In a certain region of space the electric potential is given by V=+Ax2y−Bxy2, where A = 5.00 V/m3 and B = 8.00 V/m3. Calculate the direction angle of the electric field at the point in the region that has cordinates x = 2.50 m, y = 0.400 m, and z = 0. Please explain. The answer is not 60, 120, or 30.

Two 5.00-nC charged particles are in a uniform electric field with a magnitude of 625 N/C. Each of the particles is moved from point A to point B along two different paths, labeled in Figure P26.65. a. Given the dimensions in the figure, what is the change in the electric potential experienced by the particle that is moved along path 1 (black)? b. What is the change in the electric potential experienced by the particle that is moved along path 2 (red)? c. Is there a path between the points A and B for which the change in the electric potential is different from your answers to parts (a) and (b)? Explain. FIGURE P26.65 Problems 65, 66, and 67.

At a certain distance from a charged particle, the magnitude of the electric field is 500 V/m and the electric potential is 3.00 kV. (a) What is the distance to the particle? (b) What is the magnitude of the charge?

An electric potential exists in a region of space such that V = 8x4 2y2 + 9z3 and V is in units of volts, when x, y, and z are in meters. a. Find an expression for the electric field as a function of position. b. What is the electric field at (2.0 m, 4.5 m, 2.0 m)?