MODEL Pure distilled water is a good insulator. (The conductivity of tap water is due to dissolved ions.) Thus the immersed capacitor has a dielectric between the electrodes.

SOLVE

1. From Table 26.1, the dielectric constant of water is $\kappa =80$. The presence of the dielectric increases the capacitance to

   $$C=\kappa C_0=80\times 5.0\text{nF}=400\mathrm{nF}$$

   At the same time, the voltage decreases to

   $$\text{Δ}{V}_{\text{C}}={\displaystyle \frac{(\text{Δ}{V}_{\text{C}}){}_0}{\kappa }}={\displaystyle \frac{160\text{V}}{80}}=2.0\text{V}$$

2. The presence of a dielectric does not alter the derivation leading to Equation 26.25 for the energy stored in a capacitor. Right after being disconnected from the battery, the stored energy was

   $${(U_{\text{C}})}_0=\frac{1}{2}{C}_0{(\text{Δ}{V}_{\text{C}})}_0^2=\frac{1}{2}(5.0\times {10}^{-9}{\text{ F})(160\text{ V})}^2=6.4\times {10}^{-5}\text{ J}$$

   After being immersed, the stored energy is

   $$U_{\text{C}}=\frac{1}{2}C{(\text{Δ}{V}_{\text{C}})}^2=\frac{1}{2}(400\times {10}^{-9}\text{ F})(2.0{\text{ V})}^2=8.0\times {10}^{-7}\text{ J}$$

REVIEW Water, with its large dielectric constant, has a big effect on the capacitor. But where did the energy go? We learned in Chapter 23 that a dipole is drawn into a region of stronger electric field. The electric field inside the capacitor is much stronger than just outside the capacitor, so the polarized dielectric is actually pulled into the capacitor. The “lost” energy is the work the capacitor’s electric field did pulling in the dielectric.

MODEL Model the defibrillator as a parallel-plate capacitor with a dielectric.

SOLVE

1. The capacitance of a parallel-plate capacitor in a vacuum is $C_0={ϵ}_{0}A\text{/}d$. A dielectric increases the capacitance by the factor $\kappa$, to $C=\kappa C_0$, so the area of the capacitor plates is

   $$A={\displaystyle \frac{Cd}{\kappa {ϵ}_0}}={\displaystyle \frac{(150\times {10}^{-6}\text{F})(5.0\times {10}^{-5}\text{m})}{120(8.85\times {10}^{-12}{\text{C}}^2\text{/}\text{N}{\text{m}}^2)}}=7.1{\text{m}}^2$$

   Although the surface area is very large, Figure 26.32 shows how very large sheets of very thin metal can be rolled up into capacitors that you hold in your hand.

2. The energy stored in the capacitor is

   $$U_{\text{C}}=\frac{1}{2}C(\text{Δ}{V}_{\text{C}}{)}^2=\frac{1}{2}(150\times {10}^{-6}\text{F}){\left(2100\text{V}\right)}^2=330\text{J}$$

   Because the dielectric has increased C by a factor of $\kappa$, the energy density of Equation 26.27 is increased by a factor of $\kappa$ to $u_{\text{E}}=\frac{1}{2}\kappa {ϵ}_0{E}^2$. The electric field strength in the capacitor is

   $$E={\displaystyle \frac{\text{Δ}V{}_{\text{C}}}{d}}={\displaystyle \frac{2100\text{V}}{5.0\times {10}^{-5}\text{m}}}=4.2\times {10}^7\text{V}\text{/}\text{m}$$

   Consequently, the energy density is

   $$\begin{array}{lll}{u}_{\text{E}}\hfill & =\hfill & \frac{1}{2}(120)(8.85\times {10}^{-12}{\text{C}}^2\text{/}{\text{Nm}}^2){(4.2\times {10}^7\text{V}\text{/}\text{m})}^2\hfill \\ \hfill & =\hfill & 9.4\times {10}^5{\text{J/m}}^3\hfill \end{array}$$

REVIEW 330 J is a substantial amount of energy—equivalent to that of a 1 kg mass traveling at 25 m/s. And it can be delivered very quickly as the capacitor is discharged through the patient’s chest.

MODEL Model the Geiger counter as two infinitely long, concentric, conducting cylinders. Applying a potential difference between the cylinders charges them like a capacitor; indeed, they are a cylindrical capacitor. The charge on an infinitely long conductor is infinite, but the linear charge density $\text{λ}$ is the charge per unit length (C/m), which is finite. So for a cylindrical capacitor we compute the capacitance per unit length $C=\text{λ}/\text{Δ}V$, in F/m, rather than an absolute capacitance. To avoid breakdown of the gas, the field strength at the surface of the wire—the point of maximum field strength—must not exceed the dielectric strength.

