by step.

The capacitor has a dielectric constant $$\kappa = 5$$ and a dielectric strength of $$10^6 \mathrm{~V} \mathrm{~m}^{-1}$$. The voltage rating is $$2 \mathrm{~kV}$$ and the maximum electric field allowed is $$10 \%$$ of its dielectric strength.

First, calculate the maximum electric field:

$$E_{\text{max}} = 0.1 \times 10^6 \mathrm{~V} \mathrm{~m}^{-1} = 10^5 \mathrm{~V} \mathrm{~m}^{-1}$$

Given the voltage rating, the separation between the plates (d) can be found using the relation:

$$V = E_{\text{max}} \times d$$

Rearranging for $d$:

$$d = \frac{V}{E_{\text{max}}} = \frac{2000 \mathrm{~V}}{10^5 \mathrm{~V} \mathrm{~m}^{-1}} = 0.02 \mathrm{~m}$$

Next, we use the formula for the capacitance of a parallel plate capacitor:

$$C = \kappa \epsilon_0 \frac{A}{d}$$

Where:

$$C = 60 \mathrm{~pF} = 60 \times 10^{-12} \mathrm{~F}$$
$$\kappa = 5$$
$$\epsilon_0 = 8.85 \times 10^{-12} \mathrm{~F m}^{-1}$$
$$d = 0.02 \mathrm{~m}$$
$A = \text{Area of the plates}$

Rearranging for $A$:

$$A = \frac{C \times d}{\kappa \times \epsilon_0}$$

Substituting the values:

$$A = \frac{60 \times 10^{-12} \mathrm{~F} \times 0.02 \mathrm{~m}}{5 \times 8.85 \times 10^{-12} \mathrm{~F m}^{-1}} = \frac{60 \times 0.02}{5 \times 8.85} \times 10^{-12 + 12} \mathrm{~m}^2 = \frac{1.2}{44.25} \mathrm{~m}^2$$

Simplifying further, we get:

$$A \approx 2.71 \times 10^{-2} \mathrm{~m}^2$$

Therefore, the minimum area of the plates should be:

Option B: $$2.7 \times 10^{-2} \mathrm{~m}^2$$

Capacitor

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