The Physics: Clearly q_2 will be attracted toward q_1 with Force, F_e = k (|q_1||q_2|)/r^2 where
k = 8.99xx10^9 Nm^2/C^2; q_1=3muC; q_2=-4muC
So we need to calculate r^2, we use the distance formula:
r = sqrt((x_2- x_1)^2 + (y_2-y_1)^2)
r = sqrt((-2.0- 3.5)^2 + (1.5-.5)^2)=5.59cm = 5.59xx10^-2 m
F_e = 8.99xx10^9 Ncancel(m^2)/cancel(C^2) ((3xx10^-6 * 4xx10^6)cancel(C^2))/((5.59xx10^-2 )^2 cancel(m^2))
color(red)(F_e = 35N) as stated above q_2 is getting pulled by q_1
the direction is given by the direction q_2 -> q_1
Thus the direction is:
r_(12)=(x_1-x_2)i + (y_1 - y_2)j
r_(12)=(3.5-2.0)i + (05 - 1.5)j = 5.5i - j
and the unit vector is: u_(12) = 1/5.59(5.5i - j)
and the direction angle: tan^-1 -1/5.5 = -10.3^0
The 2nd question ask where should you place q_3 = 4muC so that the force on q_2 = 0
The Physics: Given that q_2 is been pulled toward q_1 we need a force opposite that. Now since q_3 is positively charged the a Force that pulled in the opposite direction will be obtained by placing q_3 on the line of force such that q_2 somewhere between q_3 and q_1.
We calculate r_(23) from the force equation knowing it is going to be color(red)(F_e = 35N) thus
35=k(|q_2||q_3|)/r_(23)^2; r_(23)^2=8.99xx10^9 cancel(N)m^2/cancel(C^2) ((4xx10^-6 * 4xx10^6)cancel(C^2))/(35cancel(N)) = 4.1xx10^-3m; r_(23) = 6.45xx10^-2m = 6.45 cm
Now given the direction is opposite the angle we are looking for is:
theta = 180^0-10.3^0 = 169.7^0
r_(23) = 6.45cos(169.7)i + 6.45sin(169.7)j
r_(23) = -6.34i + 1.15j
Now add this to the coordinates of q_2(-2, 1.5)
and q_3 coordinates are: #q_3 (-8.34, 2.65)
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