THE UNIVERSITY OF BRITISH COLUMBIA
 
Physics 122 Assignment # 5 SOLUTIONS:
 
POTENTIAL & CAPACITANCE
 
Fri. 1 Feb. 2002 - finish by Fri. 8 Feb.

1.

CLASSICAL RADIUS OF THE ELECTRON: You are probably familiar with Einstein's famous equation E = m c². If m is the mass of an electron and E is the electrostatic potential energy required to ``assemble'' the electron from bits of charge infinitely distant from each other into a uniform spherical shell of radius r₀ and net charge e, find the numerical value of r₀ in meters.²

ANSWER:  Start with no charge, then bring successive bits dQ in to add uniformly to the charge Q of the shell. A given bit of charge acquires an electrostatic potential energy dE = k_E QdQ/r₀ in the process. Thus the total energy E required to assemble the shell is $E = k_E/r_0 \int_0^e QdQ = {1\over2} k_E e^2/r_0$. If we set this equal to m c² we get $\ds{ r_0 = {1\over2} k_E {e^2 \over m c^2 } = { (8.998 \times 10^9)(1.6022 \times 10^{-19})^2 \over 2 (9.11 \times 10^{-31})(2.998 \times 10^8)^2 } }$ or  . The conventional value (actually the Compton wavelength of the electron) is twice as big, 2.818 fm, where ``fm'' stands for ``femtometers'' or ``fermis'' (named after Enrico Fermi); both are the same as 10^-15 m.

 

2.

CAPACITOR WITH INSERT: Suppose we have a capacitor made of two large flat parallel plates of the same area A (and the same shape), separated by an air gap of width d. Its capacitance is C. Now we slip another planar conductor of width d/2 (and the same area and shape) between the plates so that it is centred halfway in between. What is the capacitance $C^\prime$ of the new system of three conductors, in terms of the capacitance C of the original pair and the other parameters given? (Neglect ``edge effects'' and any dielectric effect of air.)

ANSWER:  The original capacitance was $C = \epsz A/d$. The capacitor with the insert as shown is equivalent to two identical capacitors in series, each of which has a gap of d/4 between its plates, so that C₁ = C₂ = 4 C. The equivalent capacitance of two capacitors in series is given by $1/C^\prime = 1/C_1 + 1/C_2 = 2/4C = 1/2C$. Thus  .

 

3.

CUBIC CAPACITOR Suppose we take a roll of very thin ( 50 µm) copper sheet and a roll of 150 µm thick strontium titanate dielectric (see Table 29-2 on p. 671 of the textbook) and form a capacitor as follows: cut the sheets into strips 5 cm wide and sandwich the dielectric sheet between two sheets of copper. Then fold the sandwich back and forth to fill a cube 5 cm on each side. Assuming that we can press the layers together so that there are no empty spaces, find:

(a)

the capacitance of the resulting cube-shaped capacitor;   ANSWER:  Each layer consists of 2 sheets of Cu and 1 sheet of SrTiO₃ and so is 250 µm or $2.5 \times 10^{-4}$ m thick. The number of layers is therefore $N = 5 \times 10^{-2} / 2.5 \times 10^{-4} = 200$ and the total area is $A = N \times 0.05 \times 0.05 = 0.5$ m². The capacitance is $C = \kappa \epsz A/d$ where d = 150 µm. The dielectric constant of SrTiO₃ is (from Table 27-2) $\kappa = 310$, so $C = 310 \times 8.85 \times 10^{-12} \times 0.5 / 1.5 \times 10^{-4}$ or  .

(b)

the maximum charge it will hold without breaking down;   ANSWER:  Table 27-2 lists the dielectric strength of SrTiO₃ as 8 kV/mm, which is the same as $8 \times 10^6$ V/m. Our dielectric sheet has a thickness of $1.50 \times 10^{-4}$ m, so the breakdown voltage is 1200 V. At that potential, Q = CV gives  .¹

(c)

the total energy we can store in this small cube.   ANSWER:  The total energy stored in a capacitor is given by  .

4.

ARRAY of CAPACITORS:

The battery B supplies 6 V. The capacitances are C₁ = 2.0 µF, C₂ = 1.0 µF, C₃ = 4.0 µF and C₄ = 3.0 µF.

(a)

Find the charge on each capacitor when switch S₁ is closed but switch S₂ is still open.   ANSWER:  Let Q_i denote the charge on the $i^{\rm th}$ capacitor C_i. From charge conservation we have Q₁ = Q₃ and Q₂ = Q₄. Both pairs of capacitors in series (1 and 3; 2 and 4) must make up the full voltage: V_B = Q₁/C₁ + Q₃/C₃ = Q₂/C₂ + Q₄/C₄. Therefore V_B = Q₁[1/C₁ + 1/C₃] = Q₂[1/C₂ + 1/C₄] yielding Q₁ = Q₃ = 6/(10⁶/2.0 + 10⁶/4.0) and Q₂ = Q₄ = 6/(10⁶/1.0 + 10⁶/3.0) or and  .

(b)

What is the charge on each capacitor if S₂ is also closed?   ANSWER:  Now C₁ and C₂ are effectively just one big capacitor C₁₂ = C₁ + C₂ = 3.0 µF and similarly for C₃₄ = C₃ + C₄ = 7.0 µF. Charge conservation now requires $Q_{12} \equiv Q_1 + Q_2 = Q_{34} \equiv Q_3 + Q_4$ and the two effective capacitors in series must make up the full voltage: V_B = Q₁₂/C₁₂ + Q₃₄/C₃₄. Thus V_B = Q₁₂[1/C₁₂ + 1/C₃₄] giving $Q_{12} = Q_{34} = 6/(10^6/3.0 + 10^6/7.0) = 12.6 \times 10^{-6}$ C. Meanwhile the voltage across C₁ must be the same as that across C₂: $Q_1/C_1 = Q_2/C_2 \Longrightarrow Q_2 = Q_1(C_2/C_1) = {1\over2} Q_1 \Longrightarrow Q_{12} = {3\over2} Q_1$ or and  . Similarly, $Q_3/C_3 = Q_4/C_4 \Longrightarrow Q_4 = Q_3(C_4/C_3) = {3\over4} Q_3 \Longrightarrow Q_{34} = {7\over4} Q_3$ or and  .

5.

THUNDERCLOUD CAPACITOR: A large thundercloud hovers over the city of Vancouver at a height of 2.0 km. Between the cloud and the ground (both of which we may treat as parallel conducting plates, neglecting edge effects) the electric field is about 200 V/m. The cloud has a horizontal area of 200 km².

(a)

Estimate the number of Coulombs [C] of positive charge in the cloud, assuming that the ground has the same surface density of negative charge.   ANSWER:  The electric field between two flat plates with surface charge densities $\pm \sigma$ is given by $E = \sigma / \epsz$. Thus $\sigma = \epsz E = 8.85 \times 10^{-12} \times 200 = 1.77 \times 10^{-9}$ C/m². Over an area of $A = 200 \times 10^6 = 2 \times 10^8$ m², this gives a total charge of  .

(b)

Estimate the number of joules [J] of energy contained in the air between the cloud and the ground.   ANSWER:  The energy density stored in an electric field is given by $U/V = {1\over2} \epsz E^2 = 0.5 \times 8.85 \times 10^{-12} \times (200)^2 = 0.177 \times 10^{-6}$ J/m³. The volume between the cloud and the ground is $V = 2000 \times 2 \times 10^8 = 4 \times 10^{11}$ m³, so  .