1. Tri takasta naelektrisanja rasporeena su u prostoru kao to je pokazano na slici. Odrediti vektor sile na anelektrisanje postavljeno u tacki P. Poznato je :

Q1 = 44.28 nC = 44.28 * 10-9 CQ2 = -8.856 nC = -8.856 * 10-9 CQ3 = 12.56 nC = 12.56 * 10-9 C

tg = a/2 /a = a / 2a = = 0.5

= 26.56 tg = a /a/2 = 2a /a = 2/1

= 63.43

++ = 180

26.56+63.43+ = 180 = 90.01

r = bb2 = a2 + (a/2)2

b2 = a2 + a2 /4

b2 = 4a2 + a2 /4

b = 5a2 /4

b = a 5 / 2r = a5 / 2F1Q = 1/ 40 * Q1 *P /r2 F1Q = 1/ 4*3.14*8.85*10-12 C2/Nm2 * 44.28*10-9 C * 12.56*10-9 C / ( a5 /2)2

F1Q = 1/ 111.156*10-12 C2/Nm2 * 556.156*10-18 C2 / 5a2/4

F1Q = 1/ 111.156*10-12 C2/Nm2 * 4*556.156*10-18 C2 / 5a2

F1Q = 1/ 111.156*10-12 C2/Nm2 * 2224.62*10-18 C2 / 5a2

F1Q = 2224.62*10-18 C2/Nm2 / 555.78*10-12 C2/Nm2*a2

F1Q = 4.0*10-6 Nm2*a2

F2Q = 1/ 40 * Q2 *P /r2 F2Q = 1/ 4*3.14*8.85*10-12 C2/Nm2 * (-8.856*10-9 C) * 12.56*10-9 C / ( a5 /2)2

F2Q = 1/ 111.156*10-12 C2/Nm2 * (-111.23*10-18 C2) / 5a2/4F2Q = 1/ 111.156*10-12 C2/Nm2 * 4*(-111.23*10-18 C2) / 5a2

F2Q = 1/ 111.156*10-12 C2/Nm2 * 444.92*10-18 C2/ 5a2

F2Q = 444.92*10-18 C2 / 555.78*10-12 C2/Nm2*a2

F2Q = 0.8*10-6 Nm2*a2 F1Qx = F1*cos 26 = 4.0*10-6 Nm2*a2 * 0.89 = 3.56*10-6 Nm2*a2

F1Qy = F1*sin 26 = 4.0*10-6 Nm2*a2 * 0.43 = 1.72*10-6 Nm2*a2

F2Qx = F2 = 0.8*10-6 Nm2*a2

F2Qy = 0Fx = F1Qx + F2Qx = 3.56*10-6 Nm2*a2 + 0.8*10-6 Nm2*a2 = 4.36*10-6 Nm2*a2

Fy = F1Qy + F2Qy = 0.8*10-6 Nm2*a2 + 0 = 0.8*10-6 Nm2*a2

F = Fx2 + Fy2

F = (4.36*10-6 Nm2*a2) 2 + (0.8*10-6 Nm2*a2) 2

F = 19.0*10-12 N2m4*a4 + 0.64*10-12 N2m4*a4 F = 19.64*10-12 N2m4*a4 F = 4.43*10-12 Nm2*a2

Q1 (+)

r 2

r 1

F2Q

F1Q

P

0

a

a/2

Q3 (+)

Q2 (-)

Fr

F2Q

F1Q

y

x

F2Q

F1Q

F2Qx

F1Qy

F1Qx

y

x