Three point charges of magnitude $$5 \mu C$$, $$0.16 \mu C$$ and $$0.3 \mu C$$ are located at the vertices $$A$$, $$B$$, $$C$$ of a right angled triangle whose sides are $$AB = 3 \text{ cm}$$, $$BC = 3\sqrt{2} \text{ cm}$$ and $$CA = 3 \text{ cm}$$ and point $$A$$ is the right angle corner. Charge at point $$A$$ experiences ______ N of electrostatic force due to the other two charges.

Given three charges at the vertices of a right-angled triangle with the right angle at A, where $$q_A = 5 \, \mu C$$, $$q_B = 0.16 \, \mu C$$, and $$q_C = 0.3 \, \mu C$$. The side lengths are $$AB = 3 \text{ cm} = 0.03 \text{ m}$$, $$CA = 3 \text{ cm} = 0.03 \text{ m}$$, and $$BC = 3\sqrt{2} \text{ cm}$$.

Using Coulomb's law, the magnitude of the force on charge A due to charge B is given by $$F_{AB} = \frac{kq_Aq_B}{AB^2} = \frac{9 \times 10^9 \times 5 \times 10^{-6} \times 0.16 \times 10^{-6}}{(0.03)^2} = \frac{9 \times 10^9 \times 8 \times 10^{-13}}{9 \times 10^{-4}} = \frac{7.2 \times 10^{-3}}{9 \times 10^{-4}} = 8 \text{ N}$$.

Similarly, the force on charge A due to charge C is $$F_{AC} = \frac{kq_Aq_C}{CA^2} = \frac{9 \times 10^9 \times 5 \times 10^{-6} \times 0.3 \times 10^{-6}}{(0.03)^2} = \frac{9 \times 10^9 \times 1.5 \times 10^{-12}}{9 \times 10^{-4}} = \frac{1.35 \times 10^{-2}}{9 \times 10^{-4}} = 15 \text{ N}$$.

Since the angle at A is $$90°$$, the forces $$F_{AB}$$ and $$F_{AC}$$ are perpendicular to each other. Therefore, the net force on A is $$F_{net} = \sqrt{F_{AB}^2 + F_{AC}^2} = \sqrt{8^2 + 15^2} = \sqrt{64 + 225} = \sqrt{289} = 17 \text{ N}$$.

Hence, the electrostatic force on the charge at A is $$\textbf{17}$$ N.