Problem 1:
A two electron system has (in its center of mass frame) energy E and angular momentum L. What is the closest distance the electrons approach each other?
Solution:
 | Concepts, principles, relations
that apply to the problem: The Coulomb potential, motion in a central potential, energy conservation |
| Why do they apply? E = (1/2)m(dr/dt)2 + L2/(2mr2) + e2/r, m = me/2, e2 = qe2/(4pe0). At the distance of closest approach dr/dt = 0. |
| How do they apply? E = L2/(2mrmin2) + e2/rmin. rmin2 – e2rmin/E - L2/(2mE) = 0. rmin = e2/(2E) + (e4/(4E2) + L2/(2mE))1/2. |
| Details of the calculation: We can also write: E = mevmin2 + e2/rmin, L = 2mevmin(rmin/2). At the distance of closest approach each particle is a distance rmin/2 from the CM. vmin2 = L2/(me2rmin2), E = L2/(mermin2) + e2/rmin. |
Problem 2:
Two isolated spherical conductors of radii 3 cm and 9 cm
are charged to 1500 V and 3000 V, respectively. They are very far away from
each other.
(a) What is the total energy of the system?
(b) If we connect the two conductors by a fine wire and
wait until equilibrium is established, how much heat will be generated in the
wire?
Solution:
| Concepts, principles, relations
that apply to the problem: The energy stored in a capacitor |
| Why do they apply? Initially we have two isolate spherical capacitors. We then connect them in parallel. |
| How do they apply? (a) We have two isolated spherical capacitors. The total energy stored in the system is U = ½ (C1V12 + C2V22). For a spherical capacitor C = 4pe0R. Therefore U = 2pe0(R1V12 + R2V22) = 4.875*10-5 J. The total charge on the capacitors is Q = Q1 + Q2 = C1V1 + C2V2 = 4pe0(R1V1 + R2V2). (b) We have two capacitors in parallel with total charge Q and capacitance C = C1 + C2. The total energy stored in the system now is U’ = ½Q2/C. U’ = ½(4pe0(R1V1 + R2V2))2/(4pe0(R1 + R2)) = 2pe0(R1V1 + R2V2)2/(R1 + R2) The energy dissipated as heat will be U – U’ = 2pe0(V1-V2)2R1R2/(R1 + R2) = 2.8125*10-6 J. |
| Details of the calculation: None |
Problem 3:
A conducting sphere of radius a carrying a charge q is submerged halfway into a non-conducting dielectric liquid of dielectric constant e. The other half is in air. Will the electric field be purely radial? Explain.
Solution:
| Concepts, principles, relations
that apply to the problem: Boundary value problems, the uniqueness theorem |
| Why do they apply? There are no free charges except on the boundaries and the problem has azimuthal symmetry. |
| How do they apply? If we place the origin of our coordinate system at the center of the sphere and let the z-axis point upward, then the problem has azimuthal symmetry. We have 3 regions. Region 1: the sphere Region 2: the air Region 3: The dielectric In region 1 E = 0 and f = constant (properties of conductors). Assume that in regions 2 and 3 f(r) = B/r, with B a constant. This potential fulfills all the boundary conditions. f = constant at r = a, f --> 0 as r --> ¥, f = continuous across the air-dielectric interface, and the normal component of D is continuous across the air-dielectric interface. The uniqueness theorem guaranties that this solution is the only solution. E = -Ñf, the electric field will be purely radial. |
| Details of the calculation: None |
Problem 4:
The dielectric of a parallel plate capacitor has a permittivity that varies as e1 + ax, where x is the distance from one plate. The area of a plate is A and their spacing is s.(a) Find the capacitance.
(b) Assume e1 + ax varies from
e1 to 2e1. Find P from D and E for that case.
(c) Find the polarization charge density rp.
Solution:
| Concepts, principles, relations that apply to the
problem: Gauss' law, relationship between E, P, and D |
| Why do they apply? Assuming a surface charge density, Gauss' law for D can be used to find D, then E, then V. We then can find C = Q/V. |
| How do they apply? (a)  Gauss' law for D yields DDA = sfreeDA, D = sfree, D = eE. .C = Q/V = sfreeA/((sfree/a)ln(1+as/e1)) = aA/ln(1+as/e1) If a << e1 then ln(1+as/e1) » as/e1 - (1/2)(as/e1)2 + ... . C » e1A/s. |
| Details of the calculation: (b) 2e1 = e1 + as. a = e1/s. e = e1 +(e1/s)x. D = sfreei, E = sfreei/(e1 +(e1/s)x), P = D - e0E = sfreei(1 - [e0/(e1 +(e1/s)x)] (c) rp = -Ñ×P = -¶P/¶x = -sfreee0(e1/s)/(e1 +(e1/s)x)2 sp = P×n. At x = 0 sp = -sfree(1- e0/e1). At x = s sp = sfree(1- e0/2e1). |
Problem 5:
(a) A spherical dielectric of radius a has a uniform polarization
P in the z-direction. Show that the electric field inside the
dielectric due to the polarization is given by E = -P/(3e0).
