## Assignment 10, solutions

#### Problem 1:

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 vmax of the dielectric?  Express your answers in terms of ε0, ε, V0, w, d and m.

(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 vmax (case a) to vmax (case b) in terms of ε0 and ε.

Solution:

• Concepts:
Capacitance, energy and work
• Reasoning:
Fx = -dWmech/dx.  We must 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.
dWmech + dWbat = dU.  dWbat = dQV0 = dCV02.  dU = ½dCV02
dWmech = -½dCV02
Fx = -dWmech/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 = Fx(x - w/2), 0 < x < w.
½mvmax2 = Fxw/2.
½mv2 = ½mvmax2 - Fx(x - w), 0 < x < w.
The maximum speed of the dielectric is vmax = [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 = [vmax2 - 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 = Cw/2V0 = ½V0(ε + ε0)w2/d.
Total energy store in the capacitor:  U = ½Q02/C.
dWmech = dU = -½dC Q02/C2.
Fx = -dWmech/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.
½mvmax2 = ½Q02(1/CW/2 - 1/CW) = ½V02CW/2(1 - CW/2 /CW).
CW/2(1 - CW/2 /CW) = (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.
vmax = [¼V022 - ε02/ ε) w2/(md)]½.

(c)  vmax2(case a) /vmax2(case b) = 2ε(ε - ε0)/(ε2 - ε02) = 2ε/(ε + ε0) .

#### Problem 2:

A "dielectric" material consists of a number of brass spheres of diameter d, spaced 3d apart, in a regular lattice.  Assuming that each sphere is influenced only by the imposed external electric field, i.e. neglecting the effect of the neighboring spheres or the redistribution of induced charges, find the dielectric constant k for this material.

Solution:

• Concepts:
Polarization: P = ε0χeE  (SI units), dielectric constant: k = (1 + χe), boundary value problems
• Reasoning:
We need to find the polarization to find the dielectric constant.
• Details of the calculation:
• We need to find the dipole moment p of each sphere.
We then have P = Np,  N = 1/(3d)3.  We find the dipole moment of each sphere by finding the electric field produced by each sphere.  To find this field we solve a boundary value problem.
For a conducting sphere at the origin in an external field E = Ek we have
Φ(r,θ) = ∑n=0[Anrn + Bn/rn+1]Pn(cosθ)
outside the sphere and Φ = constant, E = 0 inside the sphere.  Choose  Φ1 = 0 inside the sphere.
Φ2 = A1'rcosθ  + (B1'/r2)cosθ  outside.
We assume all other coefficients are zero.  If we find a solution with this assumption, it is the only solution.
Boundary conditions:
(i)  Φ is continuous at r = a = d/2.  0 = A1'a + B1'/a2.
(ii)  E2 = E0 at r >> a.
-A1'cosθ = E0cosθ,  A1'sinθ = -E0sinθ,  A1' = -E0,  B1' = E0a3.
Φ2 = -E0rcosθ  + (E0a3/r2)cosθ.
E2 = -Φ2.  E2r = E0cosθ  + 2(E0a3/r3)cosθ,  E = -E0'sinθ + (E0a3/r3)sinθ.
E2 = external field + field due to polarized sphere.
Esphere_r = 2(E0a3/r3)cosθ,  Esphere_θ = (E0a3/r3)sinθ.
This is a dipole field.  For a dipole at the origin we have
E(r) = (1/(4πε0))[3(p∙r)r/r5 - p/r3],
Er = (1/(4πε0))2p cosθ/r3,  Eθ = (1/(4πε0))p sinθ/r3.
Therefore p = [(4πε0)E0a3]k = [(4πε0)E0d3/8]k.
P = Np = [(4πε0)E0/216]k, P = ε0χeE, χe = 4π/216.
κ = (1 + χe) = 1.058, ε  = ε0κ.

