Question

A ball of mass m=10g, carrying a charge q =-20μe is suspended from a string of length L= 0.8m above a horizontal uniformly charged infinite plane sheet of charge density σ = 4μe/m^2. The ball is displaced from the vertical by an angle and allowed to swing from rest.

Required:
a. Obtain the equations of motion of the charged ball based on Newtonian laws of motion.
b. Assume the displaced angle θ is small and simplify the results obtained in part (a) to obtain the frequency of oscillations of the charged ball.

Answer 1

a)
`- ( g - (q)/(m) (\sigma )/( 2 \epsilon_o) ) (sin \theta)/(R )` =
`(d^2 \theta)/(d t^2)`

b) f = 2π
`\sqrt{ (R)/( g - (q)/(m) (\sigma )/(2 \epsilon_o) ) }`

Step-by-step explanation:

a) To have the equations of motion, let's use Newton's second law.

Let's set a reference system where the x-axis is parallel to the path and the y-axis is in the direction of tension of the rope.

For this reference system the tension is in the direction of the y axis, we must decompose the weight and the electrical force.

Let's use trigonometry for the weight that is in the vertical direction down

sin θ = Wₓ / W

cos θ = W_y / w

Wₓ = W sin θ

W_y = W cos θ

we repeat for the electric force that is vertical upwards

F_{ex} = F_e sin θ

F_{ey} = F_e cos θ

the electric force is

F_e = q E

where the field created by an infinite plate is

E =
`( \sigma)/(2 \epsilon_o)`

let's write Newton's second law

Y axis

T - W_y = 0

T = W cos θ

X axis

F_{ex} - Wₓ = m a (1)

we use that the acceleration is related to the position

a = dv / dt

v = dx / dt

where x is the displacement in the arc of the curve

substituting

a = d² x /dt²

we substitute in 1

q E sin θ - mg sin θ = m
`(d^2 x)/(dt^2)`

we have angular (tea) and linear (x) variables, if we remember that angles must be measured in radians

θ = x / R

x = R θ

we substitute

sin θ (q E - mg) = m \frac{d^2 R \ theta}{dt^2}

`- ( g - (q)/(m) (\sigma )/( 2 \epsilon_o) ) (sin \theta)/(R )` =
`(d^2 \theta)/(d t^2)`

this is the equation of motion of the system

b) for small oscillations

sin θ = θ

therefore the solution is simple harmonic

θ = θ₀ cos (wt + Ф)

if derived twice, we substitute

- ( g - \frac{q}{m} \frac{\sigma }{ 2 \epsilon_o} ) \frac{\theta}{R } θ₀ cos (wt + Ф) = -w² θ₀ cos (wt + Ф)

w² =
`(g)/(R)` -
`(q)/(m) ( \sigma )/(2 \epsilon_o) (1)/(R)`

angular velocity is related to frequency

w = 2π f

f = 2π / w

f = 2π/w

f = 2π
`\sqrt{ (R)/( g - (q)/(m) (\sigma )/(2 \epsilon_o) ) }`