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°.
