Question

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?

Answer 1

`{T sin \theta}`Answer:

q = 27.16 C

Step-by-step 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 =
`(T sin \theta)/(E)`

q =
`(mg)/(cos \theta ) (sin \thrta )/(E)`

q =
`(mg)/(E) tan \theta`

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