\(\)calcule la fuerza electrostática sobre la partÃcula 3 (4.0mc)desde las dos cargas
\(f_{32}=k\ \frac{Q_3\ Q_2}{r_{32}}=\left(9x10^9\ \frac{nm^2}{c^2}\right)\left(\frac{\left(-4x10^{-6}c\right)\left(3x10^3c\right)}{\left(0,20m\right)^2}\right)=2.7n\)
\(f_{31}=k\ \frac{Q_3\ Q_1}{r_{31}}=\left(9x10^9\ \frac{nm^2}{c^2}\right)\left(\frac{\left(-4x10^{-6}c\right)\left(8x10^{-6}c\right)}{\left(0,50m\right)^2}\right)=1.15n\)