Problem Two point charges q1=+1,0 μCq_1 = +1{,}0\ \mu\text{C}q1=+1,0 μC and q2=+1,0 μCq_2 = +1{,}0\ \mu\text{C}q2=+1,0 μC are at a distance r=1,0 mr = 1{,}0\ \text{m}r=1,0 m. Calculate the magnitude of the Coulomb force (k=9,0⋅109 N\cdotpm2/C2k = 9{,}0\cdot 10^9\ \text{N·m}^2/\text{C}^2k=9,0⋅109 N\cdotpm2/C2). Solution Apply Coulomb’s law with k=9,0⋅109 N\cdotpm2/C2k = 9{,}0\cdot 10^9\ \text{N·m}^2/\text{C}^2k=9,0⋅109 N\cdotpm2/C2 and q1=q2=1,0 μC=10−6 Cq_1 = q_2 = 1{,}0\ \mu\text{C} = 10^{-6}\ \text{C}q1=q2=1,0 μC=10−6 C: F=k q1q2r2=9,0⋅109⋅(10−6)(10−6)(1,0)2=9,0⋅109⋅10−12=9,0⋅10−3 NF = k\,\frac{q_1 q_2}{r^2} = 9{,}0\cdot 10^9\cdot\frac{(10^{-6})(10^{-6})}{(1{,}0)^2} = 9{,}0\cdot 10^9\cdot 10^{-12} = 9{,}0\cdot 10^{-3}\ \text{N}F=kr2q1q2=9,0⋅109⋅(1,0)2(10−6)(10−6)=9,0⋅109⋅10−12=9,0⋅10−3 N The charges have the same sign, so the force is repulsive. F≈9,0⋅10−3 N (repulsive)\ev{F \approx 9{,}0\cdot 10^{-3}\ \text{N (repulsive)}}F≈9,0⋅10−3 N (repulsive) Links Topics: Elettrostatica Concepts: Legge di Coulomb