A parallel plate capacitor was made with two rectangular plates, each with length $$l=3\,cm$$ and breadth $$b=1\,cm.$$ The distance between the plates is$$ 3\,\mu m.$$ Out of the following, which are the ways to increase the capacitance by a factor of $$10?$$  A. $$l=30cm,$$ $$b=1cm,$$ $$d=1\mu$$ $$m$$ $$B.$$ $$l=3cm,$$  $$b=1cm,$$ $$d=30\mu  m  C.$$ $$l=6cm,$$ $$b=5cm,$$ $$d=3\mu$$ $$m  D.$$ $$l=1cm,$$ $$b=1cm, d=10\mu\ m  E.$$ $$l=5cm,$$ $$b=2cm,$$  $$d=1\mu m$$ Choose the correct answer from the options given below:

The original capacitor has $$l = 3$$ cm, $$b = 1$$ cm, $$d = 3$$ $$\mu$$m.

Capacitance formula: $$C = \frac{\epsilon_0 A}{d} = \frac{\epsilon_0 \cdot l \cdot b}{d}$$.

Original: $$C_0 = \frac{\epsilon_0 \cdot 3 \cdot 1}{3} = \epsilon_0$$ (in relative units).

We need $$C = 10C_0$$, i.e., $$\frac{l \cdot b}{d} = 10 \times \frac{3 \cdot 1}{3} = 10$$.

A: $$l = 30, b = 1, d = 1$$: $$\frac{30 \times 1}{1} = 30 \neq 10$$. No.

B: $$l = 3, b = 1, d = 30$$: $$\frac{3 \times 1}{30} = 0.1 \neq 10$$. No.

C: $$l = 6, b = 5, d = 3$$: $$\frac{6 \times 5}{3} = 10$$. Yes!

D: $$l = 1, b = 1, d = 10$$: $$\frac{1 \times 1}{10} = 0.1 \neq 10$$. No.

E: $$l = 5, b = 2, d = 1$$: $$\frac{5 \times 2}{1} = 10$$. Yes!

Options C and E give a capacitance 10 times the original.

The correct answer is Option D: C and E only.