Coulomb\'s law for the magnitude of the force F between two particles with charg
ID: 1419687 • 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 = -20.0 nC , is located atx1 = -1.745 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 = 48.5 nC placed between q1 and q2 atx3 = -1.070 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
Since Q1 is negative and to the left of Q3 the force will be to the left
Since Q2 is positive and to the right of Q3 the the force will also be to the left
Therefore you can add the forces together
Now we have F = k*Q1*Q3/r13^2 + k*Q2*Q3/r23^2
= 9 * 10^9 * 20 * 10^-9 * 48.5 * 10^-9 /(1.745 - 1.07)^2 + 9 * 10^9 * 33.5 * 10^-9 * 48.5 * 10^-9 / 1.070^2
= 3.19 * 10^-5 N
it is also direction in negative x direction