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