Two point charges, +Q and -Q of mass m, are placed on the ends of a massless rod of length L, which is fixed to a table by a pin through its center.

If the apparatus is then subjected to a uniform electric field E parallel to the table and perpendicular to the rod, find the net torque on the system of rod plus charges.
Express your answer in terms of some or all of the variables Q, m, L, and E.

To find the net torque on the system of the rod plus charges, we can consider the torque due to each individual charge and then sum them up.

Let's first consider the torque due to the positive charge, +Q, located on one end of the rod. The torque exerted by a force on an object is given by the cross product of the force and the displacement vector from the point of rotation to the point where the force is applied.

In this case, the force acting on the positive charge is given by F = Q * E, where Q is the charge and E is the electric field. The displacement vector, r, is the distance from the center of rotation (the pin) to the point where the force is applied, which is half the length of the rod, r = L/2.

The torque due to the positive charge is then given by τ = r * F * sin(θ), where θ is the angle between the displacement vector r and the force vector F. In this case, since the electric field is parallel to the table and perpendicular to the rod, the angle is 90 degrees. So sin(θ) = 1.

Therefore, the torque due to the positive charge is τ_+Q = (L/2) * (Q * E).

Similarly, the torque due to the negative charge, -Q, located on the other end of the rod, is also given by τ_-Q = (L/2) * (-Q * E) = -(L/2) * (Q * E). Note that the negative sign arises because the force is in the opposite direction for the negative charge.

Now, to find the net torque on the system, we need to sum up the torques due to both charges:

τ_net = τ_+Q + τ_-Q
= (L/2) * (Q * E) + -(L/2) * (Q * E)
= (Q * E * L/2) - (Q * E * L/2)
= 0

Therefore, the net torque on the system of the rod plus charges is zero.