The potential at the surface of a conducting sphere is given by:
\( V = \frac{Kq}{r} \).
Since the two spheres are connected, their potentials will be the same:
\[ \frac{Kq_1}{a} = \frac{Kq_2}{b}. \]
Simplifying:
\[ \frac{q_1}{q_2} = \frac{a}{b}. \]
Final Answer: \( \frac{a}{b} \).

Let \( f : \mathbb{R} \to \mathbb{R} \) be a twice differentiable function such that \[ (\sin x \cos y)(f(2x + 2y) - f(2x - 2y)) = (\cos x \sin y)(f(2x + 2y) + f(2x - 2y)), \] for all \( x, y \in \mathbb{R}. \)
If \( f'(0) = \frac{1}{2} \), then the value of \( 24f''\left( \frac{5\pi}{3} \right) \) is: