Question

A charge of 5.0 mu or micro CC is located at x = 0 cm, y = 0 cm. Another charge Q2 is located at x = 2 cm, y = 0 cm. The force on a 2 mu or micro CC charge at x = 4 cm, y = 0 cm is 78.7 N, pointing in the negative x direction. When this 2 mu or micro CC charge is positioned at x = 8.88 cm, y = 0 cm, the force on it is zero.

Determine the charge Q₂.

Answer 1

First condition:
at x=0 (q1), x= 2 (q2), x=4 (q3). The force on the third charge points in the –x direction. So q2 must be negative.

Force between Q1 and Q3 is
F1 = k(5)(2) / (4)^2 = k(0.625)
Force between Q2 and Q3 is
–F2 = k(q₂)(2) / (2)^2= kq₂(0.5)
total force is 78.7 is the algebraic sum of the two
78.7 = k(0.625) – kq₂(0.5)

Second condition:
you have charges at x=0 (q1), x= 2 (q2), x=8.88 (q3). The force on the third charge is zero. So q2 must be negative.

Force between Q1 and Q3 is
F1 = k(5)(2) / (8.88)^2 = k(0.1268)
Force between Q2 and Q3 is
–F2 = k(q₂)(2) / (6.88)^2 = kq₂(0.042)
total force is 0, is the algebraic sum of the two
0 = k(0.1268) – kq₂(0.042)

now.
0 = k(0.1268) – kq₂(0.042)

q₂(0.042) = (0.1268)
q₂ =3.01 µC