Question

An air capacitor is made from two flat parallel plates 1.50 mm apart. The magnitude of charge on each plate is 0.0180 \(\mu\)C when the potential difference is 200 V. (a) What is the capacitance? (b) What is the area of each plate? (c) What maximum voltage can be applied without dielectric breakdown? (Dielectric breakdown for air occurs at an electric-field strength of 3.0 \(\times\) 10\(^6\) V/m.) (d) When the charge is 0.0180 \(\mu\)C, what total energy is stored?

Short answer

(a) 9.0 × 10^-11 F, (b) 0.0153 m², (c) 4500 V, (d) 1.8 × 10^-6 J

Step-by-step solution

Calculate the Capacitance

The capacitance of a parallel plate capacitor can be found using the formula \( C = \frac{Q}{V} \), where \( Q \) is the charge, and \( V \) is the potential difference. Given \( Q = 0.0180 \mu C = 0.0180 \times 10^{-6} \) C and \( V = 200 \) V, the capacitance \( C \) is calculated as:\[ C = \frac{0.0180 \times 10^{-6} \text{ C}}{200 \text{ V}} = 9.0 \times 10^{-11} \text{ F} \]

Calculate the Plate Area

The capacitance for a parallel plate capacitor is also given by the formula \( C = \frac{\varepsilon_0 A}{d} \), where \( \varepsilon_0 \) is the permittivity of free space \( (8.85 \times 10^{-12} \text{ F/m}) \), \( A \) is the area, and \( d \) is the separation between plates. Rearrange the formula to find \( A \):\[ A = \frac{Cd}{\varepsilon_0} \]Substituting \( C = 9.0 \times 10^{-11} \text{ F} \), \( d = 1.50 \times 10^{-3} \text{ m} \), and \( \varepsilon_0 = 8.85 \times 10^{-12} \text{ F/m} \), we get:\[ A = \frac{9.0 \times 10^{-11} \times 1.50 \times 10^{-3}}{8.85 \times 10^{-12}} \approx 0.0153 \text{ m}^2 \]

Calculate Maximum Voltage for Dielectric Breakdown

The maximum voltage \( V_{\text{max}} \) that can be applied is calculated using the formula for electric field \( E = \frac{V}{d} \). Given the breakdown field strength \( E = 3.0 \times 10^6 \) V/m and \( d = 1.50 \times 10^{-3} \text{ m} \), solve for \( V_{\text{max}} \):\[ V_{\text{max}} = E \times d = 3.0 \times 10^6 \times 1.50 \times 10^{-3} = 4500 \text{ V} \]

Calculate the Total Energy Stored

The energy \( U \) stored in a capacitor is given by \( U = \frac{1}{2} C V^2 \). Using the previously calculated \( C = 9.0 \times 10^{-11} \text{ F} \) and \( V = 200 \text{ V} \), find \( U \):\[ U = \frac{1}{2} \times 9.0 \times 10^{-11} \times (200)^2 = 1.8 \times 10^{-6} \text{ J} \]

Key concepts

Parallel Plate Capacitor

A parallel plate capacitor comprises two conductive plates separated by a certain distance. In this scenario, the plates are 1.50 mm apart. The capacitor stores electrical energy by maintaining opposite charges on the plates.

This setup enables the calculation of capacitance using the formula:

• Capacitance, \( C \), is given by \( C = \frac{Q}{V} \) where \( Q \) is the charge on the plates and \( V \) is the voltage across them.

Capacitance is a measure of a capacitor's ability to store charge per unit voltage. For example, with a charge of 0.0180 \(\mu\)C and a voltage of 200 V, the capacitance of our parallel plate capacitor is 9.0 x 10\(^-{11}\ \text{F}\).

Additionally, capacitance can be calculated using the formula \( C = \frac{\varepsilon_0 A}{d} \), where:

• \( \varepsilon_0 \) is the permittivity of free space, \( 8.85 \times 10^{-12} \text{ F/m} \)
• \( A \) is the plate area
• \( d \) is the distance separating the plates

This alternative method is particularly useful when you know the physical dimensions and need to connect them to how much charge can be stored.

Dielectric Breakdown

Dielectric breakdown is a critical phenomenon where the insulating material between capacitor plates fails to insulate, resulting in a sudden release of stored energy. This generally happens when a particular electric field strength is exceeded.

For air, the breakdown occurs at an electric field strength of 3.0 \( \times 10^6 \) V/m. Beyond this point, the air ceases to function as an effective insulator.

To ensure safe usage of a capacitor, it's crucial to determine the maximum voltage, \( V_{\text{max}} \), the capacitor can tolerate without breaking down. For the given example, it is calculated as:

• \( V_{\text{max}} = E \times d \)

Inserting the air breakdown threshold \( E = 3.0 \times 10^6 \) V/m and the separation \( d = 1.50 \times 10^{-3} \) m yields a maximum voltage of 4500 V.

Electric Field Strength

Electric field strength is a measure of force per unit charge in the field around charged particles.It is denoted by \( E \) and is usually expressed in volts per meter (V/m).

The electric field between the plates of a parallel plate capacitor can be calculated using:

• \( E = \frac{V}{d} \)

Where \( V \) is the potential difference across the plates and \( d \) is the distance between them.

In our example, maintaining an understanding of electric field strength helps avoid exceeding conditions for dielectric breakdown, as exceeding 3.0 \( \times 10^6 \) V/m will result in an electrical failure.

Electric field intensity is essential in designing capacitors ensuring they store energy efficiently without risking breakdown.

Energy Stored in a Capacitor

The energy stored in a capacitor is crucial for various applications where retraction of stored energy is needed. The energy is stored in the electric field between the plates, and the amount depends on both the capacitance and the square of the voltage.

The formula for energy stored, \( U \), in a capacitor is:

• \( U = \frac{1}{2} C V^2 \)

Where:

• \( C \) is the capacitance
• \( V \) is the voltage across the capacitor

For the air capacitor with a charge of 0.0180 \mu C and a potential difference of 200 V, the energy stored turns out to be \( 1.8 \times 10^{-6} \) J.

This value portrays the potential of the capacitor as an energy reservoir - vital for applications like flash photography and power circuitry, where energy needs to be stored temporarily and then released rapidly.