Potential Energy and the Electric Potential Difference

[WGDawg3]

I'm back again this week with a problem:

A particle has a charge of +1.5 Micro Coulombs and moves from point A to point B, a distance of .20 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 EPE_A - EPE_B = +9.0x10^-4 J. (a) Find the magnitude and direction of the electric force that acts on the particle. (b) Find the magnitude and direction of the electric field that the particle experiences.

 

[Defennder]

What have you done regarding this? How does the force exerted by the E-field relate to its electric potential energy?

 

[Moetasim]

I would first like to clarify that potential energy and electric potential difference are related concepts, but they are not the same. Potential energy is the energy an object possesses due to its position or configuration in a system, while electric potential difference is the difference in electric potential between two points in an electric field.

Moving on to the problem at hand, we are given a particle with a charge of +1.5 Micro Coulombs moving from point A to point B, a distance of 0.20 m. We know that the particle experiences a constant electric force, and its motion is along the line of action of the force. This means that the force is acting in the same direction as the displacement of the particle.

To find the magnitude of the electric force, we can use the formula F = qE, where q is the charge of the particle and E is the electric field. Plugging in the values, we get F = (1.5x10^-6 C)(9.0x10^-4 J) = 1.35x10^-9 N. This is the magnitude of the electric force acting on the particle.

To find the direction of the electric force, we can use the right-hand rule. If we point our thumb in the direction of the electric field (from A to B), and our fingers in the direction of the particle's motion (from A to B), then the direction in which our palm is facing will be the direction of the electric force. In this case, the electric force will be pointing towards the right.

Moving on to part (b) of the problem, we are asked to find the magnitude and direction of the electric field that the particle experiences. To find the electric field, we can rearrange the formula F = qE to E = F/q. Plugging in the values, we get E = (1.35x10^-9 N)/(1.5x10^-6 C) = 9x10^-4 N/C. This is the magnitude of the electric field experienced by the particle.

To find the direction of the electric field, we can again use the right-hand rule. If we point our thumb in the direction of the electric field (from A to B), and our fingers in the direction of the force (towards the right), then the direction in which our palm is facing will be the direction of the particle's motion. In this case, the electric field will