Coulomb\'s law for the magnitude of the force F between two particles with charg
ID: 2160301 • Letter: C
Question
Coulomb's law for the magnitude of the force F between two particles with charges Q and Q^prime separated by a distance d is|F|=K rac{|Q Q^prime|}{d^2},
where K= rac{1}{4 pi epsilon_0}, and epsilon_0 = 8.854 imes 10^{-12};{ m C^2/(Ncdot m^2)} is the permittivity of free space.
Consider two point charges located on the x axis: one charge, q_1 = -12.0 nC, is located at x_1 = -1.700 m; the second charge, q_2 = 40.0 nC, is at the origin (x=0.0000).What is the net force exerted by these two charges on a third charge q_3 = 46.0 nC placed between q_1 and q_2 at x_3 = -1.125 m?
Your answer may be positive or negative, depending on the direction of the force.
Explanation / Answer
Since Q1 is negative it will excert an attractive(to the left) force on Q3 and Q2 will excert an repulsive(to the left) on Q3. Since both forces are in he same direction(to the left), the net force is the sum of these two forces. F1,3 = - k*|Q1|*|Q3|/r^2 = -1.50x10-5 N
Now, F2,3 =-k*|Q2|*|Q3|/r^2 = -1.31x10-5 N
Finally Fnet = F1,3 +F2,3 = -2.81x10-5 N