Two long, thin cylindrical conducting shells of radii r_{1}
and r_{2}, respectively, are oriented coaxially (one cylinder is
centered inside the other). The inside cylinder with radius r_{1}
carries a linear charge density λ while the outside cylinder carries an equal
and opposite linear charge density -λ.

(a) In what direction does the electric field point between the two cylinders?
Explain.

(b) Find the electric field

(i) inside the smaller cylinder,

(ii) between the cylinders,

(iii) outside the cylinders.

(c) Plot the total electric field as a function of r, i.e., as a function of
the distance from the cylinders' center.

(d) Find the potential difference between the two cylinders.

(e) The construction defines a coaxial cable which is a device useful to
transmit information. Find the capacity per unit length of this cable assuming
that there is vacuum between the two cylinders.

(f) Now assume that keeping the charge of the cylinders constant the space
between them is filled with a dielectric with dielectric constant κ. Does the
capacity per unit length change? Why?

Solution:

- Concepts:

Gauss' law - Reasoning:

The problem has the required symmetry. - Details of the calculation:

(a) Both cylinders are conductors. The electric field vector just outside the surface of a conductor is perpendicular to the surface. Here the field points in the radial direction outward, from the positively charged surface to the negatively charged surface.

(b) The problem has cylindrical symmetry, we use Gauss' law,

∫**E∙**d**A**= q_{inside}/ε_{0}.

(i)**E**= 0, the total charge enclosed is zero.

(ii) 2πrLE = λL/ε_{0}.**E**= (**r**/r) λ/(2πε_{0}r).

(iii)**E**= 0, the total charge enclosed is zero.

(c) Between the cylinders the field is proportional to 1/r.(d) The potential difference is given by

V(r) = Φ(r_{1}) - Φ(r_{2}) = -∫_{r2}^{r1}**E**(r)∙d**r**= -[λ/(2πε_{0}r)]∫_{r2}^{r1}dr/r = [λ/(2πε_{0}r)] ln(r_{2}/r_{1}).

The inner cylinder is at the higher potential.

(e) The capacitance per unit length is given by C/L = λ/V = 2πε_{0}/ln(r_{2}/r_{1}).

(f) For a given free charge density λ the electric field between the cylinders will be smaller due to the polarization charge density in the dielectric, the voltage V will be reduced and therefore C/L = λ/V will increase. C'/L = κC/L.

Find the energy necessary to move a charge Q from point A = (a,0,0) to point
B = (a,0,h), where a and h are positive constants, along a helical path
parametrized as

**r** = a cosθ **i** + a sinθ **j** + hθ/(2π) **k**,

under the influence of an electrostatic field **E** = -E_{0} **k**.

Solution:

- Concepts:

Conservative forces - Reasoning:

The electrostatic field is conservative, the work done by or against the electrostatic force does not depend on the path. - Details of the calculation:

Let V(A) = 0, then V(B) = E_{0}h, ΔU = Q(V(B) - V(A)) = QE_{0}h = energy necessary to move Q from A to B.

The displacement vector from electric dipole **p**_{1} to dipole **p**_{2} is
**r**.

(a) Calculate the electric potential energy W;

(b) Calculate the force **F**_{21} that **p**_{1} exerts on
**p**_{2}.

(c) Calculate the torque **τ**_{12} that **p**_{1} exerts on
**p**_{2}.

(d) Let **p**_{1} = (10^{-9} Cm)**k** be located at
the origin and **p**_{2} = (10^{-9} Cm)**i **at **r** =
(3 m)**i** + (4 m)**k**.

Provide numeric answers for **F**_{21} and **τ**_{12}.

Solution:

- Concepts:

The dipole field, a dipole in an external field - Reasoning:

We have one dipole in the external field of another dipole. The energy of a dipole in an external field is U = -**p∙E.** - Details of the calculation:

(a) Consider dipole**p**_{2}in the external field of dipole**p**_{1}. Choose your coordinate system such that**p**_{1}is at the origin and**p**_{2}is at**r**.

