THE UNIVERSITY OF BRITISH COLUMBIA
 
Physics 108 Assignment # 5 SOLUTIONS:
 
POTENTIAL & CAPACITANCE
 
Wed. 2 Feb. 2005 - finish by Wed. 9 Feb.

1. CLASSICAL RADIUS OF THE ELECTRON: You are probably familiar with Einstein's famous equation $E = m c^2$. 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 $r_0$ and net charge $e$, find the numerical value of $r_0$ in meters.¹

   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 $dE = k_E QdQ/r_0$ 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 $m c^2$ 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 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 $C_1 = C_2 = 4 C$. 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  .

   

   The battery $B$ supplies $12$ V. The capacitances are $C_1 = 1$ $\mu$F, $C_2 = 2$ $\mu$F, $C_3 = 4$ $\mu$F and $C_4 = 3$ $\mu$F.

   1. Find the charge on each capacitor when switch $S_1$ is closed but switch $S_2$ is still open.
      ANSWER:  Let $Q_i$ denote the charge on the $i^{\rm th}$ capacitor $C_i$. From charge conservation we have $Q_1 = Q_3$ and $Q_2 = Q_4$. Both pairs of capacitors in series (1 and 3; 2 and 4) must make up the full voltage: $V_B = Q_1/C_1 + Q_3/C_3 = Q_2/C_2 + Q_4/C_4$. Therefore $V_B = Q_1[1/C_1 + 1/C_3] = Q_2[1/C_2 + 1/C_4]$ yielding $Q_1 = Q_3 = 12/(10^6/1+ 10^6/4)$ and $Q_2 = Q_4 = 12/(10^6/2+ 10^6/3)$ or and  .
   2. What is the charge on each capacitor if $S_2$ is also closed?
      ANSWER:  Now $C_1$ and $C_2$ are effectively just one big capacitor $C_{12} = C_1 + C_2 = 3$ $\mu$F and similarly for $C_{34} = C_3 + C_4 = 7$ $\mu$F.   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: $V_B = Q_{12}/C_{12} + Q_{34}/C_{34}$.   Thus $V_B = Q_{12}[1/C_{12} + 1/C_{34}]$, giving $Q_{12} = Q_{34} = 12/(10^6/3+ 10^6/7)$ $= 25.2\times 10^{-6}$ C.   Meanwhile the voltage across $C_1$ must be the same as that across $C_2$: $Q_1/C_1 = Q_2/C_2$ $\Longrightarrow Q_2 = Q_1(C_2/C_1) = {2\over 1} Q_1$ $\Longrightarrow Q_{12} = \left( 1 + {2\over1} \right) Q_1$ or and  .   Similarly, $Q_3/C_3 = Q_4/C_4$ $\Longrightarrow Q_4 = Q_3(C_4/C_3) = {3\over 4} Q_3$ $\Longrightarrow Q_{34} = \left( 1 + {3\over 4} \right) Q_3$ or and  .

3. THUNDERCLOUD CAPACITOR: A large thundercloud hovers over the city of Vancouver at a height of $1.0$ 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 $300$ V/m. The cloud has a horizontal area of $100$ km$^2$.
   1. 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 $\sigma = \epsz E = 8.85 \times 10^{-12} \times 300= 2.656 \times 10^{-9}$ C/m$^2$. Over an area of $A = 100\times 10^6 = 10^8$ m$^2$, this gives a total charge of  .
   2. 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 $U/V = {1\over2} \epsz E^2$ $= 0.5 \times 8.85 \times 10^{-12} \times (300)^2$ $= 3.984 \times 10^{-7}$ J/m$^3$. The volume between the cloud and the ground is $V = 1000\times 10^8= 10^{11}$ m$^3$, so  .