Coulomb\'s law for the magnitude of the force F between two particles with charg
ID: 1649632 • 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 epsilon_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 = -1.720 m: the second charge, q_2 = 39.0 nC, is at the origin (x = 0.0000). What is the net force exerted by these two charges on a third charge q_3 = 48.0 nC placed between q_1 and q_2 at x_3 = -1.135 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. Force on q_3 =Explanation / Answer
force on q3 due to q1:
as q1 and q3 are of opposite sign, force is attractive in nature.
hence force is towards q1 i.e. along -ve x axis.
force magnitude=K*q1*q3/(1.72-1.135)^2
=2.52122*10^(-5) N
force on q3 due to q2:
as q2 and q3 are of same sign, force is repulsive in nature.
hence force is along -ve x axis.
force magnitude=K*q2*q3/(0-(-1.135))^2
=1.30606*10^(-5) N
so total force=2.52122*10^(-5)+1.30606*10^(-5)
=3.82728*10^(-5) N
in three significant figures,
total force on q3=3.83*10^(-5) N