Answers to homework probles Chapter 18 3) Electrons have negative charge -1.6 10^-19 C. When we remove a negative charge the remaining chrge becomes more positive. We need to remove -2 - (+3) = -5 (microCoulombs) = -5 10^-6 C This corresponds to a number of electrons N = -5 10^-6 C / (-1.6 10^-19 C) = 3.13 10^13 7) F = k (q_1*q_2)/r^2 = = 8.99 10^9 Nm^2/C^2 * (1.6 10^-19 C * 1.6 10^-19 C)/(3 10^-15 m)^2 = = 25.57 N 11) F = k (q_1*q_2)/r^2 Here we keep charge q_1 and q_2 the same and changing distace from r to 5r. The new force F_1 is F_1 = k (q_1*q_2)/(5r)^2 is 1/25 F (25 times smaller) 15) If the distance from q_1 to q_2 is x then the distance from q_2 to q_3 is 3x (one fourth and three fourth of the total). There are two forces on charge q_2 F_1 = k (q_1*q_2)/x^2 F_2 = k (q_1*q_3)/9x^2 The charge q_2 feels no net force so the two forces must be equal in size and opposite in direction F_1 = -F_2 k (q_1*q_2)/x^2 = -k (q_1*q_3)/9x^2 q_1 = -q_3/9 ---> q_3/q_1 = 9 27) E = F/q ---> F = E*q = 260 000 N/C * (-7 10^-6 C) = -1.82 N The force has negative sign i.e., it points in opposite direction than the field: due east. 31) The field points towards the charge so the charge sign has to be negative. E = - k q/r^2 ---> q = - E * r^2 / q = = - 9 10^5 N/C * (0.5 m)^2 / 8.99 10^9 Nm^2/C^2 = = - 0.25 10^-4 C 35) Electric field of the parallel plate capacitor E = q/(epsiolon_0 * Area) --> Area = q/(epsilon_0 * q) epsilon_0 = 8.85 10^-12 C^2/(Nm^2) Area = 0.15 10^-6 C /(8.85 10^-12 C^2/(Nm^2) 2.4 10^5 N/C) = = 0.0706 m^2