A 0.2 kg ball is charged up to a net charge of Q, and is then suspended on a 0.25 m string in a uniform horizontal electric field strength of your [E=Birth month N/C] , Choose an angle of your choice at which the ball comes to rest =(30, 20,45 deg) from the vertical. What is the charge on the ball?

Respuesta :

[tex]{T sin \theta}[/tex]Answer:

 q = 27.16 C

Explanation:

For this exercise we use the equations of equilibrium. Let's set a reference system with the x-axis horizontally.

X axis

            F_e - Tₓ = 0

Axis y

           T_y -W = 0

let's use trigonometry

           cos θ = T_y / T

           sin θ= Tx / T

           T_y = T cos θ

           Tₓ = T sin θ

Axis y

            T cos θ = W

            T = mg / cos θ

X axis

            F_e = T sin θ

            qE = T sin θ

            q =   [tex]\frac{T sin \theta}{E}[/tex]  

            q = [tex]\frac{mg}{cos \theta } \frac{sin \thrta }{E}[/tex]

            q = [tex]\frac{mg}{E} tan \theta[/tex]

in the exercise they indicate that the electric field is the month of birth E = 8 N / C and the angle tea = 30º

             

let's calculate

              q = 0.2 9.8 8 / tan 30

              q = 27.16 C