Coulomb\'s law for the magnitude of the force F between two particles with charg
ID: 1448032 • 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|d2, where K=140, and 0=8.854×1012C2/(Nm2) is the permittivity of free space. Consider two point charges located on the x axis: one charge, q1 = -15.0 nC , is located at x1 = -1.730 m ; the second charge, q2 = 39.0 nC , is at the origin (x=0.0000). What is the net force exerted by these two charges on a third charge q3 = 47.5 nC placed between q1 and q2 at x3 = -1.225 m ? Thanks so much in advance!
Explanation / Answer
here,
charge q1 = -15 * 10^-9 C at x1 = -1.730 m
charge q2 = +39.0 * 10^-9 C at x2 = 0 m
Charge q3 = 47.5*10^-9 C at x3 = -1.225 m
Since Q1 is negative and to the left of Q3 the force will be to the left.
also Q2 is positive and to the right of Q3 the the force will also be to the left
Therefore you can add the forces together (with left being negative)
F = -k*Q1*Q3/x13^2 - k*Q2*Q3/x23^2
F = K (-Q1*Q3/x13^2 - Q2*Q3/x23^2 )
F = 9*10^9(- (-15 * 10^-9*47.5*10^-9)/(1.730 - 1.225)^2 - (39.0 * 10^-9*47.5*10^-9)/(1.225)^2 )
F = (1.403)*10^-5 N