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Chapter 24: Problem 10

24.10. A cylindrical capacitor consists of a solid inner conducting core with radius \(0.250 \mathrm{cm},\) surrounded by an outer hollow con- ducting tube. The two conductors are separated by air, and the length of the cylinder is 12.0 \(\mathrm{cm}\) . The capacitance is 36.7 \(\mathrm{pF}\) . (a) Calculate the inner radius of the hollow tabe. (b) When the capacitor is charged to 125 \(\mathrm{V}\) , what is the charge per unit length \(\lambda\) on the capacitor?

Short Answer

Expert verified
(a) The inner radius of the hollow tube is approximately 0.251 cm. (b) The charge per unit length is 3.82 nC/m.

Step by step solution

01

Understand the Cylinder Capacitor

The cylindrical capacitor consists of a solid inner cylinder of radius \( r_1 = 0.250 \text{ cm} \) and an outer hollow tube. The length \( L \) of the capacitor is \( 12.0 \text{ cm} \). The capacitance \( C \) is given as \( 36.7 \text{ pF} \). We need to find the inner radius \( r_2 \) of the hollow cylinder.
02

Capacitance Formula for Cylindrical Capacitor

The capacitance \( C \) of a cylindrical capacitor is given by \[ C = \frac{2 \pi \varepsilon_0 L}{\ln\left(\frac{r_2}{r_1}\right)} \]where \( \varepsilon_0 = 8.854 \times 10^{-12} \text{ F/m} \) is the permittivity of free space.
03

Rearrange the Formula to Solve for r2

To find \( r_2 \), rearrange the equation:\[ \ln\left(\frac{r_2}{r_1}\right) = \frac{2 \pi \varepsilon_0 L}{C} \]and then \[ r_2 = r_1 \exp\left(\frac{2 \pi \varepsilon_0 L}{C}\right) \]
04

Calculate r2

Substitute known values into the equation. Ensure units are consistent (convert all to meters):- \( r_1 = 0.250 \text{ cm} = 0.0025 \text{ m} \)- \( L = 12.0 \text{ cm} = 0.12 \text{ m} \)- \( C = 36.7 \text{ pF} = 36.7 \times 10^{-12} \text{ F} \)Now compute:\[ r_2 = 0.0025 \exp\left(\frac{2 \times 3.1416 \times 8.854 \times 10^{-12} \times 0.12}{36.7 \times 10^{-12}}\right) \]Compute the expression inside the exponential and then find \( r_2 \).
05

Calculate the Inner Radius

Perform the calculations:\[ \frac{2 \times 3.1416 \times 8.854 \times 10^{-12} \times 0.12}{36.7 \times 10^{-12}} = 0.004072 \]\[ r_2 = 0.0025 \exp(0.004072) \approx 0.0025(1.004072) \approx 0.00251 \text{ m} \]So, the inner radius \( r_2 \approx 0.251 \text{ cm} \).
06

Charge Per Unit Length on Capacitor

Given the capacitor is charged to 125 V, use:\[ C = \frac{Q}{V} \]where \( Q \) is the total charge. The charge per unit length \( \lambda = \frac{Q}{L} \). Thus,\[ \lambda = \frac{C \cdot V}{L} \]Substitute the known values:\[ \lambda = \frac{36.7 \times 10^{-12} \cdot 125}{0.12} \text{ C/m} \]
07

Calculate Charge Per Unit Length

Perform the calculation for \( \lambda \):\[ \lambda = \frac{36.7 \times 10^{-12} \times 125}{0.12} = 3.82 \times 10^{-9} \text{ C/m} \]This is the charge per unit length on the capacitor.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Electrostatics
Electrostatics is the study of electric charges at rest. In the context of a cylindrical capacitor, electrostatic principles determine how charges are distributed between the inner and outer conducting surfaces. Capacitors store electric energy through the separation of charges across a dielectric, which in this case is air.

The electric field in a cylindrical capacitor is radially outward, and the field strength decreases with distance from the inner conductor. This is due to the geometry of the cylindrical capacitor, which dictates that the electric field lines spread out as they move away from a curved surface.
  • The inner core and the outer tube are conductors.
  • Their role is to hold equal and opposite charges.
  • The dielectric in between (air) reduces energy loss and increases capacitance.
Understanding electrostatics in capacitors helps explain the energy storage capability, the electric field distribution, and what factors can affect these properties.
Capacitance Calculation
Calculating capacitance is fundamental in understanding how well a capacitor can store charge at a given voltage. In a cylindrical capacitor, capacitance depends on the physical dimensions and the separation of the conductors.

For a cylindrical capacitor, the capacitance is given by the formula:
\[ C = \frac{2 \pi \varepsilon_0 L}{\ln\left(\frac{r_2}{r_1}\right)} \]
where:
  • \( C \) is the capacitance.
  • \( \varepsilon_0 \) is the permittivity of free space, facilitating the storage and conduction of electric field.
  • \( L \) is the length of the capacitor.
  • \( r_1 \) and \( r_2 \) are the inner and outer radii, respectively.
  • The natural logarithm \( \ln \) accounts for the geometric aspect of field distribution.
By rearranging this equation, one can determine unknown parameters, such as the radius of the outer conductor, helping in practical design and analysis of capacitors.
Charge Distribution
Charge distribution on a capacitor shows how much charge is carried on its surface. For a cylindrical capacitor, the distribution is symmetrical around the axis, which means the charges are evenly spread over the cylindrical surfaces.

