1. Let the capacitors be $C_1$ and $C_2$, with $C_2 > C_1$.

2. Equivalent capacitance in parallel: $C_p = C_1 + C_2$

3. Equivalent capacitance in series: $C_s = \frac{C_1C_2}{C_1 + C_2}$

4. Given that the parallel equivalent is $\dfrac{15}{4}$ times the series equivalent: $C_p = \frac{15}{4}C_s$

Substitute the formulas: $C_1 + C_2 = \frac{15}{4}\left(\frac{C_1C_2}{C_1 + C_2}\right)$

5. Multiply both sides by $C_1 + C_2$: $\left(C_1 + C_2\right)^2 = \frac{15}{4}C_1C_2$

6. Let $x = \frac{C_2}{C_1}$ Then $C_2 = xC_1$.

Substitute: $\left(C_1 + xC_1\right)^2 = \frac{15}{4}C_1(xC_1)$ $C_1^2(1+x)^2 = \frac{15}{4}xC_1^2$

Cancel $C_1^2$: $(1+x)^2 = \frac{15}{4}x$

7. Expand and simplify: $1 + 2x + x^2 = \frac{15}{4}x$ Multiply by 4: $4 + 8x + 4x^2 = 15x$ $4x^2 - 7x + 4 = 0$

8. Solve the quadratic: $x = \frac{7 \pm \sqrt{49 - 64}}{8} = \frac{7 \pm \sqrt{-15}}{8}$

The discriminant is negative, so there is no real solution.

Since capacitance values must be real and positive, this situation is impossible.

9. Therefore, the correct option is: $\boxed{\text{B: No Solutions}}$