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) \(0.3224 \mu \mathrm{J}\)

Step by step solution

01

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: \[A = \pi (0.05)^2 = \pi (0.0025) \mathrm{~m^2}\]

02

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: \[ C_{AB} = \frac{\epsilon_0 (\pi)(0.0025)}{2 \times 10^{-3}}\] \[ C_{CD} = \frac{\epsilon_0 (\pi)(0.0025)}{3 \times 10^{-3}}\]

03

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, \[U_{AB} = \frac{1}{2} C_{AB} (120)^2\] \[U_{CD} = \frac{1}{2} C_{CD} (120)^2\]

04

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).

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