VISUALIZE FIGURE 26.33 shows a cross section of the Geiger counter tube. We’ve chosen to let the outer cylinder be positive, with an inward-pointing electric field, but a negative outer cylinder would lead to the same answer since it’s only the field strength that we’re interested in.

FIGURE 26.33 Cross section of a Geiger counter tube.

SOLVE

1. Gauss’s law tells us that the electric field between the cylinders is due only to the charge on the inner cylinder. Thus $\vec{E}$ is the field of a long, charged wire—a field we found in Chapter 23 using superposition and again in Chapter 24 using Gauss’s law. It is

   $$\vec{E}=\left({\displaystyle \frac{\text{λ}}{2\pi {ϵ}_{0}r}}\text{,}\text{inward}\right)$$

   where $\text{λ}$ is the magnitude of the linear charge density. We need to connect this field to the potential difference between the wire and the outer cylinder. For that, we need to use Equation 26.3:

   $$\text{Δ}V=V_{\text{f}}-V_{\text{i}}=-{\displaystyle {\int }_{s_{\text{i}}}^{s_{\text{f}}}{E}_{s}ds}$$

   We’ll integrate along a radial line from $s_{\text{i}}=R_1$ on the surface of the inner cylinder to $s_{\text{f}}=R_2$ at the outer cylinder. The field component $E_s$ is negative because the field points inward. Thus the potential difference is

   $$\begin{array}{lll}\text{Δ}V\hfill & =\hfill & -{\displaystyle {\int }_{R_1}^{R_2}\left(-{\displaystyle \frac{\text{λ}}{2\pi {ϵ}_{0}s}}\right)ds={\displaystyle \frac{\text{λ}}{2\pi {ϵ}_0}}{\displaystyle {\int }_{R_1}^{R_2}{\displaystyle \frac{ds}{s}}}}\hfill \\ \hfill & =\hfill & {{\displaystyle \frac{\text{λ}}{2\pi {ϵ}_0}}\ln s|}_{R_1}^{R_2}={\displaystyle \frac{\text{λ}}{2\pi {ϵ}_0}}\ln \left({\displaystyle \frac{R_2}{R_1}}\right)\hfill \end{array}$$

   We see that the applied potential difference and the linear charge density are related by

   $${\displaystyle \frac{\text{λ}}{2\pi {ϵ}_0}}={\displaystyle \frac{\text{Δ}V}{\text{ln}\left(R_2\text{/}{R}_1\right)}}$$

   Thus the capacitance per unit length is

   $$\begin{array}{ccc}\hfill C& \hfill =\hfill & {\displaystyle \frac{\text{λ}}{\text{Δ}{V}_{\text{C}}}}={\displaystyle \frac{2\pi {ϵ}_0}{\text{ln}(R_2/R_1)}}\hfill \\ \hfill & \hfill =\hfill & {\displaystyle \frac{2\pi (8.85\times {10}^{-12}{\text{C}}^2/\text{N}{\text{m}}^2)}{\text{ln}\left(25\right)}}=17\text{pF}\text{/}\text{m}\hfill \end{array}$$

   You should convince yourself that the units of ${ϵ}_0$ are equivalent to F/m.

2. Using the above in the expression for $\vec{E}$, we find the electric field strength at distance r is

   $$E={\displaystyle \frac{\text{Δ}V}{r\text{ln}\left(R_2\text{/}{R}_1\right)}}$$

   The field strength is a maximum at the surface of the wire, where it reaches

   $$E_{\text{max}}={\displaystyle \frac{\text{Δ}V}{R_1\text{ ln}\left(R_2\text{/}{R}_1\right)}}$$

   The maximum applied voltage will bring $E_{\text{max}}$ to the dielectric strength, $E_{\text{max}}=1.0\times {10}^6\text{V}\text{/}\text{m}$. Thus the maximum potential difference between the wire and the tube is

   $$\begin{array}{ccc}\hfill \text{Δ}{V}_{\text{max}}& \hfill =\hfill & R_1{E}_{\text{max}}\ln \left({\displaystyle \frac{R_2}{R_1}}\right)\hfill \\ \hfill & \hfill =\hfill & (5.0\times {10}^{-4}\text{m})(1.0\times {10}^6\text{V}\text{/}\text{m})\ln \left(25\right)\hfill \\ \hfill & \hfill =\hfill & 1600\text{ V}\hfill \end{array}$$

REVIEW This is the maximum possible voltage, but it’s not practical to operate right at the maximum. Real Geiger counters operate with typically a 1000 V potential difference to avoid an accidental breakdown of the gas. If a high-speed charged particle from a radioactive decay then happens to pass through the tube, it will collide with and ionize a number of the gas atoms. Because the tube is already very close to breakdown, the addition of these extra ions and electrons is enough to push it over the edge: A breakdown of the gas occurs, with a spark jumping across the tube. The “clicking” sounds of a Geiger counter are made by amplifying the current pulses associated with the sparks.