Answer:
The __law__ conservation
of state that either energy may produced nor destroyed
Answer:
W = 2.74 J
Explanation:
The work done by the charge on the origin to the moving charge is equal to the difference in the potential energy of the charges.
This is the electrostatic equivalent of the work-energy theorem.

where the potential energy is defined as follows

Let's first calculate the distance 'r' for both positions.

Now, we can calculate the potential energies for both positions.

Finally, the total work done on the moving particle can be calculated.

Answer:
The magnetic field through the wire must be changing
Explanation:
According to Faraday's law, the induced emf, ε in a metallic conductor is directly proportional to the rate of change of magnetic flux,Φ through it. This is stated mathematically as ε = dΦ/dt.
Now for the wire, the magnetic flux through it is given by Φ = ABcosθ where A = cross-sectional area of wire, B = magnetic field and θ = angle between A and B.
So, dΦ/dt = dABcosθ/dt
Since A and B are constant,
dΦ/dt = ABdcosθ/dt = -(dθ/dt)ABsinθ
Since dθ/dt implies a change in the angle between A and B, since A is constant, it implies that B must be rotating.
So, <u>for an electric current (or voltage) to be produced in the wire, the magnetic field must be rotating or changing</u>.