Coulomb\'s law for the magnitude of the force F between two particles with charg
ID: 1407668 • 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=140 , and 0=8.854×10^12 C^2/(Nm^2) is the permittivity of free space.
Consider two point charges located on the x axis: one charge, q1 = -12.5 nC , is located at x1 = -1.740 m ; the second charge, q2 = 37.0 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 = 48.0 nC placed between q1 and q2 at x3 = -1.100 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 force due to both charges will be in same direction
net force on q3
F3 = F1 + F2
F3 = 9*10^9 * 48 *10^-9 * 10^-9 *(12.5/(1.74 - 1.1)^2 + 37/1.1^2)
solving for F3
F3 = 2.64 *10^-5 N
the force on charge q3 is 2.64 *10^-5 N