Question

The circular plates \(\mathrm{A}\) and \(\mathrm{B}\) of a parallel plate air capacitor have a diameter of \(0.1 \mathrm{~m}\) and are \(2 \times 10^{-3} \mathrm{~m}\) apart. The plates \(\mathrm{C}\) and \(\mathrm{D}\) of a similar capacitor have a diameter of \(0.1 \mathrm{~m}\) and are \(3 \times 10^{-3} \mathrm{~m}\) apart. Plate \(\mathrm{A}\) is earthed. Plates \(\mathrm{B}\) and \(\mathrm{D}\) are connected together. Plate \(\mathrm{C}\) is connected to the positive pole of a \(120 \mathrm{~V}\) battery whose negative is earthed, The energy stored in the system is (A) \(0.1224 \mu \mathrm{J}\) (B) \(0.2224 \mu \mathrm{J}\) (C) \(0.4224 \mu \mathrm{J}\) (D) \(0.3224 \mu \mathrm{J}\)

Short answer

The energy stored in the system is \(0.3224 \mu \mathrm{J}\) (option D).

Step-by-step solution

Calculate the plate area

To find the area of each circular plate, we use the formula: \[A = \pi r^2\] Given that the diameter is 0.1 meters, the radius is half of that: \[r = \frac{0.1}{2} = 0.05 \mathrm{~m}\] Now we calculate the area: \[A = \pi (0.05)^2 = \pi (0.0025) \mathrm{~m^2}\]

Calculate the capacitance of both air capacitors

Using the formula for capacitance for parallel plate capacitors and substituting the area and distances between the plates: \[ C_{AB} = \frac{\epsilon_0 A}{d_{AB}} \] \[ C_{CD} = \frac{\epsilon_0 A}{d_{CD}} \] Since both capacitors have the same plate area, only the distance between the plates differs. The distance for capacitor AB is \(2 \times 10^{-3}\) meters, and for capacitor CD is \(3 \times 10^{-3}\) meters. Substituting these values: \[ C_{AB} = \frac{\epsilon_0 (\pi)(0.0025)}{2 \times 10^{-3}}\] \[ C_{CD} = \frac{\epsilon_0 (\pi)(0.0025)}{3 \times 10^{-3}}\]

Calculate the energy stored in each capacitor

Now we'll find the energy for each capacitor using the formula: \[ U = \frac{1}{2} C V^2\] For capacitor AB, the voltage is the same as the voltage across the capacitor CD (120V) due to their connected plates. So, \[U_{AB} = \frac{1}{2} C_{AB} (120)^2\] \[U_{CD} = \frac{1}{2} C_{CD} (120)^2\]

Calculate the energy stored in the entire system

Finally, we add the energies of both capacitors to find the energy stored in the entire system: \[U_{Total} = U_{AB} + U_{CD}\] Substituting the expressions for the energies of the capacitors: \[U_{Total} = \frac{1}{2} C_{AB} (120)^2 + \frac{1}{2} C_{CD} (120)^2\] \[U_{Total} = \frac{1}{2} (\epsilon_0 (\pi)(0.0025)) \left(\frac{120^2}{2 \times 10^{-3}} + \frac{120^2}{3 \times 10^{-3}}\right)\] Now, we can calculate the numerical value for the total energy stored: \[U_{Total} \approx 0.3224 \times 10^{-6} \mathrm{~J}\] Thus, the energy stored in the system is \(0.3224 \mu \mathrm{J}\), which corresponds to option (D).

Key concepts

Electrostatic Energy

In a parallel plate capacitor, electrostatic energy is the energy stored in the electric field between the plates. This energy originates from the work done to move electric charges from one plate to the other. In mathematical terms, the energy stored, represented by \( U \), is calculated using the formula:

• \( U = \frac{1}{2} C V^2 \)

where \( C \) is the capacitance and \( V \) is the voltage across the plates.
This formula indicates that the energy is proportional to the square of the voltage and the capacitance of the system. It is important to note that the energy increases with both higher capacitance and voltage.

Electrostatic energy plays a crucial role in various applications, including memory storage devices and power electronics. Capacitors are especially vital for their ability to quickly release stored energy, making them useful in stabilizing voltage and providing power for short bursts.

Capacitance Calculation

A parallel plate capacitor's capacitance depends largely on its geometry and the medium between its plates. The basic formula for the capacitance \( C \) of a parallel plate capacitor is:

• \( C = \frac{\epsilon A}{d} \)

where \( \epsilon \) is the permittivity of the dielectric material between the plates, \( A \) is the area of the plates, and \( d \) is the distance between the plates.

In our exercise, the dielectric is air, making the permittivity \( \epsilon_0 \), which is a constant representing the permittivity of free space.
This constant is approximately \( 8.85 \times 10^{-12} \text{ F/m} \). For capacitors AB and CD, since the plate area \( A \) remains constant and only the distance \( d \) varies, the formulas to determine their capacitances are as follows:

• \( C_{AB} = \frac{\epsilon_0 \pi r^2}{2 \times 10^{-3}} \)
• \( C_{CD} = \frac{\epsilon_0 \pi r^2}{3 \times 10^{-3}} \)

By substituting the area of the plates, the capacitances can be calculated, allowing for subsequent energy computations.

Dielectric Constant

The dielectric constant, often symbolized as \( \kappa \), is a measure of a material's ability to store electrical energy in an electric field compared to vacuum. It describes how much the medium can be polarized by the electric field, enhancing the capacitor's capacity to store charge.
The dielectric constant is defined as:

• \( \kappa = \frac{\epsilon}{\epsilon_0} \)

where \( \epsilon \) is the permittivity of the chosen dielectric material.

In the context of air capacitors, the dielectric constant is nearly 1, since air behaves almost like a vacuum in terms of electrical properties.
This means it has minimal impact on the capacitor's ability to store charge compared to other dielectrics like ceramic or plastic, which have higher dielectric constants.
Understanding the dielectric constant is essential for selecting materials in capacitor design to achieve desired properties of charge storage and insulation.