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A \(3.350-\mathrm{m}\) -long cylindrical capacitor consists of a solid conducting core with a radius of 1.20 \(\mathrm{mm}\) and an outer hollow conducting tube with an inner radius of 2.00 \(\mathrm{mm}\) . The two conductors are separated by air and charged to a potential difference of 6.00 \(\mathrm{V}\) . Calculate (a) the charge per length for the capacitor; (b) the total charge on the capacitor; (c) the capacitance; (d) the energy stored in the capacitor when fully charged.

Short Answer

Expert verified
The charge per length is found using electric field integral; total charge involves multiplying by length. Capacitance is derived from geometric formula, and energy follows utilizing obtained capacitance.

Step by step solution

01

Finding Charge per Length

To find the charge per unit length (\(\lambda\)) of the cylindrical capacitor, we use the formula for the electric field between two cylindrical conductors:\[E = \frac{\lambda}{2\pi\varepsilon_0 r}\]where \(\varepsilon_0\) is the permittivity of free space (\(8.85 \times 10^{-12} \, \text{C}^2/\text{N} \cdot \text{m}^2\)), and \(r\) is the radial distance from the axis, but since we have the potential difference (\(V\)) used across the capacitors, we use,\[V = \int_{r_1}^{r_2} E \, dr = \frac{\lambda}{2\pi\varepsilon_0} \ln \frac{r_2}{r_1}\]Given is \(V = 6\, \text{V}\), \( r_2 = 2 \times 10^{-3} \, \text{m} \), and \( r_1 = 1.2 \times 10^{-3} \, \text{m} \). Solve \[ 6 = \frac{\lambda}{2\pi(8.85 \times 10^{-12})} \ln \frac{2 \times 10^{-3}}{1.2 \times 10^{-3}} \] to get \( \lambda \).
02

Calculating Total Charge

The total charge (\(Q\)) on the cylinder can be found by multiplying the charge per unit length (\(\lambda\)) by the length of the capacitor (\(L\)),\[ Q = \lambda \cdot L \]Given length \( L = 3.350 \, \text{m} \), substitute \(\lambda\) to find \(Q\).
03

Determining Capacitance

The formula for the capacitance per unit length of a cylindrical capacitor is given by,\[ C' = \frac{2\pi\varepsilon_0}{\ln\left(\tfrac{r_2}{r_1}\right)} \]Thus, the total capacitance \(C\) is \[ C = C' \cdot L \]Substitute the known values to find \(C\).
04

Calculating Stored Energy

The energy stored (\(U\)) in a capacitor is given by,\[ U = \frac{1}{2} C V^2 \]Substituting the capacitance \(C\) and the potential difference \(V = 6\, \text{V}\), compute \(U\).

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Key Concepts

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

Charge per Unit Length
In a cylindrical capacitor setup, understanding the charge distribution is crucial. Here, the charge per unit length is denoted by \( \lambda \). It represents how much charge is stored for each meter of the capacitor's length.
For a cylindrical capacitor, the electric field \( E \) between the cylinders is directly related to this charge to length ratio. By integrating along the path from the inner cylinder to the outer cylinder, we relate the electric field to the potential difference \( V \) between the two cylinders:
  • \( E = \frac{\lambda}{2\pi\varepsilon_0 r} \)
  • \( V = \int_{r_1}^{r_2} E \, dr = \frac{\lambda}{2\pi\varepsilon_0} \ln \frac{r_2}{r_1} \)
By knowing the potential difference, we can rearrange this relationship to solve for \( \lambda \). The given parameters like the radii \( r_1 \) and \( r_2 \), and the voltage \( V \), allow us to calculate the charge per unit length. This charge value is crucial for understanding the capacitor's ability to store charge over its length.
Capacitance Calculation
The capacitance of a cylindrical capacitor is a measure of its ability to store charge relative to the potential difference across it. For a cylindrical geometry, the capacitance per unit length \( C' \) can be calculated using:
  • \( C' = \frac{2\pi\varepsilon_0}{\ln\left(\frac{r_2}{r_1}\right)} \)
This formula derives from the relationship between electric field and charge distribution in cylindrical coordinates.
To obtain the total capacitance \( C \) of the entire cylinder, we multiply the capacitance per unit length by the total length \( L \) of the cylinder:
  • \( C = C' \cdot L \)
Thus, by substituting the known lengths and radii into these formulas, we determine the total capacitance, which reflects the cylinder's efficiency in storing electrical energy.
Energy Stored in Capacitor
The energy stored in a capacitor is an important characteristic because it dictates how much energy the device can provide when connected to a circuit. The energy \( U \) stored in the capacitor is given by:
  • \( U = \frac{1}{2} C V^2 \)
This expression comes from integrating the work done to charge the capacitor. Here, \( C \) is the total capacitance, and \( V \) is the potential difference applied across the capacitor.
By calculating the product of these values and considering the factor of \( \frac{1}{2} \), we get the total energy stored. This tells us the potential energy available for performing work in electrical circuits, hence crucial for assessing the capacitor's performance in practical scenarios.
Potential Difference
The potential difference \( V \) across a capacitor indicates how much work is done by an electric field to move a unit charge from one conductor to another.
This potential difference is a central part of what allows capacitors to store energy. In this cylindrical setup, the relationship between potential difference, charge, and capacitance is vividly highlighted in the formulation:
  • \( V = \int_{r_1}^{r_2} E \, dr = \frac{\lambda}{2\pi\varepsilon_0} \ln \frac{r_2}{r_1} \)
Here, the potential difference not only helps in determining charge distribution but is also essential to compute the stored energy. Knowing \( V \) assists in understanding the efficiency of the capacitor in releasing stored energy. This voltage is provided as a design specification and allows for assessing how capacitors interact with connected electrical circuits.

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Most popular questions from this chapter

A \(10.0-\mu \mathrm{F}\) parallel-plate capacitor with circular plates is connected to a 12.0 -V battery. (a) What is the charge on each plate? (b) How much charge would be on the plates if their separation were doubled while the capacitor remained connected to the battery? (c) How much charge would be on the plates if the capacitor were connected to the \(12.0-\mathrm{V}\) battery after the radius of each plate was doubled without changing their separation?

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.

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 \(\mathrm{mm}\) . (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 \(12.0 \mathrm{mm},\) 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 poster board? Explain.

A 10.0 -\muF parallel-plate capacitor is connected to a 12.0 -V battery. After the capacitor is fully charged, the battery is disconnected without loss of any of the charge on the plates. (a) A voltmeter is connected across the two plates without discharging them. What does it read? (b) What would the voltmeter read if (i) the plate separation were doubled; (ii) the radius of each plate were doubled but their separation was unchanged?

A parallel-plate capacitor with only air between the plates is charged by connecting it to a battery. The capacitor is then disconnected from the battery, without any of the charge leaving the plates. (a) A voltmeter reads 45.0 \(\mathrm{V}\) when placed across the capacitor. When a dielectric is inserted between the plates, completely filling the space, the voltmeter reads 11.5 \(\mathrm{V}\) . What is the dielectric constant of this material? (b) What will the voltmeter read if the dielectric is now pulled partway out so it fills only one-third of the space between the plates?

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