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esercizio3E



$Q_1=-2E-6C$
$Q_2=Q_4=+5E-6C$
$Q_3=3E-6C$

$r_(12)^2=(Delta_x^2+Delta_y^2)=(0,40m)^2=(4/10)^2m^2=16/100m^2=0.16m^2$
$r_(14)^2=(Delta_x^2+Delta_y^2)=(0,40m)^2=(4/10)^2m^2=16/100m^2=0.16m^2$
$r_(13)^2=(Delta_x^2+Delta_y^2)=(0,40m)^2+(0,40m)^2=2*(4/10)^2m^2=2*16/100m^2=32/100m^2=0.32m^2$

$F_(12)=K (Q_1 Q_2)/r_(12)^2= (9*10^9 (Nm^2)/C^2) *( (-2*10^(-6))C * (+5*10^(-6))C )/(0.16m^2)=(-90*10^(9-12))/(0.16)N=(-90*10^(-3))/(16/100) N=(-9*10^(-2+2)/16)N=(-9/16)N=-0.56N$ direzione sinistra
$F_(12x)=F_(12)=-0.56N$


$F_(14)=K (Q_1 Q_2)/r_(14)^2= (9*10^9 (Nm^2)/C^2) *( (-2*10^(-6))C * (+5*10^(-6))C )/(0.16m^2)=(-90*10^(9-12))/(0.16)N=(-90*10^(-3))/(16/100) N=(-9*10^(-2+2)/16)N=(-9/16)N=-0.56N=-0.56N$ direzione basso
$F_(14y)=F_(14)=-0.56N$


$F_(13)=K (Q_1 Q_3)/r_(13)^2= (9*10^9 (Nm^2)/C^2) *( (-2*10^(-6))C * (+3*10^(-6))C )/(0.32m^2)=(-54*10^(9-12))/(0.32)N=(-54*10^(-3))/(32/100) N=(-54*10^(-3+2)/32)N=(-54/32)*10^(-1)N=-1.68*10^(-1)N=-0.168N$ direzione diagonale

$F_(13x)=F_(13)cos(45°)=-0.168sqrt(2)/2N=0.236/2N=0,118N$
$F_(13y)=F_(13)sin(45°)=-0.168sqrt(2)/2N=0.236/2N=0,118N$

$F_(1x)=F_(12x)+F_(13x)=0.56N+0.118N=0,678N$
$F_(1y)=F_(12y)+F_(13y)=0.56N+0.118N=0,678N$

$F=sqrt(F_(1x)^2+F_(1y)^2)=sqrt(0,678^2N^2+0,678^2N^2)=sqrt(2*0,678^2N)=0,678sqrt(2)N=0,678*1.41N=0,9559N '=' 0,96 N = 9.6*10^(-1)N$ OK!



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