PROBLEMA 1
1.- Calcular la fuerza electrostática neta sobre la carga \(Q_3\) debido a las cargas \(Q_1\) y \(Q_{2\ }\)
Solución
Formula: \(F_{3\ 2}=K\frac{Q_3\ Q_2}{r_{3\ 2}^2}\)
Sustituimos la forma con los datos dados en la imagen:
\(F_{3\ 2}\) = \(9x10^9\ \frac{Nm^2}{C^2}\) = \(\frac{\left(65x10^{-6}C\right)\left(50x10^{-6}C\right)}{\left(0.3m\right)^2}=325N\)
\(F_{3\ 1}=K\ \frac{Q_{3\ }Q_1}{r_{3\ 1}^2}=\ 9x10^9\)\(\frac{\left(65x10^{-6}C\right)\left(50x10^{-6}C\right)}{\left(0.3m\right)^2}=325N\) \(\frac{Nm^2}{C^2}=\)
\(\frac{\left(65x10^{-6}C\right)\left(-86x10^{-6}C\right)}{\left(0.6m\right)^2}\ =-139.75N\ \)
\(F_{3\ 2}=K\frac{Q_3\ Q_2}{r_{3\ 2}^2}\)
\(Fx_{3\ 1}=\left(140N\right)\left(\cos\ 30^o\right)=121.2\ N\)
\(Fy_{3\ 1}=F_{3\ 1}\ sen\ 30^o=-\left(110\ N\right)Sen\ 30^o=-70\ N\ \)
\(F_y=F_{3\ 2}+F_{3\ 1}y=330\ N\ -\ 70\ N=260\ N\)
\(F=\sqrt{F^2x+F^2y}=\sqrt{\left(121.2\ N\right)^2}+\left(260\right)^2=286.8\ N\)
\(\tan\theta=\frac{Fy}{Fx}=\frac{260\ N}{121.2}=2.14\)
\(\theta=\tan^{-1}=64.95^o\)
\(F=\sqrt{F^2x+F^2y}=\sqrt{\left(121.2\ N\right)^2}+\left(260\right)^2=286.8\ N\)