 # An electron is shot towards a target that is negatively charged. While the electron is traveling, th spazzter08dyk2n 2022-05-19 Answered
An electron is shot towards a target that is negatively charged. While the electron is traveling, the target makes an abrupt move towards the electron. While the information that the target moved is traveling from the target to the electron, the electron behaves like an electron that is moving towards a target that is in the original position.
How can energy be conserved when an electron that is moving towards a nearby charge behaves like it was moving towards a far away charge? Seems we end up with electron being at 2 meters distance from the target, while the electron had enough energy to travel to at most 4 meters distance from the target.
It also seems to me that "moving the target requires energy" is not a solution to this problem.
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Suppose the target is an infinite plane with constant charge density $\sigma$. It will not radiate when you move it because the electric field is constant everywhere. Suppose the test charge $q$ is small enough that its radiation is negligible.
The electric field of the plane is $2\pi \sigma$ in the direction perpendicular to the plane.
The charge begins a distance $d$ from the plane. The potential energy in the system is $-2\pi \sigma qd$ (define to be zero when $d=0$). The force on the charge is $2\pi \sigma q$ and by Newton's third law there is an equal and opposite force on the plane. (We are assuming there is no radiation, so momentum of the charge carriers is conserved, and Newton's third law holds.)
The charge also has some kinetic energy $T$.
We move the plane towards the charge a distance $\mathrm{\Delta }d$. This takes energy because the force of the charge on the plane does negative work. The energy required is the force multiplied by the distance, or $2\pi \sigma q\mathrm{\Delta }d$.
The new distance of the charge from the plane is $d-\mathrm{\Delta }d$, so the new potential energy is $-2\pi \sigma q\left(d-\mathrm{\Delta }d\right)$. The potential energy has increased by $2\pi \sigma q\mathrm{\Delta }d$, exactly the amount of work that had to be put into the system to move the plane. The charge still has kinetic energy $T$, so energy is conserved in that the change in energy of the system is equal to the energy that was used to move the plane.