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A cylindrical capacitor has an inner conductor of radius 2.2 mm and an outer conductor of radius 3.5 mm. The two conductors are separated by vacuum, and the entire capacitor is 2.8 m long. (a) What is the capacitance per unit length? (b) The potential of the inner conductor is 350 mV higher than that of the outer conductor. Find the charge (magnitude and sign) on both conductors.

Short Answer

Expert verified
The capacitance per unit length is \(3.62 \times 10^{-11} \text{ F/m}\). The inner conductor has a charge of \(3.56 \times 10^{-11} \text{ C}\), and the outer has \(-3.56 \times 10^{-11} \text{ C}\).

Step by step solution

01

Understanding the Capacitance Formula for a Cylindrical Capacitor

The formula for capacitance per unit length (C') of a cylindrical capacitor is given by: \[ C' = \frac{2\pi\varepsilon_0}{\ln(b/a)} \]where \( \varepsilon_0 \) is the permittivity of free space, \( a \) is the radius of the inner conductor, and \( b \) is the radius of the outer conductor. In this problem, \( a = 2.2 \text{ mm} = 2.2 \times 10^{-3} \text{ m} \) and \( b = 3.5 \text{ mm} = 3.5 \times 10^{-3} \text{ m} \). \( \varepsilon_0 \approx 8.85 \times 10^{-12} \text{ F/m} \).
02

Substitute Values to Compute Capacitance Per Unit Length

Substitute the given values into the capacitance formula: \[ C' = \frac{2\pi \times 8.85 \times 10^{-12}}{\ln(3.5 \times 10^{-3} / 2.2 \times 10^{-3})} \]Calculate the natural logarithm and simplify the expression to find \( C' \).
03

Solve for Capacitance Per Unit Length

First, compute the ratio inside the logarithm:\[ \ln(3.5/2.2) \approx \ln(1.5909) \approx 0.463 \]Next, plug this into the equation to solve for \( C' \):\[ C' \approx \frac{2\pi \times 8.85 \times 10^{-12}}{0.463} \approx 3.62 \times 10^{-11} \text{ F/m} \].
04

Compute Charge from Potential Difference

The potential difference \( V \) is 350 mV, or 0.35 V. Use the formula for charge per unit length \( q' = C' \cdot V \), where \( C' \) is the capacitance per unit length:\[ q' = (3.62 \times 10^{-11}) \cdot (0.35) = 1.27 \times 10^{-11} \text{ C/m} \].
05

Determine Total Charge on Conductors

The total charge \( Q \) on each conductor for the entire 2.8 m length of the capacitor is found using \( Q = q' \cdot L \), where \( L = 2.8 \text{ m} \):\[ Q = (1.27 \times 10^{-11}) \cdot 2.8 \approx 3.56 \times 10^{-11} \text{ C} \].
06

Conclude on Magnitude and Sign of Charge

The magnitude of the charge on the inner conductor is \( 3.56 \times 10^{-11} \text{ C} \), and it is positive because the potential of the inner conductor is higher. The outer conductor has the same magnitude of charge but negative sign, so \(-3.56 \times 10^{-11} \text{ C} \).

