Introductory Physics Volume Two

1.7 More Examples 25
The net force is to be zero:
1 Q(3Q)
4πɛ 0 x 2 − 1 Q(2Q)
4πɛ 0 (1 − x) 2 = 0
−→ 3(1 − x) 2 − 2x 2 = 0
Solving the resulting quadratic equation yields two answers: x = 5.45m
and x = 0.55m. The latter value is the answer, since we already argued
that the position must lie between the two larger charges.
⊙ Do This Now 1.4
Analyze the configuration in the previous example, except replace the +3Q
charge with a −3Q charge.
(a) −0.18N (b)4.45m to the right of the +2Q charge.
Example
Four charges are located at the corners of a square as shown in the
diagram below. A fifth charge is located at the center of the square.
For the charge values indicated compute the net electric force on the
charge at the center of the square.
25cm
+4μC
+1μC
+4μC
Draw a force diagram:
+4μC
-1μC
+2μC
+1μC
F +2
45 o
y
x
F -1
F +1
45 o
F +4
-1μC
+2μC
The distances between each corner charge and the center charge are
the same:
√ (a ) 2 ( a
) √
2 2
d = + =
2 2 2 a,