Field of**p**_{1}at**r**:**E**(_{1}**r**) = (1/(4πε_{0}))[3(**p**_{1}**∙r**)**r**/r^{5}-**p**_{1}/r^{3}].

Energy of**p**_{2}in this field: U(**r**) = -**p**_{2}**∙E**(_{1}**r**) = (1/(4πε_{0}))[**p**_{1}**∙p**_{2}/r^{3}- 3(**p**_{1}**∙r**)(**p**_{2}**∙r**)/r^{5}].

(b)**F**_{21}(**r**) = -**∇**U(**r**) = -(1/(4πε_{0}))[**∇**(**p**_{1}**∙p**_{2}/r^{3}) - 3**∇**((**p**_{1}**∙r**)(**p**_{2}**∙r**)/r^{5})].

**p**_{1}and**p**_{2}are constant**, ∇**(**p**_{1}**∙p**_{2}) = 0,**∇**(**p**_{i}**∙r**)= (**p**_{i}**∙∇**)**r**=**p**_{i}.

**F**_{21}(**r**) = -(1/(4πε_{0}))[(**p**_{1}**∙p**_{2})**∇**(1/r^{3}) - 3[(**p**_{1}**∙r**)(**p**_{2}**∙r**)**∇**(1/r^{5}) +**p**_{2}(**p**_{1}**∙r**)/r^{5}+**p**_{1}(**p**_{2}**∙r**)/r^{5}]].

**∇**(1/r^{3}) = -3**r**/r^{5}.**∇**(1/r^{5}) = -5**r**/r^{7}.

**F**_{21}(**r**) = (3/(4πε_{0}r^{5}))[(**p**_{1}**∙p**_{2})**r**+**p**_{2}(**p**_{1}**∙r**) +**p**_{1}(**p**_{2}**∙r**)] - 5(**p**_{1}**∙r**)(**p**_{2}**∙r**)(**r**/r^{2})].

(c)**τ**_{12}(**r**) =**p**_{2}×**E**(_{1}**r**) = (1/(4πε_{0}))[3(**p**_{1}**∙r**)(**p**_{2}×**r**)/r^{5}- (**p**_{2}×**p**_{1})/r^{3}].

(d)**I**n SI units:**p**_{1}**∙p**_{2}= 0.**p**_{1}**∙r**= 4*10^{-9}.**p**_{2}**∙r**= 3*10^{-9}.**p**_{2}×**r**= -4*10^{-9}**j**.**p**_{2}×**p**_{1}= -10^{-18}**j**.

**F**_{21}= (3/(4πε_{0}5^{5}))[(4*10^{-18}C^{2}m^{3})**i**+ (3*10^{-18}C^{2}m^{3})**k**] - 12*10^{-18}C^{2}m^{3}(3**i**/5 + 4**k**/5))]

= -(3/(4πε_{0}5^{5}))[(3.2*10^{-18}C^{2}m^{3})**i**+ (6.6*10^{-18}C^{2}m^{3})**k**].

= -[(2.8*10^{-11}N)**i**+ (5.7*10^{-11}N)**k**].**τ**_{12}= (1/(4πε_{0}5^{5}))[-36*10^{-18}**j**+ 25*10^{-18}**j**]Nm = (1/(4πε_{0}5^{5}))[-11*10^{-18}**j**] Nm = -3.17*10^{-11}Nm**j**.

Suppose the input voltages V_{1}, V_{2} and V_{3} in the circuit
shown can
assume values of either 0 or 1 (0 means ground). There are thus 8 possible
combinations of input voltages. List the V_{out} for each of these possibilities.

