1. Problem statement: Do charges +3μC aur -5μC jo 2 meter door hain, unke darmiyan electric force find karna hai. 2. Formula: Coulomb's law ke mutabiq force ka formula hai: $$F = k \frac{q_1 q_2}{r^2}$$ Jahan: - $k = 9 \times 10^9$ Nm$^2$/C$^2$ (Coulomb's constant) - $q_1 = 3 \times 10^{-6}$ C (charge 1) - $q_2 = -5 \times 10^{-6}$ C (charge 2) - $r = 2$ m (distance) 3. Calculation: Substitute karte hain values formula mein: $$F = 9 \times 10^9 \times \frac{(3 \times 10^{-6})(5 \times 10^{-6})}{2^2}$$ 4. Simplify karte hain: $$F = 9 \times 10^9 \times \frac{15 \times 10^{-12}}{4} = 9 \times 10^9 \times 3.75 \times 10^{-12}$$ 5. Final value: $$F = 33.75 \times 10^{-3} = 0.03375 \text{ N}$$ 6. Interpretation: Force positive nahi hai kyun ke charges opposite hain, is liye force attractive hoga. Final answer: Force ki value 0.03375 N hai aur yeh attractive force hai.