(8%) Problem 1: The electric field in a certain region is given by the function = Ak cos (kx) cos (by) Absin (kx) sin (by) where A = 5.01 N m/C, k = 0.251 m, and b = 1.03 m. The points in the figure use values x = 1.25 m and y = 2.04 m. (0,y) (xy) (0,0) (x,0) 14% Part (a) Enter an expression for the change in electric potential from point (0,0) to point (x1,0) in terms of A, k, b, and x. V(x1,0)-V(0,0) = - A sin(kx1) Correct! 14% Part (b) Enter an expression for the change in potential from point (x1,0) to point (x1,y) in terms of A, k, b, x and y. V(x11)-V(x1,0) A (sin(kx) - sin(kx1) cos(by)) Correct! 14% Part (c) Enter an expression for the change in potential from point (0,0) to point (x1,y1), along the path that passes through (x1,0) V(x1,y1)-V(0,0)=-A sin(kx) cos(by) Correct! 14% Part (d) Enter an expression for the change in electric potential from point (0,0) to point (0, y) in terms of A, k, b, and y. V(0,1)-(0,0)=0 Correct! 14% Part (e) Enter an expression for the change in electric potential from point (0,y) to point (x1,y) in terms of A, k, b, x and y. V(x1,y1)-V(0,y)-A cos(by) sin(kx) Correct! 14% Part (f) Enter an expression for the change in potential from point (0,0) to point (x1,y), along the path that passes through (0,y). V(x1,y1)-V(0,0) = -A cos(by) sin(kx) Correct! 14% Part (g) Calculate the change in potential between (0,0) and (x1,y1), in joules, along the path that passes through (0,1). V(x1,y1)-V(0,0) =