Solution:

- Concepts:

Kirchhoff's rules - Reasoning:

We can find the currents I_{1}, I_{2}, I_{3}and I_{4}using Kirchhoff's rules. The junction rule states that the sum of the currents entering a junction must equal the sum of the currents leaving that junction. The loop rule states that the sum of the potential differences around any closed circuit loop must be zero. - Details of the calculation:

Redraw the circuit.

V_{3}- 2R(I_{3}- I_{2}) - 2R(I_{3}- I_{4}) = 0.

(V_{2}- V_{3}) - 2R(I_{2}- I_{1}) - R(I_{2}- I_{4}) - 2R(I_{2}- I_{3}) = 0.

(V_{1}- V_{2}) - 2RI_{1}- R(I_{1}- I_{4})] - 2R(I_{1}- I_{2}) = 0.

2R(I_{4}- I_{3}) + R(I_{4}- I_{2}) + R(I_{4}- I_{1}) + 2RI_{4}= 0.

V_{out}= 2I_{4}R.

We want to solve this system of linear equations for I_{4}R.

We obtain I_{4}R = (4V_{1}+ 2V_{2}+ V_{3})/24.

Therefore V_{out }= (4V_{1}+ 2V_{2}+ V_{3})/12.

V_{out}for the 8 possible combinations of input voltages is given in the table below.

V _{1}V _{2}V _{3}V _{out}--- V _{1}V _{2}V _{3}V _{out}0 0 0 0 1 0 0 1/3 0 0 1 1/12 1 0 1 5/12 0 1 0 1/6 1 1 0 ½ 0 1 1 1/4 1 1 1 7/12

A conical surface (an empty ice-cream cone) carries a uniform surface charge
σ. The height of the cone is a, as is the radius of the top.

(a) Find the electrical potential at point P (vertex of the cone).

(b) Find the electrical potential at point Q (center of the top of the
cone).

(c) Find the potential difference between points P (the vertex) and Q (the
center of the top).

∫_{ }x dx/(ax^{2} + bx + c)^{½} = (ax^{2}
+ bx + c)^{½}/a - (b/(2a))∫ dx/(ax^{2} + bx + c)^{½
}∫ dx/(ax^{2} + bx + c)^{½} = a^{-½} ln(2a^{ ½}
(ax^{2} + bx + c)^{½} + 2ax + b)

Solution:

- Concepts:

The electrostatic potential - Reasoning:

We find the electrostatic potential Φ(**r**) due to a charge distribution using the principle of superposition. - Details of the calculation:

(a) For a surface charge distribution Φ(**r**) = [1/(4πε_{0})] ∫_{A' }dA' σ(**r'**)/|**r**-**r**'|.

Divide the cone into rings. Note that θ = π/4, sinθ = cosθ = 2^{-½}.

Φ(P) = Φ(0,0,0) = [1/(4πε_{0})]∫_{A' }dA' σ(**r'**)/|**r**-**r**'| = [1/(4πε_{0})] ∫_{0 }^{a√2}2πr sinθ dr σ/r

= [σ√2/(4ε_{0})]∫_{0 }^{a√2}dr = σa/(2ε_{0}).

(b) Φ(Q) = Φ(0,0,a) = [1/(4πε_{0})] ∫_{0 }^{a√2}2πr sinθ dr σ/l(r)

= [σ√2/(4ε_{0})]∫_{0 }^{a√2}dr r/(r^{2}+ a^{2}- 2ar cosθ)^{½ }= [σ√2/(4ε_{0})]∫_{0 }^{a√2}dr r/(r^{2}+ a^{2}- ar√2)^{½}

= [σa/(4ε_{0})]ln((2 + √2)^{2}/2) = [σa/(2ε_{0})]ln((2 + √2)/√2) = [σa/(2ε_{0})]ln(1 + √2).

Note: ln((2 + √2)^{2}/2) = ln(((2 + √2)/√2)^{2}) = 2ln(1 + √2).

(c) ΔV = Φ(P) - Φ(Q) = [σa/(2ε_{0})](1 - ln(1 + √2)).