The Earth's electric field creates a potential that increases 100 V for every meter of altitude. If an object of charge +4.5 mC and mass 68 g falls a distance of 1.0 m from rest under the influence of the Earth's electric and gravitational fields, what is its final kinetic energy?

Two parallel metal plates separated by 20 cm are connected across a 12 V potential difference. An electron is released from rest at a location 10 cm from the negative plate. When the electron arrives at a distance 5 cm from the positive plate, how much kinetic energy has the electron gained? Hint: First you will need to calculate the value of the electric field between the plates, then the potential difference between the initial and final positions. Now use the Work–Energy Theorem. Standard units!

A negatively charged particle of mass 8.0 x 10-13 kg is traveling rightward between two plates separated by a distance d = 50 cm, as shown below. The particle is launched with a speed of 1.0 x 104 m/s at the left plate, and strikes the right plate at a speed of 5.0 x 10³ m/s. The left plate is at electric potential V₁ = +1.0 V, while the right plate is at electric potential V₂ = -5.0 V. Assume the electric field between the plates is uniform, and ignore the effect of gravity. 2 (a) (b) V₁ d 7 V₂ Does the potential energy of the particle increase or decrease as it moves from the left plate to the right plate? Is the electric field between the plates directed rightward or leftward? Justify your answers briefly. Show clearly that the electric field between the plates has magnitude 12 V/m (or 12 N/C). Then find the charge of the particle (with correct magnitude and sign). (Note: You may obtain the charge using either energy considerations or a "force and kinematics" approach.

If a charge of -3.0 *(10^-6) C were allowed to fall through a potential difference of +500V, the change in potential energy for the charge would be?