Question

1. Calculate the magnitude of the net force on particle 1 due to the other two. Calculate the direction of the net force on particle 1 due to the other two.

2. Calculate the net force on particle 2 due to the other two. Calculate the direction of the net force on particle 2 due to the other two.

3. Calculate the magnitude of the net force on particle 3 due to the other two. Calculate the direction of the net force on particle 3 due to the other two.

<Ch 21 HW Problem 21.13 Constants I Periodic Table Three charged particles are placed at the corners of an equilateral triangle of side 1.20 m (see (Figure 1)). The charges are 1-6.7 uC, Q2 8.1 pC, and 03 4.7 C Figure 1 of 1 01 1.20 m 0- Type here to search

Answer 1

Q = 6.7 Mc Q, = -8,1 MC Q2 = -4.7 MC Q, &Q2 Magnitude of force between I FQjQ l= kaia = 8.9920' X 6.7810*x 8-1410-6 (1.22 - 0.3388 N of force between Q, & Q₂ Magnitude KQ,Q3 | FQ, Qg) = 7 : 8.99 X 10' x 67816 4.7X166 (102) 0.1966 M Magnitude Q & Qz of force between ffalfa - 22 8.99x10°x 8-1710*x4.7716 (112) = 0.2377 N

For FQ3 Foi = -fo Sinzo î - FQ, cos3o) + FQ Sin 30 i - For cos30 } -0.3384 Singo î - 0.3388 (0)30 ſ + 0.1966 Singo i - 0-196666s 30 = -0.0711 î - 0.4637 ģ magnitude TFQla 160.0711)' + (-0.4637 je imao - 0.469 N direction. -0.4637 fano -0.0711 Q = 81.28° 81.28° Wort to negative x Counter clock wise direction direction along 81.289

de = fa, cosbo? + Fa, sinto ] + Faz ? = 0.3388 Coyou î +0.3388 Simbo + 0.2377 î 0.4071 ĉ + 0.2934 N Magnitude - 0.4071)2 + (0-2974€ N = 0.502 angle tano 0.2934 0.407 1 0 - 35.789 35.78 won't ta gxis along counter clock wise 33.780

@fast - Fo costo î + fa, Sinbo ŷ - FQ ? - - 0.1966 C 6o + 0.1964 Sinbo j -0.2377 î -0.336 î + 0.1903 Ĵ N. Magnitude : JC-0.336)2+60-1903) N angle tano = 0.1703 0.336 Q = 26.87' 26.87 wirt negative x aiki's along clockwise y 26.87