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/m2?

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

The point charge starts at a certain height and with zero initial velocity. As it falls, its gravitational potential energy is converted into kinetic energy. At the point where it achieves maximum speed, all of its potential energy has been converted into kinetic energy.

Let's denote the initial height as h and the length of the rod as L. When the rod is horizontal, the point charge is at height h, and when it makes an angle of 30 degrees with respect to the vertical, the length of the opposite side is L*sin(30°) = L/2.

The gravitational potential energy of the point charge is given by mgh, where m is the mass, g is the acceleration due to gravity, and h is the height. Here, m = 100 g = 0.1 kg and g = 9.8 m/s^2.

The kinetic energy of the point charge at its maximum speed is given by (1/2)mv^2, where v is the speed. Since all the potential energy is converted into kinetic energy at this point, we can equate the two expressions:

mgh = (1/2)mv^2

Simplifying, we get:

gh = (1/2)v^2

Now, we need to find the height h in terms of L. Considering the right triangle formed by the rod and the vertical line, we can use the sine function:

sin(30°) = h / L

Simplifying, we have:

h = L/2

Now, substituting this into the previous equation:

g(L/2) = (1/2)v^2

Simplifying further:

gL = v^2

The acceleration due to gravity, g, is approximately 9.8 m/s^2. The length of the rod, L, is given as 10 cm, which is equal to 0.1 m. The speed, v, can be determined based on the given conditions.

Using the equation for speed in uniform circular motion, v = ωR, where ω is the angular velocity and R is the radius of rotation.

Since the rod makes an angle of 30 degrees with the vertical, the angle between the rod and the horizontal is 60 degrees. The angular velocity can be found using the relationship:

v = ωR

R = L/2, the distance from the center of rotation to the point charge, and v is the maximum speed of the point charge.

Therefore, the value of v is ω(L/2).

Now we can substitute this into the previous equation:

g(L) = ω^2(L/2)^2

Simplifying, we have:

4g = ω^2L^2

Substituting the known values:

4 * 9.8 = ω^2 * 0.1^2

39.2 = ω^2 * 0.01

ω^2 = 39.2 / 0.01

ω^2 = 3920

Taking the square root:

ω = √3920

To find the speed, we can use the relationship v = ωR:

v = √3920 * (L/2)

Substituting the given value for L:

v = √3920 * 0.05

v ≈ 31.36 m/s

Now, substituting this value of v into the equation gL = v^2:

9.8 * 0.1 = 31.36^2

0.98 = 984.4096

σ is the linear charge density of the sheet, which is the charge per unit length. It is related to the surface charge density δ by the equation σ = δL. Since the sheet has uniform charge density, the charge on the sheet per unit area A is δA, where A is the area of the sheet which is L × d, where d is the width of the sheet. Therefore, σ = δL = (δA) / d. Since σ is given in C/m^2, we need to convert the units.

Let's assume d = 1 m (since the width is not provided). We can now find δ using the equation σ = δL:

δ = σ / L

Substituting the given values:

δ = 0.98 C/m^2 / 0.1 m

δ = 9.8 C/m^2

Therefore, the value of σ in C/m^2 is approximately 9.8 C/m^2.