Coulomb\'s law for the magnitude of the force F between two particles with charg
ID: 1331223 • 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 = -17.0 nC , is located at x1 = -1.655 m ; the second charge, q2 = 31.5 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 q1 and q2 at x3 = -1.155 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
Incorrect Answers
4.22 x 10 ^-6
2.02 X 10^-5
PLEASE MAKE SURE IT IS RIGHT AND CHECK WORK
Explanation / Answer
let,
q1=-17 nC at x1=-1.655 m
q2=31.5 nC at x2=0
q3=50.5 nc at x3=-1.155 m
now,
F23=k*q2*q3/(x3-x2)^2
=9*10^9*31.5*10^-9*50.5*10^-9/(-1.155-0)^2
=1.0732*10^-5 N
and
F13=k*q1*q3/(x1-x2)^2
=9*10^9*17*10^-9*50.5*10^-9/(-1.155-(-1.655))^2
=1.0732*10^-5 N
=3.0906*10^-5 N
net force on charge q3 is,
F=F13+F23
=1.0732*10^-5+3.0906*10^-5
=4.164*10^-5 N