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 $$V=\frac{\lambda }{2\pi {\epsilon }_0}\ln \left(\frac{r_2}{r_1}\right)$$ where $\lambda$ is the charge per unit length, ${\epsilon }_0$ 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 $\lambda$: $$\lambda =\frac{2\pi {\epsilon }_{0}V}{\ln (r_2/r_1)}$$ Substitute $V=6.00V$, $r_1=1.20\times {10}^{-3}m$, $r_2=2.00\times {10}^{-3}m$ and ${\epsilon }_0=8.85\times {10}^{-12}{C}^2/N\cdot m^2$:$$\lambda =\frac{2\pi (8.85\times {10}^{-12})(6.00)}{\ln (2.00/1.20)}\approx 2.68\times {10}^{-11}C/m$$.

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 $L=0.350m$, and using the $\lambda$ calculated:$$Q=(2.68\times {10}^{-11}C/m)(0.350m)\approx 9.38\times {10}^{-12}C$$.

Determine the capacitance (C)

The capacitance of a cylindrical capacitor is given by:$$C=\frac{2\pi {\epsilon }_{0}L}{\ln (r_2/r_1)}$$Substitute the given values $L=0.350m$, $r_1=1.20\times {10}^{-3}m$, $r_2=2.00\times {10}^{-3}m$, and ${\epsilon }_0=8.85\times {10}^{-12}{C}^2/N\cdot m^2$:$$C=\frac{2\pi (8.85\times {10}^{-12})(0.350)}{\ln (2.00/1.20)}\approx 1.56\times {10}^{-12}F$$.

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 $V=6.00V$:$$U=\frac{1}{2}(1.56\times {10}^{-12})({6.00}^2)\approx 2.81\times {10}^{-11}J$$.

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:

• $$\lambda =\frac{2\pi {\epsilon }_{0}V}{\ln (r_2/r_1)}$$

In this formula:

• ${\epsilon }_0$ 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 $V=6.00V$, $r_1=1.20\times {10}^{-3}m$, 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:

• $$C=\frac{2\pi {\epsilon }_{0}L}{\ln (r_2/r_1)}$$

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.
• ${\epsilon }_0$ is the permittivity of free space, as before.

Using our example values, such as $L=0.350m$, 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 $6.00V$.

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.