(b) A large capacitor in vacuum has parallel circular plates of radius R
separated by a distance d (d<<R). The capacitor is charged to a
potential difference V and disconnected from the source. Find the energy
stored in the capacitor.
(c) Subsequently, a small sphere of radius a (a<<d) and
dielectric constant K is placed in the center of the capacitor between
the plates. Find the electric field inside the dielectric.
(d) Will the capacitor have less, more, or the same energy than before the
dielectric was inserted? Explain!
Solution:
| Concepts, principles, relations that apply to the
problem: Boundary value problems,  is
the general solution to Laplace's equation for problems with azimuthal
symmetry. |
| Why do they apply? Ñ2f = 0 inside and outside the sphere since rf = 0 inside and outside the sphere. We are looking for a solution for Ñ2f = 0, subject to the boundary conditions. |
| How do they apply? (a) Place the center of the sphere at the origin. Then rb = -Ñ×P = 0, sb = P×en = Pcosq. We have a sphere with a surface charge density proportional to cosq. We solve for the potential and the field using boundary value methods. Assume f1 = A0 + A1rcosq inside, f2 = B0'/r + (B1'/r2)cosq outside. We assume all higher-order coefficients are zero. If we find a solution with this assumption, it is the only solution. Boundary conditions: (i) f is continuous at r = a. A0 = B0'/a, A1a = A1'a + B1'/a2. (ii) The radial component of D is continuous at r = a. (There are no free charges.) E1 = -Ñf1, E1r = -A1cosq, E1q = A1sinq, E1 = -A1k. D1 = e0E1 + P E2 = -Ñf2. E2r = B0'/r2 + 2(B1'/r3)cosq, E2q = (B1'/r3)sinq, D2 = e0E2. -e0A1cosq + Pcosq = e0B0'/a2 + 2e0(B1'/a3)cosq. -e0A1cosq + P = 2e0(B1'/a3), B0' = A0 = 0. -e0A1cosq + P = 2e0A1. A1 = P/3e0. E1 = -A1k = P/3e0. |
| Details of the calculation: (b) U = (1/2)CV2, C = e0A/d = e0pR2/d, U = (1/2)(e0pR2/d)V2. (c) The dielectric sphere is placed in a previously uniform electric field E0. (E0 = V/d) The field inside the sphere will be Ein = -Ñfin = 3E0/(K+2) from previously solved problem. (d) The capacitor is disconnected from the battery. Q is constant. The capacitance increases when a dielectric is inserted. The energy stored in the capacitor (U = Q2/2C) is proportional to 1/C for a capacitor with constant Q. The energy stored in the capacitor decreases. |
Problem 6:
A parallel-plate capacitor is connected to a battery which maintains a
potential difference V0 between its plates. A slab of
dielectric constant K is inserted between the plates, completely filling the
space between them.
(a) Show that the battery does an amount of work q0V0(K -
1) during the insertion process, if q0 is the charge on the capacitor
plates before the slab is inserted.
(b) How much work is done by mechanical forces on the slab when it is inserted
between the plates? Is this work done on, or by, the agent inserting the
slab?
Solution:
| Concepts, principles, relations that apply to the
problem: Capacitance, energy conservation |
| Why do they apply? We must find the work done by external agents as a dielectric is inserted into a capacitor and the capacitance changes. |
| How do they apply? (a) before insertion: Q0 = C0V0, C0 = e0A/d. after insertion: Q = CV0, C = eA/d. Q > Q0. Work done by the battery: Wbat = (Q - Q0)V0 =(C - C0)V02 = (e - e0)V02A/d = (Ae0/d )(K-1)V02 = Q0V0(K-1). (b) Total energy store in the capacitor before insertion: U0 = (1/2)C0V02. Total energy store in the capacitor after insertion: U = (1/2)CV02. DU = U - U0 = (1/2)V02(C - C0) = (1/2)Q0V0(K-1). Wmech + Wbat = DU, Wmech = DU - Wbat = -(1/2)Q0V0(K-1). The agent inserting the slab does negative work, work is being done on the agent inserting the slab, the slab is being pulled in. |
| Details of the calculation: |
Problem 7:
Regarding the Earth and a cloud layer 800 m above the Earth as the plates of a capacitor, calculate the capacitance if the cloud layer has an area of 1 km2. If an electric field of 3x106 N/C makes the air break down and conduct electricity, (that is, cause lightning,) what is the maximum charge (in C) the cloud can hold?
Solution:
| Concepts, principles, relations that apply to the
problem: Capacitance C = Q/V, parallel plate capacitors, C = e0A/d. |
| Why do they apply? We are asked to regard the Earth and a cloud layer as the plates of a capacitor. |
| How do they apply? C = e0A/d = e0106m2/800m. Vmax = Emax*d = 3*106N/C*800m = 2.4*109J/C. Qmax = C*Vmax = 26.55 C |
| Details of the calculation: None |