Susan D. Kunk Phyllis J. Fleming August 8, 2002 April 29, 2003