A point charge of charge 1 mC and mass 100 g is attached to a non-conducting massless rod of length 10 cm. The other end of the rod is attached to a two-dimensional sheet with uniform charge density σ and the rod is free to rotate. The sheet is parallel to the y-z plane (i.e. it's a vertical sheet). I lift the point charge so the rod is horizontal and release it. I observe that the point charge achieves its maximum speed when the rod makes an angle of 30∘ with respect to the vertical. What is σ in C/m^2?

Answer please

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To find the value of σ, the uniform charge density of the sheet in C/m^2, we can start by analyzing the forces acting on the point charge when the rod makes an angle of 30∘ with respect to the vertical.

When the rod is at an angle of 30∘, the force due to gravity can be resolved into two components:
1. The component parallel to the rod: mg * sin(30∘)
2. The component perpendicular to the rod: mg * cos(30∘)

The force due to the electric field produced by the charged sheet will exert a torque on the rod, causing it to rotate. Since the rod is in equilibrium at its maximum speed, the torques due to gravity and the electric field must balance each other.

The torque due to gravity about the point of rotation (assumed to be the end of the rod) is given by: τ_gravity = (mg * sin(30∘)) * (length of the rod) * sin(30∘)

The torque due to the electric field can be calculated as follows:
1. The electric field at the location of the point charge due to the sheet is given by: E = (σ / (2 * ε0))
(σ is the charge density and ε0 is the permittivity of free space)
2. The force experienced by the point charge due to this electric field is given by: F = q * E
(q is the charge on the point charge)

Since the point charge is in equilibrium and moving at its maximum speed, the torque due to the electric field is balanced by the gravitational torque. Therefore, we can equate the torque equations:

(mg * sin(30∘)) * (length of the rod) * sin(30∘) = (q * E) * (length of the rod)

Substituting the known values:
(m * g * sin(30∘)) * (l) * sin(30∘) = (q * (σ / (2 * ε0))) * (l)

Given:
Charge on the point charge, q = 1 mC = 1 * 10^(-3) C
Mass of the point charge, m = 100 g = 100 * 10^(-3) kg
Length of the rod, l = 10 cm = 10 * 10^(-2) m
Angle, θ = 30∘

g = acceleration due to gravity ≈ 9.8 m/s^2
ε0 = permittivity of free space ≈ 8.85 * 10^(-12) C^2 / N * m^2

Plugging in these values into the equation, we can solve for σ:

[(100 * 10^(-3) kg * 9.8 m/s^2 * sin(30∘)) * (10 * 10^(-2) m) * sin(30∘)] = [1 * 10^(-3) C * (σ / (2 * 8.85 * 10^(-12) C^2 / N * m^2)) * (10 * 10^(-2) m)]

Simplifying the equation gives:

[(100 * 10^(-3) kg * 9.8 m/s^2 * 0.5) * (10 * 10^(-2) m) * 0.5] = [1 * 10^(-3) C * (σ / (2 * 8.85 * 10^(-12) C^2 / N * m^2)) * (10 * 10^(-2) m)]

Solving for σ:

σ = [(100 * 10^(-3) kg * 9.8 m/s^2 * 0.5) * (10 * 10^(-2) m) * 0.5 * (2 * 8.85 * 10^(-12) C^2 / N * m^2)] / (1 * 10^(-3) C * (10 * 10^(-2) m))

Evaluating this expression will give you the value of σ in C/m^2.