For an electron moving in a direction of the electric field its potential energy _______ and its electric potential _________.
Wouldn't the blanks be "increases" and "decreases"? because PE=mgh while V=kq/r
increases for the first.
decreases on the second.
Actually, the correct answers are "decreases" and "decreases".
When an electron moves in the direction of the electric field, its potential energy decreases. The formula for potential energy in this case is given by PE=qV, where q is the charge of the electron and V is the electric potential. Since the electron is moving in the direction of the electric field, it experiences a decrease in potential energy.
Similarly, the electric potential also decreases in this scenario. The electric potential is given by V=kq/r, where k is the electrostatic constant, q is the charge of the electron, and r is the distance from the electron to the source of the electric field. As the electron moves towards the source of the electric field, the distance r decreases, leading to a decrease in the electric potential.
So, to summarize, when an electron moves in the direction of the electric field, its potential energy and electric potential both decrease.
For an electron moving in the direction of an electric field, its potential energy decreases and its electric potential increases.
The potential energy (PE) of a charged particle in an electric field is given by the formula PE = qV, where q is the charge of the particle and V is the electric potential. As the electron moves in the same direction as the electric field, it experiences a decrease in potential energy.
On the other hand, the electric potential (V) at a point in an electric field is given by the formula V = kQ/r, where k is the Coulomb's constant, Q is the magnitude of the source charge, and r is the distance from the source charge. As the electron moves closer to the source charge or in the direction of the electric field, the distance (r) decreases, resulting in an increase in the electric potential.