Problem 1:

A static charge distribution produces a radial electric field
E = (A/r²)exp(-br) e_r.
where A and b are constants.
(a)  What is the charge density?  Sketch it.
(b)  What is the total charge Q?

Solution:

• Concepts:
  Gauss' law
• Reasoning:
  ∇∙E = -∇²Φ = ρ/ε₀.  For the given field, the expression is not well defined at r = 0. 
  We can handle the divergence at r = 0 by using ∇∙(r/r³) = 4πδ(r).
  We can also use the integral form of Gauss' law.
• Details of the calculation:
  (a)  E(r)4πr² = Q_inside/ε₀ in SI units.
  Q_inside = 4πε₀Aexp(-br).
  ρ(r) = (1/(4πr²))(dQ_inside /dr) = -(ε₀/r²)bAexp(-αr), except at r = 0.
  As r --> 0 Q_inside --> 4πε₀A, there is a point charge at the origin.  It is surrounded by a charge cloud of opposite sign.
  Therefore ρ(r) = -(ε₀/r²)bA₀exp(-αr) + 4πε₀Aδ(r).
  
  Or:
  ∇∙E = Aexp(-br)∇∙(r/r³) + (r/r³)∙∇(Aexp(-br)= ρ(r)/ε₀.
  ∇∙(r/r³) = 4πδ(r).
  ρ(r) = ρ(r) = 4πε₀A[δ(r) -  bexp(-br)/(4πr²)].
  
  (b)  As r --> infinite, Q_inside --> 0.  The total charge is zero.

Problem 2:

One half of the region between the plates of a spherical capacitor of inner and outer radii a and b is filled with a linear isotropic dielectric of permittivity ε₁ and the other half has permittivity ε₂, as shown in figure.  If the inner plate has total charge Q and the outer plate has total charge -Q, find
(a)  the electric displacements D₁ and D₂ in the region of ε₁ and ε₂,
(b)  the electric fields in the region of ε₁ and ε₂,
(c)  and the total capacitance of this system.

Solution:

• Concepts:
  Capacitors with dielectrics
• Reasoning:
  The electric field is confined to the region between the spheres.  The spheres are equipotentials.   Assume that the electric field points in the r-direction and has spherical symmetry.
  This implies that E₁ = E₂ = E = (A/r²) e_r, from Gauss/ law.
• Details of the calculation:
  (a)  D_i(r) = ε_iE = ε_i(A/r²),  D_i(a) = σ_if(a) = Q_i(a)/(2πa²).  Q_i = ε_i(A/a²)2πa² = 2πε_iA.
  2π(ε₁ + ε₂)A = Q,  A = Q/(2π(ε₁ + ε₂)).
  D₁ = ε₁Q/(2π(ε₁ + ε₂)r²) e_r,  D₂ = ε₂Q/(2π(ε₁ + ε₂)r²) e_r. 
  D∙n = 0, there is no bound surface charge density at the dielectric-dielectric interface.
  
  (b)  E = Q/(2π (ε₁ + ε₂)r²) e_r.
  
  (c) C = Q/ΔV.  ΔV = ∫_a^bE(r)dr =  [Q/(2π(ε₁ + ε₂))]∫_a^b(1/r²)dr.
  ΔV = [Q/(2π(ε₁ + ε₂))](1/a - 1/b) = Q(b - a)/(2πab(ε₁ + ε₂)).
  C = 2πab(ε₁ + ε₂)/( b - a).

Problem 3:

A parallel plate capacitor has square plates of width w and plate separation d.  A square dielectric also of width w and thickness d, with permittivity ε and mass m, is inserted between the plates a distance x into the capacitor and held there.  The plates are connected to a battery with battery voltage V₀.

(a)  Derive a formula for the force exerted on the dielectric as a function of x   Neglect edge effects. 
Assume that the battery stays connected and the dielectric is released from rest at x = w/2.  Describe its subsequent motion.
What is the range of the dielectric's motion, and within that range, what is the dielectric's speed as a function of position x? 
What is the maximum speed v_max of the dielectric?  Express your answers in terms of ε₀, ε, V₀, w, d and m.
(Neglect friction, ohmic heating, radiation.)

(b)  Now assume that the dielectric is released from rest at x = w/2 after the battery has been disconnected.
Derive a formula for the force exerted on the dielectric for w/2 < x < w.  Neglect edge effects. 
What is the maximum speed of the dielectric?  Express your answer in terms of ε₀, ε, V₀, w, d and m.

(c)  Find the ratio v_max(case a) to v_max(case b) in terms of ε₀ and ε.

