Introductory Physics Volume Two

26 Electric Field 1.7
where a = 25cm. To find the net force, add all the forces using vector
addition:
⃗F net = îF x + ĵF y ,
where
F x = (F +4 − F +2 − F +1 − F −1 ) sin(45 ◦ )
F y = (−F +4 + F +2 − F +1 − F −1 ) cos(45 ◦ )
Use Coulomb’s law to compute the magnitude of each force:
F +4 = (9 × 109 )(4 × 10 −6 )(4 × 10 −6 )
( √2 ) 2
= 4.6N
2 (.25)
F +1 = F −1 = (9 × 109 )(1 × 10 −6 )(4 × 10 −6 )
( √2
2 (.25) ) 2
= 1.15N
F +2 = (9 × 109 )(2 × 10 −6 )(4 × 10 −6 )
( √2
2 (.25) ) 2
= 2.3N
Finally, compute the components of the net force:
F x = (4.6N − 2.3N − 1.15N − 1.15N) sin(45 ◦ ) = 0
F y = (−4.6N + 2.3N − 1.15N − 1.15N) cos(45 ◦ ) = −3.25N
So, the net force points straight down with a magnitude of 3.25N.
Example
Consider the configuration of three charges along a line as shown below.
Assuming the only forces acting are mutual electric forces between the
charges, show that the configuration is in equilibrium, meaning that
the net force on each charge is zero.
a
a
+4q -q
+4q
Here is a force diagram for all the charges:
+4q -q
+4q
The net force on the center charge must be zero, since the two forces
are in opposite directions and caused by equal charges that are the
same distance away. One down, two to go. Use the force law to write