Question

Forces in a Three-Charge System 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=14πϵ0, and ϵ0=8.854×10−12C2/(N⋅m2) is the permittivity of free space. Consider two point charges located on the x axis: one charge, q1 = -17.5 nC , is located at x1 = -1.680 m ; the second charge, q2 = 32.5 nC , is at the origin (x=0.0000)
What is the net force exerted by these two charges on a third charge q3 = 55.0 nC placed between q1 and q2 at x3 = -1.085 m in N

Answer 1

Answer:

Explanation:

Given the charges.

Q1=-17.5nC. Negative charge

Q2=32.5nC. Positive charge

Q3=55nC. Positive charge

Q1 is at a distance of -1.68m on the x-axis

Q2 is at the origin i.e at 0m

Q3 is at between Q1 and Q2 at -1.085m on the x-axis

It shows that,

Q1 is at -1.085+1.68 =0.595m from Q3

Also, Q2 is at 1.085m from Q3.

K=9×10^9Nm²/C²

We need to find the net force on Q3

Then we need F13 and F23

Firstly F13

Between Q1 and Q3

There will be attraction i.e, Q3 will move to the negative direction of the x axis, then F13 will be in negative direction

So,

F13=kQ1Q3/r²

F13=9E9×17.5E-9×55E-9/0.595²

F13=2.45×10^-5N

In vector form

F13=—2.45×10^-5N i

Now, we need F23,

This will the force of repulsion because they are both positive charge, the the charge Q3 will move to the negative direction of x axis, since Q2 is at the origin and Q3 is at negative x axis. So, F23 will be negative

F23=kQ2Q3/r²

F23=9E9×32.5E-9×55E-9/1.085²

F23=1.367×10^-5N

In vector form

F23=—1.367×10^-5N i

Then the net force is given as

Fnet = F13+F23

Fnet=—2.45×10^-5Ni—1.367×10^-5Ni

Fnet=—3.82×10^-5N i

Magnitude for the Fnet is

Fnet=3.82×10^-5N.

And the direction

θ= arctan(y/x).

y=0 and x=3.82×10-5

θ= arctan(0/-3.82E-5)

θ=arctan(-0)

θ= 0. in the negative direction, i.e 180°.