Coulomb\'s law for the magnitude of the force F between two particles with charg
ID: 1650304 • Letter: C
Question
Coulomb's law for the magnitude of the force F between two particles with charges Q and Q' separated by a distance d is |F| = K |QQ'|/d^2. where K = 1/4 pi epsilon_0, and elementof_0 = 8.854 times 10^-12 C^2/(N middot m^2) is the permittivity of free space. Consider two point charges located on the x axis: one charge, q_1 = -20.0 nC, is located at x_1 = -1.660 m: the second charge, q_2 = 37 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 = 47.5 nC placed between q_1 and q_2 at x_3 = -1.245 m? Your answer may be positive or negative, depending on the direction of the force. Express your answer numerically in newtons to three significant figures.Explanation / Answer
given q1=-20.0nc
q2=37.0nc
q3=47.5nc
q1,q2,q3 are located at points x1=-1.660m,x2=0.0m(origin),x3=-1.245m
the net electrostatic force on q3 is
F=F1+F2
F1=(kq1xq3)/(1.660-1.245)^2
=(9x10^9x20x10^-9x47.5x10^-9)/0.415^2
=4.96x10^-5N
F2=kq3xq2/(1.245^2)
=(9x10^9x37x47.5x10^-18)/(1.245^2)
=1.02x10^-5N
F=(4.96+1.02)10^-5N
=5.98x10^-5N
the net force on q3 acts towards negative xaxis
F=-5.98x10^-5N