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The value of q is and the value of r is 75.0 cm. Note that, in this question, you are only asked to find the magnitude of the net force, but you should also think about the direction of the net force. What is the magnitude of the net force on the ball of mass m that is located on the positive y-axis, because of the other four balls

Respuesta :

Answer:

F_net = - 0.365 N (Down-ward direction)

Explanation:

Given:

- Value of r = 0.75 m

- Charges on x axis are -2*q

- Charge +q on origin

- Charge on - y axis is -2q

- Charge on + y axis is +q

- q = 5.00 * 10^-6 C

Find:

-What is the magnitude of the net force on the ball of mass m that is located on the positive y-axis, because of the other four balls?

Solution:

- Force due to each of the two charges on x axis:

                                  F_x = k*(-2*q)*(+q) / r*^2

                                  r* = sqrt(2)*r

                                  F_x = -k*q^2 / r^2 (Down-wards)

- Force due to +q charge on origin:

                                  F_y = k*(+q)*(+q) / r^2

                                  F_y = + k*q^2 / r^2 (Up-wards)

- Force due to -2*q charge on y-axis:

                                  F_-2y = k*(-2*q)*(+q) / 4*r^2

                                  F_-2y = - k*q^2 / 2*r^2 (Downwards-wards)

- Total net Force on charge +q on + y-axis:

                                  2*F_x*sin(45) + F_y + F_-2y = F_net

                                 -sqrt(2)*k*q^2 / r^2 + k*q^2 / r^2 - k*q^2 / 2*r^2 = F_net

                                 (0.5-sqrt(2))*k*q^2 / r^2 = F_net

                                  F_net =  (0.5-sqrt(2))*(8.99*10^9)*(5*10^-6)^2 / 0.75^2

                                  F_net = - 0.365 N

                                 

                                 

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