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?

To find the value of σ, we need to use the principle of conservation of energy.

When the point charge is released and starts to rotate, the total mechanical energy of the system remains constant. At the maximum speed, all of the potential energy has been converted into kinetic energy.

Let's define the initial and final positions of the point charge as follows:
- Initial position: The rod is horizontal, meaning it is perpendicular to the vertical sheet. At this position, the point charge has the maximum potential energy.
- Final position: The rod makes an angle of 30 degrees with respect to the vertical. At this position, the point charge has the maximum kinetic energy.

The potential energy (U) of the charge at the initial position is given by the equation:
U_initial = q * V_initial

Where:
q = charge of the point charge = 1 mC (1e-3 C)
V_initial = electric potential at the initial position

Since the sheet has a uniform charge density (σ), the electric potential at a distance (d) from the sheet can be calculated using the equation:
V = k * σ * d

Where:
k = Coulomb's constant = 9 x 10^9 N m^2 / C^2
d = distance from the sheet

Substituting the values into the equation for the initial potential energy, we get:
U_initial = (1e-3 C) * (9 x 10^9 N m^2 / C^2) * d

At the final position, the kinetic energy (K) of the charge is given by the equation:
K_final = (1/2) * m * v^2

Where:
m = mass of the point charge = 100 g (0.1 kg)
v = maximum speed of the point charge

The kinetic energy at the final position is equal to the potential energy at the initial position:
K_final = U_initial

Substituting the appropriate equations and values, we have:
(1/2) * (0.1 kg) * v^2 = (1e-3 C) * (9 x 10^9 N m^2 / C^2) * d

Since we know that the rod length is 10 cm, we can relate the distance (d) with the angle (θ) using trigonometry:
d = 0.1 m * sin(θ)

Substituting this into the equation, we have:
(1/2) * (0.1 kg) * v^2 = (1e-3 C) * (9 x 10^9 N m^2 / C^2) * (0.1 m * sin(30°))

Simplifying and solving for v^2, we get:
v^2 = (1e-3 C) * (9 x 10^9 N m^2 / C^2) * (0.1 m * sin(30°)) / (0.05 kg)

Finally, taking the square root of both sides, we can find the value of v.

Once we have the value of v, we can substitute it back into the equation for d to find the electric potential at the initial position (V_initial). Then, using V_initial, we can find the value of σ using the equation V_initial = k * σ * d.

Therefore, follow these steps:
1. Calculate the value of v using the equation above.
2. Calculate the value of d using d = 0.1 m * sin(30°).
3. Calculate the value of V_initial using V_initial = (v^2) / (9 x 10^9 N m^2 / C^2).
4. Calculate the value of σ using the equation V_initial = k * σ * d, where k = 9 x 10^9 N m^2 / C^2.

By following these steps, you can find the value of σ in C/m^2.