Particle 1 has a mass of m₁ = 4.00 × 106 kg, while particle 2 has a mass of m₂ = 7.00 × 106 kg. Each has the same electric charge. These particles are initially held at rest, and the two-particle system has an initial electric potential energy of 0.150 J. Suddenly, the particles are released and fly apart because of the repulsive electric force that acts on each one (see the figure). The effects of the gravitational force are negligible, and no other forces act on the particles. At one instant following the release, the speed of particle 1 is measured to be v₁ = 172 m/s. What is the electric potential energy at this instant?

In the image provide these exists two parallel plates call A and B the length of both being 5 centimeters and both separated from each other with a distance of 4 centimeters. An electron enters 1 cm above plate B and it is traveling horizontally at speed of 1x10^6 m/s. VA = 20 V and VB = 10 V are the plate potentials, respectively. A) What is the electrical field (vector) between the plates? B) What is the electric potential V at the entry point of the electron and its initial potential energy? C) The electron eventually find the speed one of the plates, find out out witch one and the speed of the electron before the impact. D) What is the location along the place where the electron makes impact (take x =0 as the left edge of the plates). OFocus 81 F 0.

The figure below shows a small, charged sphere, with a charge of q = +41.0 nC, that moves a distance of d = 0.177 m from point A to point B in the presence of a uniform electric field E of magnitude 265 N/C, pointing right.   (a) What is the magnitude (in N) and direction of the electric force on the sphere?     (b) What is the work (in J) done on the sphere by the electric force as it moves from A to B?     (c) What is the change of the electric potential energy (in J) as the sphere moves from A to B? (The system consists of the sphere and all its surroundings.)     (d) What is the potential difference (in V) between A and B?

A particle has a charge of +3.50 µC and moves from point A to point B, a distance of 0.130 m. The particle experiences a constant electric force, and its motion is along the line of action of the force. The difference between the particle's electric potential energy at A and B is EPEA - EPEB = +9.30 x 10-4 J. (a) Find the magnitude of the electric force that acts on the particle. (b) Find the magnitude of the electric field that the particle experiences. (a) Number i (b) Number Units Units