Coulomb\'s law for the magnitude of the force F between two particles with charg
ID: 1405298 • 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 = -18.5 nC , is located at x1 = -1.675 m ; the second charge, q2 = 36.5 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 = 48.5 nC placed between q1 and q2 at x3 = -1.200 m ?
Your answer may be positive or negative, depending on the direction of the force.
Explanation / Answer
Here ,
as the force between two chares is given as
F = k * q1*q2/d^2
as the force due to both charges will be in the same direction
net force on q3 = F1 + F2
Fnet = 9*10^9 * 48.5 *10^-9 * (- 18.5 *10^-9/(1.675 - 1.20)^2 - 36.5 *10^-9/1.20^2)
Fnet = -4.69 *10^-5 N
the net force acting on charge q3 is -4.69 *10^-5 N