Question

A 0.350-m-long cylindrical capacitor consists of a solid conducting core with a radius of 1.20 mm and an outer hollow conducting tube with an inner radius of 2.00 mm. The two conductors are separated by air and charged to a potential difference of 6.00 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

(a) Charge per length \( \lambda \approx 2.68 \times 10^{-11} \, C/m \). (b) Total charge \( Q \approx 9.38 \times 10^{-12} \, C \). (c) Capacitance \( C \approx 1.56 \times 10^{-12} \, F \). (d) Energy stored \( U \approx 2.81 \times 10^{-11} \, J \).

Step-by-step solution

Identify the formula for voltage difference (V) in a cylindrical capacitor

In a cylindrical capacitor, the electric potential difference \( V \) between the core and outer tube is given by the formula \[ V = \frac{\lambda}{2\pi\varepsilon_0} \ln\left(\frac{r_2}{r_1}\right) \] where \( \lambda \) is the charge per unit length, \( \varepsilon_0 \) is the permittivity of free space \((\approx 8.85 \times 10^{-12} \, C^2/N\cdot m^2)\), \( r_1 \) is the radius of the inner core, and \( r_2 \) is the radius of the outer hollow conductor.

Calculate charge per length (\( \lambda \))

Rearrange the formula from Step 1 to solve for charge per unit length \( \lambda \): \[ \lambda = \frac{2\pi\varepsilon_0 V}{\ln(r_2/r_1)} \] Substitute \( V = 6.00 \, V \), \( r_1 = 1.20 \times 10^{-3} \, m \), \( r_2 = 2.00 \times 10^{-3} \, m \) and \( \varepsilon_0 = 8.85 \times 10^{-12} \, C^2/N\cdot m^2 \):\[ \lambda = \frac{2\pi (8.85 \times 10^{-12})(6.00)}{\ln(2.00/1.20)} \approx 2.68 \times 10^{-11} \, C/m \].

Calculate the total charge (Q) on the capacitor

The total charge \( Q \) is the charge per unit length \( \lambda \) times the length of the capacitor \( L \):\[ Q = \lambda \times L \] Given \( L = 0.350 \, m \), and using the \( \lambda \) calculated:\[ Q = (2.68 \times 10^{-11} \, C/m)(0.350 \, m) \approx 9.38 \times 10^{-12} \, C \].

Determine the capacitance (C)

The capacitance of a cylindrical capacitor is given by:\[ C = \frac{2\pi\varepsilon_0 L}{\ln(r_2/r_1)} \]Substitute the given values \( L = 0.350 \, m \), \( r_1 = 1.20 \times 10^{-3} \, m \), \( r_2 = 2.00 \times 10^{-3} \, m \), and \( \varepsilon_0 = 8.85 \times 10^{-12} \, C^2/N\cdot m^2 \):\[ C = \frac{2\pi (8.85 \times 10^{-12})(0.350)}{\ln(2.00/1.20)} \approx 1.56 \times 10^{-12} \, F \].

Calculate the energy stored (U)

The energy stored in the capacitor when fully charged can be calculated using:\[ U = \frac{1}{2} C V^2 \]Substitute \( C = 1.56 \times 10^{-12} \, F \) and \( V = 6.00 \, V \):\[ U = \frac{1}{2} (1.56 \times 10^{-12})(6.00^2) \approx 2.81 \times 10^{-11} \, J \].

Key concepts

Charge per Length in a Cylindrical Capacitor

In the context of capacitors, especially cylindrical ones, it is important to understand the term "charge per length." This refers to how much electric charge is distributed along a given length of the cylindrical surface. The symbol \( \lambda \) is commonly used to represent this quantity.

The cylindrical capacitor features a solid core and an outer hollow tube. To find the charge per length \( \lambda \), we use the formula:

• \[ \lambda = \frac{2\pi\varepsilon_0 V}{\ln(r_2/r_1)} \]

In this formula:

• \( \varepsilon_0 \) signifies the permittivity of free space, approximately \( 8.85 \times 10^{-12} \, C^2/N\cdot m^2 \).
• \( V \) is the potential difference between the core and tube.
• \( r_1 \) and \( r_2 \) are the radii of the inner core and outer tube, respectively.

To solve an example from a given problem, you might substitute specific values such as \( V = 6.00 \, V \), \( r_1 = 1.20 \times 10^{-3} \, m \), and \( r_2 = 2.00 \times 10^{-3} \, m \). This gives a charge per length \[ \lambda \approx 2.68 \times 10^{-11} \, C/m \].

Understanding charge per length lays a foundation for calculating total charge and other properties of capacitors.

Capacitance of a Cylindrical Capacitor

Capacitance is a measure of the ability of a capacitor to store an electrical charge. In cylindrical capacitors, it depends on the geometry and material properties. For a cylindrical capacitor, the formula for capacitance \( C \) is:

• \[ C = \frac{2\pi\varepsilon_0 L}{\ln(r_2/r_1)} \]

Here:

• \( L \) is the length of the cylindrical capacitor.
• Again, \( r_1 \) and \( r_2 \) are the radii of the inner and outer parts, respectively.
• \( \varepsilon_0 \) is the permittivity of free space, as before.

Using our example values, such as \( L = 0.350 \, m \), you calculate the capacitance to be approximately \[ C \approx 1.56 \times 10^{-12} \, F \].

The capacitance indicates how much charge the capacitor can hold when a particular voltage is applied. This makes capacitance a central quantity in the analysis of electrical circuits.

Energy Stored in a Capacitor

The energy stored in a capacitor is crucial for understanding how capacitors function in circuits. This stored energy can be calculated using the formula:

• \[ U = \frac{1}{2} C V^2 \]

Where:

• \( C \) is the capacitance found earlier, \( 1.56 \times 10^{-12} \, F \) in our drumroll example.
• \( V \) is the potential difference, given as \( 6.00 \, V \).

Substituting these values into the formula, the energy stored \( U \) is \[ 2.81 \times 10^{-11} \, J \].

This energy is essentially the potential energy that can be released into a circuit to do work, such as lighting a bulb or powering a device. It highlights the utility of capacitors in storing and releasing energy quickly when needed.