#### Problem 3:

In the circuit shown in the figure R1 = 6 Ω, R2 = 4 Ω, and R3 = 2Ω .
(a)  Find the currents I1, through resistor R1, I2, through resistor R2, and I3, through resistor R3, in the circuit shown in the Figure.
(b)  Indicate in what direction each current flows. Give the results in terms of the junctions labeled in the circuit.
For example, you need to say whether I1 goes from b to c or from c to b.
(c)  Find the potential difference between junctions b and c.  Clearly indicate which junction has the higher potential.

Solution:

• Concepts:
Kirchhoff's rules
• Reasoning:
We use one junction and two loop rules to solve for the three unknown currents.
• Details of the calculation:
Choose directions for the currents as shown.
(a) Junction c:  I1 + I2 = I3.
Loop bcdab:  10 V - R1I1 - R3I3 = 0.
Loop befcb:  - R2I2 - 14 V + R1I1 - 10 V = 0.
I1 = 2 A, I2 = -3 A, I3 = -1A.
(b)  I1 flows from b to c,  I2 flows from e to b, and I3 flows from a  to d.
(c)  ΔV = 10V - I1R1 = -2 V, b is at higher potential than c,  Vb - Vc = 2 V.

#### Problem 4:

The space between the plates of a parallel-plate capacitor (see figure) is filled with two slabs of linear dielectric material.  Each slab has thickness s, so the total distance between the plates is 2s.  Slab 1 has a dielectric constant of 2, and slab 2 has a dielectric constant of 1.5.  The free charge density on the top plate is σ and on the bottom plate −σ.
(a)  Find the electric displacement D in each slab.
(b)  Find the electric field E in each slab.
(c)  Find the polarization P in each slab.
(d)  Find the potential difference between the plates.
(e)  Find the location and amount of all the bound charge.
(f)  Now that you know all the charge (free and bound), recalculate the field in each slab, and compare with your answer to (b).

Solution:

• Concepts:
Gauss' law for D,  E = D/ε,  D = ε0E + P,  ρtrue + ρbound = ρtotal,  ∇∙P = -ρboundP·n = σbound
• Reasoning:
Gauss' law can be used to find D.  Given D, we find E, P, V, and the charge densities.
• Details of the calculation:
(a)  Let x-axis point downward and let the upper plate be at x = 0.
Gauss' law for DD = σ i inside both slabs.
(b)  E1 = D1 = σ/(2ε0) iE2 = D2 = 2σ/(3ε0) i.
(c)  P = D - ε0EP1 = σ/(2ε0) i,  P2 = σ/(3ε0) i,
(d)  ΔV = V(0) - V(2s) = E1s + E2s = 7σs/(6ε0).
(e)  ρbound = 0 in each slab, the volume charge density is zero in each slab.
Surface charge σbound on surface of dielectric 1 at x = 0:
E1 = σtotal0 = (σ + σbound)/ε0 = σ/(2ε0).  σbound =  -σ/2.
Surface charge σbound on surface of dielectric 1 at x = s:  σbound = σ/2.
Surface charge σbound on surface of dielectric 2 at x = 2s:
E2 = -σtotal0 = (-σ + σbound)/ε0 = -2σ/(3ε0).  σbound =  σ/3.
Surface charge σbound on surface of dielectric 1 at x = s:  σbound = -σ/3.
Total surface charge at the boundary at x = s:  σbound = σ/6.
(f)  Gauss' law for E:
E
1 = σtotal0 i = (σ - σ/2)/ε0 i = σ/(2ε0) iE2 = (-σ + σ/3)/ε0 (-i) = 2σ/(3ε0) i.
The results agree with what we found in (b).

#### Problem 5:

Five 1 Ω resistors are connected as shown in the figure.  The resistance in the conducting wires (fully drawn lines) is negligible.  Determine the resulting resistance R between A and B.

Solution:

• Concepts:
Resistors in series and parallel
• Reasoning:
The circuit has enough symmetry so that we can analyze it like a simple circuit with resistors in series and parallel.
• Details of the calculation:
The system of resistors can be redrawn as show.
We have two branches.  The upper branch has a resistance of 1 Ω (two sets of parallel 1 Ω resistors connected in series, and the lower branch has a resistance of 1 Ω.  The resistance R between A and B is R = 0.5 Ω.