Question

A parallel plate capacitor is to be constructed which can store q = 10 µC charge at V = 1000 volt. The minimum plate area of the capacitor is required to be A₁ when space between the plates has air. If a dielectric of constant K = 3 is used between the plates the minimum plate area required to make such a capacitor is A₂. The breakdown field for the dielectric is 8 times that of air. Find $\frac{A_1}{7A_2}$.

Answer 1

24/7

Answer 2

Let $C$ be the required capacitance. $C = \frac{q}{V} = \frac{10 \times 10^{-6} \, C}{1000 \, V} = 10^{-8} \, F$.

For a parallel plate capacitor with plate area $A$ and separation $d$ with a dielectric of constant $K$, the capacitance is $C = \frac{\epsilon_0 K A}{d}$. The electric field between the plates is $E = \frac{V}{d}$. The dielectric can withstand a maximum electric field $E_{max}$ before breakdown. Thus, $E \le E_{max}$, which means $\frac{V}{d} \le E_{max}$. This gives a minimum allowed plate separation $d_{min} = \frac{V}{E_{max}}$.

To achieve the required capacitance $C$ with the minimum possible plate area $A_{min}$, we must use the minimum possible separation $d_{min}$. $C = \frac{\epsilon_0 K A_{min}}{d_{min}} = \frac{\epsilon_0 K A_{min}}{V/E_{max}}$. Solving for $A_{min}$: $A_{min} = \frac{C V}{\epsilon_0 K E_{max}}$.

Case 1: Air between plates ($K_1 = 1$). The minimum plate area is $A_1$. Let the breakdown field for air be $E_{max,1}$. $A_1 = \frac{C V}{\epsilon_0 K_1 E_{max,1}} = \frac{C V}{\epsilon_0 (1) E_{max,1}} = \frac{C V}{\epsilon_0 E_{max,1}}$.

Case 2: Dielectric between plates ($K_2 = 3$). The minimum plate area is $A_2$. Let the breakdown field for the dielectric be $E_{max,2}$. $A_2 = \frac{C V}{\epsilon_0 K_2 E_{max,2}}$.

We are given that the breakdown field for the dielectric is 8 times that of air, so $E_{max,2} = 8 E_{max,1}$.

Now, we find the ratio $\frac{A_1}{A_2}$: $\frac{A_1}{A_2} = \frac{\frac{C V}{\epsilon_0 E_{max,1}}}{\frac{C V}{\epsilon_0 K_2 E_{max,2}}} = \frac{K_2 E_{max,2}}{E_{max,1}}$. Substitute the values $K_2 = 3$ and $E_{max,2} = 8 E_{max,1}$: $\frac{A_1}{A_2} = \frac{3 \times (8 E_{max,1})}{E_{max,1}} = 3 \times 8 = 24$.

The question asks for the value of $\frac{A_1}{7A_2}$. $\frac{A_1}{7A_2} = \frac{1}{7} \times \frac{A_1}{A_2} = \frac{1}{7} \times 24 = \frac{24}{7}$.

The final answer is $\frac{24}{7}$.