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Key Concepts

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

Capacitance Per Unit Length
Cylindrical capacitors, interestingly, have their capacitance expressed in terms of length. This is because they are often quite long, and understanding capacitance in this way gives us a better idea of the capacitor's behavior over its length. The capacitance per unit length (denoted as \( C' \)) for a cylindrical capacitor is a measure of how well it can store an electric charge along each meter of its length.
To find \( C' \), we use the formula: \[ C' = \frac{2\pi\varepsilon_0}{\ln(b/a)} \]where:\
  • \( \varepsilon_0 \) is the permittivity of free space, approximately \( 8.85 \times 10^{-12} \) F/m.
  • \( a \) is the radius of the inner conductor.
  • \( b \) is the radius of the outer conductor.
In our exercise, the inner radius \( a \) is 2.2 mm and the outer radius \( b \) is 3.5 mm. These need to be converted to meters when plugged into the formula.
By calculating, you'll see that \( C' \approx 3.62 \times 10^{-11} \) F/m, which tells us that for each meter of length, the capacitor can hold \( 3.62 \times 10^{-11} \) Farads of charge. This is an essential characteristic used in design and application.
Potential Difference
The potential difference in a cylindrical capacitor is the voltage between the inner and outer conductors. It's a crucial aspect as it directly influences how much charge the capacitor can hold. This potential difference is measured in volts and, in our scenario, the inner conductor is 350 mV, or 0.35 V, more positive than the outer conductor.
We utilize this voltage difference to further calculate the charge distribution. By the relation \( V = q'/C' \), where \( V \) is the potential difference, \( q' \) is the charge per unit length, and \( C' \) is the capacitance per unit length, we can rearrange to determine \( q' \):\[ q' = C' \cdot V \]Using \( C' \approx 3.62 \times 10^{-11} \) F/m from our previous calculation and the given \( V = 0.35 \) V, we find \( q' \approx 1.27 \times 10^{-11} \) C/m. This indicates how much charge is distributed along each meter of the capacitor given the existing voltage difference.
Charge Distribution
Charge distribution in a cylindrical capacitor describes how electric charge is positioned along its length. The formula \( Q = q' \cdot L \), where \( Q \) is the total charge, \( q' \) is charge per unit length, and \( L \) is the total length of the capacitor, allows us to calculate the charge managed by the entire capacitor.
In the given problem, the capacitor is 2.8 meters long. Using the previously calculated \( q' \approx 1.27 \times 10^{-11} \) C/m, we find the total charge \( Q \approx 3.56 \times 10^{-11} \) C.
The charge distribution is symmetric. It means the inner conductor has a positive charge of \( +3.56 \times 10^{-11} \) C, while the outer conductor has an equal and opposite charge, \( -3.56 \times 10^{-11} \) C. This negative charge on the outer conductor balances the positive charge, ensuring the entire capacitor is electrically neutral. This understanding of charge distribution is vital for effectively designing and using capacitors in electronic circuits.

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Most popular questions from this chapter

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

A parallel-plate capacitor has capacitance \(C\) = 12.5 pF when the volume between the plates is filled with air. The plates are circular, with radius 3.00 cm. The capacitor is connected to a battery, and a charge of magnitude 25.0 pC goes onto each plate. With the capacitor still connected to the battery, a slab of dielectric is inserted between the plates, completely filling the space between the plates. After the dielectric has been inserted, the charge on each plate has magnitude 45.0 pC. (a) What is the dielectric constant \(K\) of the dielectric? (b) What is the potential difference between the plates before and after the dielectric has been inserted? (c) What is the electric field at a point midway between the plates before and after the dielectric has been inserted?

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 sheets 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 mm. (a) If the sheets of paper measure 22 \(\times\)28 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 12.0 mm, instead of the paper. What area of aluminum foil will she need for her plates to get her 1.0 nF of capacitance? (c) Suppose she goes high-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 Teflon than of posterboard? Explain.

A parallel-plate vacuum capacitor has 8.38 J of energy stored in it. The separation between the plates is 2.30 mm. If the separation is decreased to 1.15 mm, what is the energy stored (a) if the capacitor is disconnected from the potential source so the charge on the plates remains constant, and (b) if the capacitor remains connected to the potential source so the potential difference between the plates remains constant?

Three capacitors having capacitances of 8.4, 8.4, and 4.2 \(\mu\)F are connected in series across a 36-V potential difference. (a) What is the charge on the 4.2-\(\mu\)F capacitor? (b) What is the total energy stored in all three capacitors? (c) The capacitors are disconnected from the potential difference without allowing them to discharge. They are then reconnected in parallel with each other, with the positively charged plates connected together. What is the voltage across each capacitor in the parallel combination? (d) What is the total energy now stored in the capacitors?

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