Two small charges with a mass of 1.0 g each are connected by 2.0-cm-long insulating rod of negligible mass. Each ball has the same magnitude of charge of 10 nC, but with opposite sign. The rod is held in a 1.0 × 104 N/C uniform electric field at the angle of 30°, then released to rotate around its center of mass.

What is the dipole moment of the rod?
Calculate the magnitude of the torque exerted on the rod by the electric field.
Obtain the moment of inertia around the center of mass.
What is the initial angular acceleration when the rod is released?

To find the dipole moment of the rod, we need to calculate the product of the charge magnitude and the distance between the charges. The dipole moment (p) is given by the equation:

p = q * d

where q is the charge magnitude and d is the distance between the charges. In this case, both charges have the same magnitude (10 nC) and the distance between them is 2.0 cm. Converting the distance to meters:

d = 2.0 cm = 0.02 m

Now, plugging the values into the equation:

p = (10 nC) * (0.02 m)
p = 0.2 nC*m

The dipole moment of the rod is 0.2 nC*m.

To calculate the magnitude of the torque exerted on the rod by the electric field, we use the equation:

τ = p * E * sin(θ)

where τ is the torque, p is the dipole moment, E is the electric field magnitude, and θ is the angle between the dipole moment vector and the electric field vector. In this case, the electric field magnitude (E) is given as 1.0 × 10^4 N/C and the angle (θ) is 30°. Plugging in the values:

τ = (0.2 nC*m) * (1.0 × 10^4 N/C) * sin(30°)
τ ≈ 0.1 N*m

The magnitude of the torque exerted on the rod by the electric field is approximately 0.1 N*m.

To obtain the moment of inertia around the center of mass, we need to consider the equation for the moment of inertia of an object. For a rod rotating around its center of mass, the moment of inertia (I) can be calculated using the formula:

I = (1/12) * m * L^2

where m is the mass of the rod and L is the length of the rod. In this case, the mass of each charge (m) is given as 1.0 g, which is equivalent to 0.001 kg, and the length of the rod (L) is given as 2.0 cm, which is equivalent to 0.02 m. Plugging in the values:

I = (1/12) * (0.001 kg) * (0.02 m)^2
I ≈ 1.67 × 10^-6 kg*m^2

The moment of inertia around the center of mass is approximately 1.67 × 10^-6 kg*m^2.

To find the initial angular acceleration when the rod is released, we can use the following equation derived from Newton's second law for rotational motion:

τ = I * α

where τ is the torque exerted on the object, I is the moment of inertia, and α is the angular acceleration. Rearranging the equation to solve for α:

α = τ / I

Plugging in the values:

α = (0.1 N*m) / (1.67 × 10^-6 kg*m^2)
α ≈ 6 × 10^4 rad/s^2

The initial angular acceleration when the rod is released is approximately 6 × 10^4 rad/s^2.