An infinite uniformly-charged sheet with a = -2 x 10^-10 C/m2 occupies the yz plane.

A deuteron1 is created at a point on the +x axis 5 m from the sheet. The deuteron is
produced with zero kinetic energy and is free to move.
a) What is the location of the deuteron at the instant its kinetic energy
is 5.428 x 10~18 J?
(b) Would the work required to move the deuteron back to its original
position be done by the system or an external agent? Justify your answer.

To determine the location of the deuteron at the instant its kinetic energy is 5.428 x 10^-18 J, we can use the principle of conservation of mechanical energy. The initial potential energy of the deuteron is due to the electric field created by the uniformly-charged sheet, and its final kinetic energy is given.

(a) Let's first determine the potential energy of the deuteron at the given location on the x-axis. The potential energy of a point charge in an electric field is given by U = q * V, where q is the charge and V is the electric potential.

Since the sheet is infinitely large, the electric potential is constant on one side and is zero on the other side, forming an electric field between them. For a uniformly charged sheet, the electric potential at a distance y from the sheet is given by V = σ / (2ε₀), where σ is the charge density on the sheet and ε₀ is the permittivity of free space.

Using the given charge density σ = -2 x 10^-10 C/m² and the distance 5 m, we can calculate the electric potential V:
V = (-2 x 10^-10 C/m²) / (2ε₀)

Next, we can calculate the potential energy U = q * V. Since the deuteron has zero charge (q = 0), its potential energy at that point is zero.

Now, the total mechanical energy of the deuteron is the sum of its potential energy and kinetic energy: E = U + K, where E is the total mechanical energy, U is the potential energy, and K is the kinetic energy.

Given that the total mechanical energy is 5.428 x 10^-18 J and the potential energy is zero, we can rearrange the equation to solve for the kinetic energy: K = E - U.

Substituting the values, we can find the kinetic energy:
K = (5.428 x 10^-18 J) - 0

Therefore, the kinetic energy is 5.428 x 10^-18 J.

(b) To answer whether the work required to move the deuteron back to its original position would be done by the system or an external agent, we need to consider the forces acting on the deuteron.

The electric field created by the uniformly charged sheet exerts a force on the charged deuteron. Since the electric field is perpendicular to the movement of the deuteron when it is displaced from its original position, the work done by the electric field is zero.

If an external agent wants to move the deuteron back to its original position, it would need to exert a force opposite to the electric field. The external agent would do work against the electric field to move the deuteron. Therefore, the work required to move the deuteron back to its original position would be done by an external agent.

Justification:
- The work done by a force is positive if it is in the same direction as the displacement.
- The work done by a force is negative if it is in the opposite direction as the displacement.
- The work done by a force is zero if it is perpendicular to the displacement.

In this scenario, the work done by the electric field is zero, indicating that it is not responsible for moving the deuteron back to its original position. The external agent needs to do work against the electric field to return the deuteron to its original position.