Question

The position of two charged particles are held fixed in an xy−plane. The
charges are q1 = +3.0 mC, at x1 = 3.5 cm and y1 = 0.50 cm, and q2 = −4.0
mC, x2 = 2.0 cm, y2 = 1.5 cm.
a) Find the magnitude and direction of the electrostatic force on particle 2
due to particle 1.
b) At what x and y coordinates should a third particle of charge q3 = 4.0 mC
be placed such that the net electrostatic force on particle 2 due to particles
1 and 3 is zero?

Answer 1

The position of two charged particles are held fixed in an xy−plane. The
charges are q1 = +3.0 mC, at x1 = 3.5 cm and y1 = 0.50 cm, and q2 = −4.0
mC, x2 = 2.0 cm, y2 = 1.5 cm.

The distance between q1andq2 is
r12 =√[(x2-x1)2 +(y2 -y1)2 ] = 0.018027m
The force is F =k q1q2 /r122
=102248520.71 N
angle θ =arc tan [( y2 -y1) / (x2-x1)]
= arc tan [ (1.5 -0.5 ) / ( 2 -3.5) ]
= 146.3 degrees (CCW from +x axis) or -33.69o CW

c) a third charge q3 = +4.0 mC be placed such that the net electrostatic force on particle 3 due to particles 1 and 2 is zero
so k q2q3 /r2 = k q1q2 /r122
r =r12√ ( q3 /q1 ) = 0.0208 m
the positions are
x3 = x2 - rcos - 33.69 = 0.2693 cm
y3 = y2 - rsin - 33.69 = 2.65 cm