Question

If a small particle with electric charge q = - 3.5 times 10^-4 Coulombs is placed at the point with coordinates (-4.0m, 2.1m, 2 2m) where the x, y and z components of the electric field vector are E_x = -10.5 Newtons/Coulomb. E_y = 10 5 Newtons/Coulomb, E_z = - 8 0 Newtons/Coulomb. what is the force acting on the particle? How is the electric potential energy of a charged particle with charge q at a given point related to the electric potential, or the voltage, at that point? Give a formula. Equipotential surfaces are imaginary surfaces over which the electric potential in a given problem is constant Electric field lines are directed curves such that at each point in space the curve is directed along the electric field. Explain how (geometrically) the electric field lines are related to the equipotential surfaces. The figure at the right shows a cross-sectional view of a conducting metal fin (above) at 10.000 Volts and a flat metal ground plane at 0 Volts Sketch in your best estimate of how the field lines and the equipotentials would look. Represent field lines as lines with arrow heads on them to indicate direction, and the equipotentials as dashed curves Draw at least five field lines and three equipotentials.

Answer 1

2) Force = Fx = qEx = 3.675*10^-3 N
Force = Fy = qEy = -3.675*10^-3 N
Force = Fz = qEz = 2.8*10^-3 N
F = 5.9*10^-3 N

3) Electric potential energy = Electric potential*charge

U = V*q

4) The electric field lines are always perpendicular to the equipotential surfaces.