contestada

A point charge q1 = -2.1 ?C is located at the origin of a co-ordinate system. Another point charge q2 = 6.7 ?C is located along the x-axis at a distance x2 = 6.7 cm from q1.
What is F12,x, the value of the x-component of the force that q1 exerts on q2?
i got the answer which is -28.20N
Charge q2 is now displaced a distance y2 = 2.8 cm in the positive y-direction. What is the new value for the x-component of the force that q1 exerts on q2?
i got the answer which is -22.17N
3)
A third point charge q3 is now positioned halfway between q1 and q2. The net force on q2 now has a magnitude of F2,net = 8.005 N and points away from q1 and q3. What is the value (sign and magnitude) of the charge q3?

Respuesta :

carlosego
answer 1
 By definition of the coulomb law we have that the force between two charges is given by
 F12 = K (q1 * q2) / (r ^ 2)
 where
 K = 9 * 10 ^ 9 [N.m ^ 2 / C ^ 2] (constant)
 q1 = load 1 [C]
 q2 = load 2 [C]
 r = distance between charges [m]
 Substituting the values we have
 F12 = (9000000000) * ((- 2.10E-06) * (6.70E-06) / ((0.067) ^ 2))
 F12 = -28.2N
 
 answer 2

 In this case the value of r will be given by
 ROOT ((0.067) ^ 2 + (0.028) ^ 2) = 0.0726 m
 Then, the net force will be given by
 F12 = (9000000000) * ((- 2.10E-06) * (6.70E-06) / ((0.0726) ^ 2))
 F12 = -24.01
 The horizontal component of this force is
 F12x = F12 (x / r) = - 24.01 (0.067 / 0.072)
 F12x = -22.2N
 The answer is -22.2N


 answer 3
 For this case we have by sum of forces that
 F2N = F12 + F32
 where
 F2N = 8,005 N
 F12 = -24.01 N
 Clearing F32
 F32 = F2N + F12 = (8.005) - (- 24.01) = 32.01N
 Then, by definition
 F32 = K (q2 * q3) / (r ^ 2)
 clearing q3
 q3 = (F32 (r ^ 2)) / (K * q2)
 substituting
 q3 = ((32.01) ((0.0726 / 2) ^ 2)) / ((9000000000) * (6.70E-06))
 q3 = 1.4E-06 C
 the answer is 1.4E-06 C (+)

Otras preguntas