Final answer:
An electron moves toward regions of higher potential in an electric field because it is negatively charged. The force on an electron in an electric field of 12.0 N/C to the right results in a force of 1.92× 10⁻¹⁸ N to the left.
Step-by-step explanation:
An electron moving in an electric field
When an electron is placed in an electric field, it experiences a force due to its negative charge. The electric field is defined from the perspective of a positive charge, which means it points from regions of higher potential to regions of lower potential. Since an electron has a negative charge, it will experience a force in the opposite direction of the electric field, hence moving toward regions of higher potential.
As for the statement regarding an electron initially moving to the right, without further context, we cannot determine the truthfulness; it's neither inherently true nor false.
In the case where an electron has an initial velocity of 5.00 × 106 m/s and is placed in a uniform electric field of 2.00 × 105 N/C, the electron will be accelerated opposite to the direction of the electric field. This is because the direction of the force on the electron is opposite the direction of the electric field due to the electron's negative charge.
Regarding the specific question about the force on an electron in an electric field of 12.0 N/C to the right, the force on the electron would be to the left. Given the electron's charge (which is approximately -1.6× 10-19 C), we can calculate the force using Coulomb's law (F = qE). The resulting force would be -1.92× 10-18 N, pointing to the left (option d).