Coulomb\'s law for the magnitude of the force F between two particles with charg
ID: 1405867 • Letter: C
Question
Coulomb's law for the magnitude of the force Fbetween two particles with charges Q and Qseparated 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 = -17.0 nC , is located atx1 = -1.650 m ; the second charge,q2 = 33.5 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 = 50.5 nC placed between q1and q2 at x3 = -1.165 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
Here ,
as the charge 50.5 nC lies between the charges ,
the force acts on the charge in the same direction because of both charges
Fnet = F1 + F2
Fnet = 9*10^9 * 50 *10^-9 * 10^-9 *( - 17/(1.650 - 1.165)^2 - 33.5/1.165^2)
calculating
Fnet = -4.36 *10^-5 N
the net force acting on the charge is -4.36 *10^-5 N