Question

A parallel-plate air capacitor is made by using two plates 12 cm square, spaced 3.7 mm apart. It is connected to a 12-V battery. (a) What is the capacitance? (b) What is the charge on each plate? (c) What is the electric field between the plates? (d) What is the energy stored in the capacitor? (e) If the battery is disconnected and then the plates are pulled apart to a separation of 7.4 mm, what are the answers to parts (a)-(d)?

Short answer

(a) 28.64 pF; (b) 343.68 pC; (c) 3.24 kV/m; (d) 2.06 μJ; (e) (a) 14.32 pF, (b) 343.68 pC, (c) 6.48 kV/m, (d) 4.12 μJ.

Step-by-step solution

Calculating Capacitance (Initial Setup)

First, we need to find the capacitance of the initial setup where the separation is 3.7 mm. The formula for the capacitance of a parallel-plate capacitor is \( C = \frac{\varepsilon_0 \cdot A}{d} \), where \( \varepsilon_0 \) is the permittivity of free space \( (8.85 \times 10^{-12} \, \text{F/m}) \), \( A \) is the area of one plate, and \( d \) is the separation between the plates. Here, \( A = (0.12)^2 \) m\(^2\), and \( d = 0.0037 \) m. Substituting these values in, we calculate \( C \).

Calculating Initial Charge on Plates

Using the capacitance \( C \) found in the previous step, we can find the charge \( Q \) on each plate using the relation \( Q = C \cdot V \), where \( V = 12 \) V is the battery voltage. Substitute the values to find \( Q \).

Calculating Electric Field Between Plates (Initial Setup)

The electric field \( E \) between the plates can be calculated using \( E = \frac{V}{d} \), where \( V = 12 \) V is the voltage and \( d = 0.0037 \) m is the initial separation between the plates. Compute \( E \) using these values.

Calculating Initial Energy Stored

The energy \( U \) stored in the capacitor is given by \( U = \frac{1}{2} C V^2 \). Using the capacitance from step 1 and the voltage \( V = 12 \) V, calculate \( U \).

Capacitance after Battery Disconnection and Plate Separation

Now, with the battery disconnected and the plates pulled to \( d = 0.0074 \) m, recalculate capacitance \( C = \frac{\varepsilon_0 \, A}{d} \) using the new separation \( d = 0.0074 \) m. The area \( A \) remains the same.

Charge After Changing Plate Separation

Since the battery is disconnected, the charge \( Q \) remains the same as initially calculated. Note that \( Q \) does not change after the separation.

Electric Field After Changing Plate Separation

With the charge constant, recalculate the electric field using \( E = \frac{Q}{\varepsilon_0 \, A} \). Substitute the unchanged charge \( Q \), \( \varepsilon_0 \), and the area \( A \) to find the new \( E \).

Energy Stored After Changing Plate Separation

The energy in the capacitor changes because the capacitance changes. Calculate the new energy using \( U = \frac{Q^2}{2C} \) with the new capacitance from step 5 and the unchanged charge \( Q \).

Key concepts

Parallel-plate capacitor

A parallel-plate capacitor is one of the simplest types of capacitors, consisting of two conductive plates separated by a thin insulation layer known as the dielectric. In our exercise, the plates are separated by air, which serves as the dielectric medium. The function of this setup is to store electric energy when connected to a power source such as a battery. The stored energy depends on the area of the plates, the separation distance between them, and the dielectric material used.
The idea is that when a voltage is applied, one plate accumulates a positive charge and the other a negative charge, creating an electric field between them. With a parallel-plate capacitor, the larger the surface area of the plates and the smaller the distance between them, the more energy it can store. This makes the design quite efficient for several applications, including energy storage, filtering, and in various electronic circuits.

Capacitance calculation

Capacitance is a measure of a capacitor's ability to store charge. For a parallel-plate capacitor, the capacitance is calculated using the formula: \[ C = \frac{\varepsilon_0 \cdot A}{d} \] where \( C \) is the capacitance, \( \varepsilon_0 \) is the permittivity of free space \((8.85 \times 10^{-12} \, \text{F/m})\), \( A \) is the area of one plate, and \( d \) is the separation between the plates.
For the given problem, the area \( A \) is \((0.12)^2\) square meters and the initial separation \( d \) is 0.0037 meters. By substituting these values into the formula, you can calculate the initial capacitance.
This capacity reflects how much charge the plates can hold for every volt of electric potential difference applied, establishing a core function of capacitors in storing and releasing charge as needed in circuits.

Electric field between plates

The electric field \( E \) between the plates of a parallel-plate capacitor is a uniform field, calculated using the formula: \[ E = \frac{V}{d} \] where \( V \) is the voltage applied across the plates, and \( d \) is the distance between them.
For the initial setup with a 12-V battery and a separation of 0.0037 meters, this formula helps compute the strength of the electric field. The electric field represents the force per unit charge that would be experienced by a charge placed within it.
When the separation is increased while keeping the charge constant (as when the battery is disconnected), the voltage changes and consequently, the electric field can be recalibrated. This variability is essential in understanding capacitor behavior under different physical configurations.

Energy stored in capacitor

The energy stored in a capacitor can be calculated by the equation: \[ U = \frac{1}{2} C V^2 \] where \( U \) is the stored energy, \( C \) is the capacitance, and \( V \) is the voltage across the capacitor. In our scenario, this formula allows us to calculate how much energy is stored initially with the given capacitance and voltage.
Upon disconnecting the battery and altering the plate separation, the voltage across the capacitor changes, although the charge remains the same. The energy stored is then recalculated with the modified capacitance using the alternative expression: \[ U = \frac{Q^2}{2C} \] This highlights how energy is influenced by changes in configuration and the potential energy shift due to altered geometric or material properties. Understanding these concepts emphasizes the dynamic nature of energy storage in capacitors, vital for managing power in various applications.