Coulomb\'s law for the magnitude of the force F between two particles with charg
ID: 1527786 • 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 = -12.0 nC , is located at x1 = -1.725 m ; the second charge, q2 = 34.0 nC , is at the origin (x=0.0000). Part A What is the net force exerted by these two charges on a third charge q3 = 45.0 nC placed between q1 and q2 at x3 = -1.205 m ? Your answer may be positive or negative, depending on the direction of the force.
Explanation / Answer
here,
q1 = - 12.0 nC
q1 = -12.0 * 10^-9 C
x1 = - 1.725m
q2 = 34. 0nC
q2 = 34.0* 10^-9 C
x2 = 0
q3 = 45nC
q3 = 45 * 10^-9 C
x3 = - 1.205m
the net force exerted by these two charges on a third charge F= K * q1 * q3/(x1-x3)^2 + K * q2 * q3/(x3)^2
Put value in this formula
So after putting in Value
F= 9 * 10^9 *12 * 10^-9* 45 * 10^-9/(0.45)^2 + 9 * 10^9 *34 * 10^-9* 45 * 10^-9/(1.205)^2
F = 3.34 * 10^-5 N
the net force exerted by these two charges on a third charge q3 = 45 nC is 3.34 * 10^-5 N towards the charge q1