Solution:

• Concepts:
  Capacitance, energy and work
• Reasoning:
  F_x = -F_mech = -dW_mech/dx.  We  find the work done by external agents as a dielectric is inserted into a capacitor and the capacitance changes.
• Details of the calculation:
  Capacitance, energy and work
  (a)  Capacitance as a function of x:  C = εwx/d  + ε₀w(w - x)/d,  0 < x < w.
  C = εw(2w - x)/d  + ε₀w(x - w)/d,  w < x < 2w.
  C = ε₀w²/d, x < 0 and x > 2w.
  Total energy store in the capacitor:  U = ½CV₀².
  As the position of the dielectric changes, the voltage across the capacitor plates stays the same.
  dW_mech + dW_bat = dU.  dW_bat = dQV₀ = dCV₀².  dU = ½dCV₀². 
  Therefore dW_mech = -½dCV₀². 
  F_x = -dW_mech/dx = ½V₀²dC/dx = ½V₀²(ε - ε₀)w/d  for 0 < x < w.
  The dielectric is pulled towards the right, into the capacitor for 0 < x < w.
  Similarly, the dielectric is pulled towards the left, into the capacitor for w < x < 2w.
  The force has the same constant magnitude.
  
  The dielectric is released from rest at x = w/2.
  The dielectric will oscillate about the equilibrium position x = w.  The motion is periodic, but not simple harmonic.  The restoring force has constant magnitude and points towards the right for w/2 < x < w and towards the left for w < x < 3w/2.
  
  Work - kinetic energy theorem:  ½mv² = F_x(x - w/2), 0 < x < w.
  ½mv_max² = F_xw/2.
  ½mv² = ½mv_max² - F_x(x - w), 0 < x < w.
  The maximum speed of the dielectric is v_max = [½V₀²(ε - ε₀)w²/(md)]^½.
  The speed as a function of position is v = [V₀²(ε - ε₀)w (x - w/2)/(md)]^½ for  w/2 < x < w,
  v = [v_max² - V₀²(ε - ε₀)w (x - w)/(md)]^½ = [V₀²(ε - ε₀)w (3w/2 - x)/(md)]^½  for  w < x < 3w/2.
  
  (b)  Now, as the position of the dielectric changes, the charge on the capacitor plates stays the same, Q₀ = C_w/2V₀ = ½V₀(ε + ε₀)w²/d.
  Total energy store in the capacitor:  U = ½Q₀²/C.
  dW_mech = dU = -½dC Q₀²/C².
  F_x = -dW_mech/dx = ½(Q₀²/C²) dC/dx = ½(Q₀²/C²) (ε - ε₀)w/d  for 0 < x < w.
  The dielectric is pulled towards the right, into the capacitor for 0 < x < w.
  Similarly, the dielectric is pulled towards the left, into the capacitor for w < x < 2w.
  But the force now does not have constant magnitude, since C depends on the position x.
  
  The dielectric is released from rest at x = w/2.
  ½mv_max² = ½Q₀²(1/C_W/2 - 1/C_W) = ½V₀²C_W/2(1 - C_W/2/C_W).
  C_W/2(1 - C_W/2 /C_W) = (½(ε + ε₀)w²/d)(1 - (½(ε + ε₀)w²/d)/(εw²/d))
  = ((ε + ε₀)w²/d)(ε - ε₀)/ε = ¼(w²/d)(ε² - ε₀²)/ε.
  v_max = [¼V₀² (ε² - ε₀²/ ε) w²/(md)]^½.
  
  (c)  v_max²(case a) /v_max²(case b) = 2ε(ε - ε₀)/(ε² - ε₀²) = 2ε/(ε + ε₀).

Problem 4:

A long straight wire is connected to an ideal battery.  When the room temperature is 20.0 ^oC, the established temperature of the wire is 22.0 ^oC.
The wire is disconnected and one-third of it is cut off.  The remaining piece of the wire is then connected to the same battery in the same room.
What is the new established temperature of the wire?
Assume Newton's law of cooling applies, i.e. the rate at which a system transfers heat to its environment is proportional to the temperature difference between the system and its environment.

Solution:

• Concepts:
  Energy dissipated by a resistance, Newton's law of cooling, equilibrium
• Reasoning:
  Let l be the length of the wire that is connected to a voltage source, V.
  The wire has resistance that is proportional to its length, R = cl, c = proportional constant.
  The rate at which energy is dissipated by the wire is P = V²/R = V²/cl.
  Assume Newton's law of cooling applies.
  We therefore have dQ/dt = kl(T_wire - T_room), k = proportional constant.
  dQ/dt is proportional to (T_wire - T_room) (Newton's law of cooling).  dQ/dt must also be proportional to the surface area, which is proportional to the length of the wire.
• Details of the calculation:
  In equilibrium P = dQ/dt, V² = ckl²(T_wire - T_room).
  Let T₁ be T_wire for l = l₁ and T₂ be T_wire for l = l₂ = 2l₁/3.
  Then (T₁ - T_room) = (4/9) (T₂ - T_room).
  (T₂ - T_room) = (9/4)*2 ^oC = 4.5 ^oC, T₂ = 24.5 ^oC.