Could you answer it with python? The question is completed. The second picture, it seems to me it covers all the steps,but I can't find the correct final answer.

Exercise 4. Electrostatic force due to a dipole. An electric dipole is created. The positive point charge q1 2.0 x 10-18 C is placed at the point (0.0, 0.01) m and the negative point charge 92 -2.0 x 10-18 C is placed at the point (0.02,-0.01) m. Find the resulting force exerted on a test point charge C3 1.0 x 10-18 C located at the point (0.03, 0.04) m by the dipole. (Ans -1.60 x 10-221 + 1.28 x 10-29; N) Exercise 5 Basic Numpy Import Numpy into your program with the line import numry as an fnet = -f12 + f13 # Exercise 4 Dipole # enter the charges in C q1 = 2.0e-18 q2 = -2.00-18 q3 = 1.0e-18 # enter the positions in m xc1 = 0.0 XC2 = 0.02 XC3 = 0.03 yc1 = 0.01 yc2 = -0.01 yc3 = 0.04 113 = np.sqrt((xc3 - xc1)**2 + (yc3 - yc1)**2) r23 = np.sqrt((xc3 - Xc2)**2 + (yc3 - Yc2)**2) ang13 = np.arctan2((yc3 - yc1), (xc3 - xc1)) ang23 = np.arctan2((yc3 - yc2), (xc3 - XC2)) ke = 8.99e9 # calculate f13x, f13y, f23x, f23y f13x = ke*mt.fabs(q1)"mt.fabs(93)*np.cos(ang13)/r13**2 f13y = ke*mt.fabs(q1)*mt.fabs(q3)*np.sin(ang13)/r13**2 f23x = ke*mt.fabs(q2)*mt.fabs(93)*np.cos(ang23)/r23**2 f23y = ke"mt.fabs(q2)*mt.fabs(93)*np.sin(ang23)/r23**2 # find total force as q1 and q3 are positive then q3 is repelled from q1 # as q2 is negative and q3 is positive then q3 is attracted to q2 # q3 moves away in positive x, y directions from q1 and towrds q2 in # negative x, y directions fnetx = f13x - f23x fnety = f13y - f23y print("Net force is {0:5.2e)] + {1:5.2e)j".format(fnetx, fnety), "N")