one of the most important types of vector fields in physics and engineering is the electric field E. If we place an electric charge at the point o= (xo, yo, zo) in space, it generates a vector force field E(x, y, z) given by: (x, y, z)= q x-xo in the sense E= -VU. x- 2-20 y - Yo x-xo||x- 13 where q is the electric charge, and so is a constant representing the permittivity of air, and |x| represents the magnitude of the vector a, i,e., |x| = x + y + 2. -200/13), (sometimes, the magnitude is denoted by ||.||, that is, || || = x + y + z). a) Prove that E is a conservative force field with the potential energy: U(x, y, z)= q 4|2 - 20 b) Two electric charges, q and -q, are located at the points (1,0,0) and (-1,0,0), respectively. Find the electric field E and the associated potential energy U at an arbitrary point (x, y, z) that is not (1,0,0) or (-1,0,0). The electric field and potential energy follow the principle of superposition, where they result from the addition of two fields and potentials. Use the following code to visualize the electric field and contour lines of the potential in the xy-plane. Attach the image to your solution: x=-2:0.1:2; y=x'; u=(x-1)./((x-1).^2+y. 2).(3/2)-(x+1)./((x+1).^2+y. -2). (3/2); v=y./((x-1).^2+y. 2). (3/2)-y./((x+1).^2+y.^2).^(3/2); streamslice (x,y,u,v,0.5) axis tight equal hold on z=1./sqrt contour (x,y,z, [-2 -1 1 2], 'color', 'r') pl = nsidedpoly (1000, 'Center', [1 0], 'Radius', 0.12); p2 nsidedpoly (1000, 'Center', [-1 0], 'Radius', 0.12); plot (p1, 'FaceColor', 'b') plot (p2, 'FaceColor', 'r') You can observe that the stream lines are perpendicular to the contour curves which repre- sents equipotential curves. ((x-1).^2+y.^2)-1./sqrt((x+1).^2+y.^2); =