Consider the combination of two capacitors $$C_1$$ and $$C_2$$, with $$C_2 > C_1$$, when connected in parallel, the equivalent capacitance is 10 times the equivalent capacitance of the same connected in series. Calculate the ratio of capacitors, $$\frac{C_2}{C_1}$$.

Let the two capacitors be $$C_1$$ and $$C_2$$ with $$C_2 > C_1$$. When connected in parallel, the equivalent capacitance is $$C_p = C_1 + C_2$$. When connected in series, the equivalent capacitance is $$C_s = \frac{C_1 C_2}{C_1 + C_2}$$.

We are given that $$C_p = 10 C_s$$, so $$C_1 + C_2 = 10 \cdot \frac{C_1 C_2}{C_1 + C_2}$$.

This gives $$(C_1 + C_2)^2 = 10 C_1 C_2$$.

Let $$r = \frac{C_2}{C_1}$$. Then $$(1 + r)^2 = 10r$$, which expands to $$r^2 + 2r + 1 = 10r$$, or $$r^2 - 8r + 1 = 0$$.

Using the quadratic formula: $$r = \frac{8 \pm \sqrt{64 - 4}}{2} = \frac{8 \pm \sqrt{60}}{2} = 4 \pm \sqrt{15}$$.

Since $$C_2 > C_1$$, we need $$r > 1$$. We have $$4 + \sqrt{15} \approx 4 + 3.87 = 7.87 > 1$$ and $$4 - \sqrt{15} \approx 0.13 < 1$$. Therefore $$\frac{C_2}{C_1} = 4 + \sqrt{15}$$.

Hence the correct answer is Option 1: $$4 + \sqrt{15}$$.