Question

A parallel plate capacitor with a plate area of \(12.0 \mathrm{~cm}^{2}\) and air in the space between the plates, which are separated by \(1.50 \mathrm{~mm}\), is connected to a 9.00-V battery. If the plates are pulled back so that the separation increases to \(2.75 \mathrm{~mm}\), how much work is done?

Short answer

Answer: To find the work done, follow these steps: 1. Calculate the capacitance before pulling the plates apart, \(C_1\). 2. Calculate the capacitance after pulling the plates apart, \(C_2\). 3. Calculate the energies stored in the capacitor before and after pulling the plates apart, \(U_1\) and \(U_2\). 4. Calculate the work done, \(W = U_2 - U_1\). Substitute the given values into the equations from the solution steps, and find the work done.

Step-by-step solution

Find the capacitance before pulling the plates apart

The capacitance of a parallel plate capacitor can be calculated using the formula: $$ C = \epsilon_0 \frac{A}{d} $$ where \(C\) is the capacitance, \(\epsilon_0\) is the vacuum permittivity (\(\approx 8.85 x 10^{-12} F/m\)), \(A\) is the plate area, and \(d\) is the plate separation. Given the initial plate separation of \(d_1=1.50\,mm\) and the plate area \(A=12.0\,cm^2\), we can find the initial capacitance, \(C_1\): $$ C_1 = \epsilon_0\frac{A}{d_1} = (\varepsilon_0) \left(\frac{0.12\,\mathrm{m}^2}{0.0015\,\mathrm{m}}\right) $$

Calculate the capacitance after pulling the plates apart

Now, we need to calculate the new capacitance, \(C_2\), when the plate separation increases to \(d_2=2.75\,mm\): $$ C_2 = \epsilon_0\frac{A}{d_2} = \left( \varepsilon_0 \right) \left(\frac{0.12\,\mathrm{m}^2}{0.00275\,\mathrm{m}}\right) $$

Calculate the energies in the capacitor before and after pulling the plates apart

The energy stored in a capacitor can be calculated using the formula: $$ U = \frac{1}{2}CV^2 $$ where \(U\) is the energy stored in the capacitor, and \(V\) is the voltage across the capacitor. In our case, the voltage remains constant at \(9.00 V\). Before pulling the plates apart, the energy stored in the capacitor, \(U_1\), is: $$ U_1=\frac{1}{2}C_1 V^2 $$ After pulling the plates apart, the energy stored in the capacitor, \(U_2\), is: $$ U_2=\frac{1}{2}C_2 V^2 $$

Calculate the work done

The work done, \(W\), is the difference in the energies stored in the capacitor before and after pulling the plates apart: $$ W = U_2 - U_1 = \frac{1}{2}C_2V^2 - \frac{1}{2}C_1V^2 $$ Calculate the numerical values for \(C_1\) and \(C_2\) using the equations from Steps 1 and 2, and substitute them into the equation above to find the work done.

Key concepts

Capacitance Calculation

Capacitance is a critical concept in the study of electric fields and circuits, representing a device's capacity to store electrical charge. For a parallel plate capacitor, which consists of two conducting plates separated by an insulating material, the capacitance is determined using the formula:

c = \frac{\varepsilon_0 A}{d}, where we have:

• \(C\) as the capacitance,
• \(\varepsilon_0\) as the vacuum permittivity, roughly \(8.85 \times 10^{-12} F/m\),
• \(A\) as the plate area, and
• \(d\) as the distance between the plates.

Using the given dimensions of the capacitor and rearranging our initial plate area from square centimeters to square meters, we can calculate the initial capacitance before the plates are pulled apart. If the separation changes, as in the exercise where it increases, the capacitance will decrease since it is inversely proportional to the distance between the plates. This fundamental relationship helps to understand how modifications in the capacitor's physical arrangement affect its ability to store electric charge.

Energy Stored in a Capacitor

Understanding the energy stored in a capacitor is essential for many areas of electronics and physics. The energy, often denoted as \(U\), is the amount of electrical potential energy saved within the capacitor due to the accumulated charge. The formula to calculate the energy stored in a parallel plate capacitor is given by:

\(U = \frac{1}{2}CV^2\), where:

• \(U\) is the energy,
• \(C\) is the capacitance, and
• \(V\) is the voltage across the capacitor.

When the voltage applied to the capacitor remains constant, as in our example, any change in capacitance will result in a change in stored energy. Specifically, an increase in plate separation reduces capacitance, and therefore, less energy is stored. By inserting the calculated capacitances before and after changing the distance between the plates into this formula, we can determine the energy stored in each scenario.

Work Done by Electric Forces

Work done by electric forces in the context of capacitors is a crucial aspect of electromechanics. The work is the energy required to change the separation between the plates of a capacitor when there is electric potential present. In our exercise, the work done is the energy difference between the initial and final states of the capacitor:

\(W = U_2 - U_1\), which simplifies to \(W = \frac{1}{2}C_2V^2 - \frac{1}{2}C_1V^2\).Here, \(W\) stands for work done, \(U_1\) and \(U_2\) for the initial and final stored energies respectively. Voltage \(V\) remains unchanged as the plates are separated. This equation encapsulates the concept that work done by the electric forces is equivalent to the change in energy within the capacitor system. To find the amount of work done, we substitute our previously calculated values for the initial and final capacitances into the work equation, knowing that the increase in plate separation results in energy being expended, as the capacitor's ability to store charge is diminished. The result of this calculation reveals how much work was done during the process.