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 \(9.00-\mathrm{V}\) battery. If the plates are pulled back so that the separation increases to \(2.75 \mathrm{~mm}\) how much work is done?

Short answer

Based on the given problem, the work done in pulling the plates of the capacitor apart is \(7.61\times10^{-10}\;\mathrm{J}\).

Step-by-step solution

Calculate the initial capacitance

We will use the formula for the capacitance of a parallel plate capacitor: \(C = \frac{\epsilon_0A}{d}\) where \(C\) is capacitance, \(\epsilon_0\) is the vacuum permittivity, \(A\) is the plate area, and \(d\) is the separation between the plates. Given values: \(A = 12.0 \mathrm{~cm}^{2} = 1.2\times10^{-3} \mathrm{~m}^{2}\) \(d_{1} = 1.50 \mathrm{~mm} = 1.50\times10^{-3} \mathrm{~m}\) So, the initial capacitance is: \(C = \frac{(8.854\times10^{-12} \mathrm{~F/m})\times(1.2\times10^{-3} \mathrm{~m}^{2})}{1.50\times10^{-3} \mathrm{~m}} = 7.084\times10^{-12} \mathrm{~F}\)

Calculate the initial charge on the capacitor

With the initial capacitance, we can find the initial charge stored on the capacitor using the formula: \(Q = CV\) where \(Q\) is charge, \(C\) is capacitance, and \(V\) is voltage. Given voltage, \(V = 9.00\; \mathrm{V}\), we can compute the initial charge: \(Q = (7.084\times10^{-12}\;\mathrm{F})(9.00\;\mathrm{V}) = 63.76\times10^{-12}\;\mathrm{C}\)

Calculate the final voltage across the capacitor

The charge on a capacitor is conserved when it is disconnected from any voltage source. Thus, as the plates are pulled apart, the charge remains the same. We can calculate the final capacitance and voltage as the separation distance increases. Given final separation distance: \(d_{2} = 2.75 \mathrm{~mm} = 2.75\times10^{-3} \mathrm{~m}\) The final capacitance is: \(C_{2} = \frac{(8.854\times10^{-12} \mathrm{~F/m})\times(1.2\times10^{-3} \mathrm{~m}^{2})}{2.75\times10^{-3} \mathrm{~m}} = 3.863\times10^{-12} \mathrm{~F}\) The final voltage across the capacitor when the separation increases is: \(V_{2} = \frac{Q}{C_{2}}\) \(V_{2} = \frac{63.76\times10^{-12}\;\mathrm{C}}{3.863\times10^{-12} \mathrm{~F}} = 16.51\; \mathrm{V}\)

Calculate the work done in pulling the plates apart

Now, we will calculate the change in the energy stored in the capacitor and find the work done. The energy stored in a capacitor is: \(E=\frac{1}{2}CV^2\) The initial energy stored in the capacitor is: \(E_{1}=\frac{1}{2}(7.084\times10^{-12}\;\mathrm{F})(9.00\;\mathrm{V})^2 = 2.874\times10^{-9}\;\mathrm{J}\) The final energy stored in the capacitor after separation increases is: \(E_{2}=\frac{1}{2}(3.863\times10^{-12}\;\mathrm{F})(16.51\;\mathrm{V})^2 = 2.113\times10^{-9}\;\mathrm{J}\) Since the battery is connected to the capacitor, we must consider the energy change in the battery as well. The work done on the plate can be expressed as: \(W = E_2 - E_1\) \(W = 2.113\times10^{-9}\;\mathrm{J} - 2.874\times10^{-9}\;\mathrm{J} = -7.61\times10^{-10}\;\mathrm{J}\) The result is negative since the energy used to pull again the capacitor's force is stored back into the battery due to its negative terminal. Hence, the work done in pulling the plates of the capacitor apart is \(7.61\times10^{-10}\;\mathrm{J}\).

Key concepts

Capacitance

Capacitance is a fundamental concept in the study of capacitors, which are devices that store electrical energy. It concerns the ability of a system to store an electric charge. For a parallel plate capacitor, which consists of two conductive plates separated by an insulating material (dielectric), the capacitance is calculated using the formula:
\[ C = \frac{\epsilon_0 A}{d} \]where:

• \( C \) is the capacitance in farads (F)
• \( \epsilon_0 \) is the vacuum permittivity, a constant equal to \( 8.854 \times 10^{-12} \; \text{F/m} \)
• \( A \) is the area of one of the plates in square meters (\( \text{m}^2 \))
• \( d \) is the distance between the plates in meters (m)

This formula indicates that larger plate areas and smaller distances between the plates result in higher capacitance, allowing more electric charge to be stored. In the given exercise, the initial and final separations of the plates influence the capacitance, showcasing direct application of this concept.

Electric Charge

Electric charge in a capacitor refers to the quantity of electric charge stored on the plates when a voltage is applied across them. It is determined by the formula:
\[ Q = CV \]Here, \( Q \) is the charge in coulombs (C), \( C \) is the capacitance in farads, and \( V \) is the voltage in volts (V). When the plates are connected to a battery, as in the given scenario, they accumulate charges of equal magnitude and opposite sign.
As per the original exercise, the charge stays constant when the capacitor is disconnected from the battery. However, when increasing the separation between the charged plates, the capacitance decreases, but the stored charge remains unaltered. This conservation of charge is crucial, as it directly affects the resulting voltage across the capacitor when plate separation changes.

Work Done on Capacitor

The work done on a capacitor when its configuration is changed, such as increasing the separation between parallel plates, is related to the change in energy stored in the capacitor. The energy stored in a capacitor is given by:
\[ E = \frac{1}{2} C V^2 \]This formula illustrates that the energy output is directly proportional to the capacitance and the square of the voltage across it. In the context of the exercise, the initial and final energies are calculated using the respective initial and final capacitances and voltages.
The work done is essentially the change in the stored energy, represented as:
\[ W = E_2 - E_1 \]where \( E_1 \) is the initial energy and \( E_2 \) is the final energy. The negative result shown in the exercise indicates that work is done against the electric force when separating the plates, and energy is taken from the system, which can be stored elsewhere, such as in the connected battery. The computation of work done helps us understand the energy transformations occurring within the capacitor system during such physical alterations.