A parallel plate capacitor has air between the plates, which are circular with radius 0.07 m. The separation gap is 0.375 mm. What is the maximum energy that can be stored in this capacitor if air-breakdown is to be avoided? The breakdown threshold field is approximately 3 × 10⁶ N/C.

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The maximum energy stored in a capacitor without causing air breakdown can be calculated using the formula for the energy stored in a capacitor, $U = \frac{1}{2}CV^2$, where $C$ is the capacitance and $V$ is the voltage. The breakdown voltage $V_{bd}$ can be found from the breakdown field $E_{bd}$ and the separation $d$, using $V_{bd} = E_{bd} \times d$. The capacitance $C$ of a parallel plate capacitor is given by $C = \epsilon_0 \frac{A}{d}$, where $\epsilon_0$ is the permittivity of free space ($8.85 \times 10^{-12} \, \text{F/m}$), $A$ is the area of the plates, and $d$ is the separation between the plates. The area $A$ of a circular plate is $\pi r^2$, where $r$ is the radius of the plate.Given:- Radius $r = 0.07 \, \text{m}$- Separation $d = 0.375 \times 10^{-3} \, \text{m}$ (converted from mm to m)- Breakdown threshold field $E_{bd} = 3 \times 10^6 \, \text{N/C}$First, calculate the breakdown voltage $V_{bd}$:$$V_{bd} = E_{bd} \times d = 3 \times 10^6 \times 0.375 \times 10^{-3} = 1125 \, \text{V}$$Next, calculate the area $A$:$$A = \pi r^2 = \pi \times (0.07)^2 \approx 0.0154 \, \text{m}^2$$Then, calculate the capacitance $C$:$$C = \epsilon_0 \frac{A}{d} = 8.85 \times 10^{-12} \times \frac{0.0154}{0.375 \times 10^{-3}} \approx 3.65 \times 10^{-10} \, \text{F}$$Finally, calculate the maximum energy $U$:$$U = \frac{1}{2}CV^2 = \frac{1}{2} \times 3.65 \times 10^{-10} \times (1125)^2 \approx 2.3 \times 10^{-4} \, \text{J}$$Therefore, the maximum energy that can be stored in the capacitor without causing air breakdown is approximately $2.3 \times 10^{-4} \, \text{J}$, which corresponds to choice B.

A Parallel Plate Capacitor Has Air Between the Plates, Which