Charge per unit length, represented by \( \lambda \), helps quantify how much charge exists over a certain length of the capacitor's surface. This is calculated using:
\[ \lambda = \frac{C \cdot V}{L} \]
where:
  • \( \lambda \) is the charge per unit length.
  • \( C \) is the capacitance.
  • \( V \) is the potential difference (voltage).
  • \( L \) is the physical length of the capacitor.
Charge distribution is important for understanding how energy is stored within the capacitor, influencing its total energy capacity and efficiency. Proper charge distribution ensures consistent and reliable performance in electronic circuits.

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Most popular questions from this chapter

24.53. Electronic flash units for cameras contain a capacitor for storing the energy used to produce the flash. In one such unit, the flash lasts for \(\frac{1}{6 / 5}\) s with an average light power output of \(2.70 \times 10^{5} \mathrm{W}\) . (a) If the conversion of electrical energy to light is 95\(\%\) efficient (the rest of the energy goes to thermal energy), how much energy must be stored in the capacitor for one flash? (b) The capacitor has a potential difference between its plates of 125 \(\mathrm{V}\) when the stored energy equals the value calculated in part (a). What is the capacitance?

24.11. A cylindrical capacitor has an inner conductor of radius 1.5 \(\mathrm{mm}\) and an outer conductor of radius 3.5 \(\mathrm{mm}\) . The two conductors are separated by vacuum, and the entire capacitor is 2.8 \(\mathrm{m}\) long. (a) What is the capacitance per unit length? (b) The potential of the inner conductor is 350 \(\mathrm{mV}\) higher than that of the outer conductor. Find the charge (magnitude and sign) on both conductors.

24.74. The parallel-plate air capacitor in Fig. 24.40 consists of two horizontal conducting plates of equal area \(A .\) The bottom plate rests on a fixed support, and the top plate is suspended by four springs with spring constant \(k,\) positioned at each of the four corners of the top plate as shown in the figure. When uncharged, the plates are separated by a distance \(z_{0}\) . A battery is connected to the plates and produces a potential difference \(V\) between them. This causes the plate separation to decrease to \(z\) . Neglect any fringing effects. (a) Show that the electrostatic force between the charged plates has a magnitude \(\epsilon_{0} A V^{2} / 2 z^{2}\) . (Hint: See Exercise \(24.29 . )\) (b) Obtain an expression that relates the plate separation \(z\) to the potential difference \(V\) . The resulting equation will be cubic in \(z\) . (c) Given the values \(A=0.300 \mathrm{m}^{2}, z_{0}=1.20 \mathrm{mm}, k=25.0 \mathrm{N} / \mathrm{m},\) and \(V=120 \mathrm{V},\) find the two values of \(z\) for which the top plate will be in equilibrium. (Hint: You can solve the cubic equation by plugging a trial value of \(z\) into the equation and then adjusting your guess until the equation is satisfied to three significant figures. Locating the roots of the cubic equation graphically can help you pick starting values of \(z\) for this trial-and error procedure. One root of the cubic equation has a nonphysical negative value. (d) For each of the two values of \(z\) found in part (c), is the equilibrium stable or unstable? For stable equilibrium a small displacement of the object will give rise to a net force tending to return the object to the equilibrium position. For unstable equilibrium a small displacement gives rise to a net force that takes the object farther away from equilibrium.

24.21. Two parallel-plate vacuum capacitors have areas \(A_{1}\) and \(A_{2}\) and equal plate spacings \(d .\) Show that when the capacitors are connected in parallel, the equivalent capacitance is the same as for a single capacitor with plate area \(A_{1}+A_{2}\) and spacing \(d\) .

24.40. A budding electronics hobbyist wants to make a simple 1.0 -nF capacitor for tuning her crystal radio, using two sheets of aluminum foil as plates, with a few shects of paper between them as a dielectric. The paper has a dielectric constant of 3.0, and the thickness of one sheet of it is 0.20 \(\mathrm{mm}\) . (a) If the sheets of paper measure \(22 \times 28 \mathrm{cm}\) and she cuts the aluminum foil to the same dimensions, how many sheets of paper should she use between her plates to get the proper capacitance? (b) Suppose for convenience she wants to use a single sheet of posterboard, with the same dielectric constant but a thickness of \(120 \mathrm{mm},\) instead of the paper. What area of aluminum foil will she need for her plates to get her 1.0 \(\mathrm{nF}\) of capacitance? (c) Suppose she goes hight-tech and finds a sheet of Teflon of the same thickness as the posterboard to use as a dielectric. Will she need a larger or smaller area of Tellon than of posterboard? Explain.
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