35.

(a) Eq. 22-1 gives

F

12

= k

q

1

q

2

d

2

=

8.99

× 10

9

N

· m

2

C

2

 

20.0

× 10

−6

C



2

(1.50 m)

2

= 1.60 N .

(b) A force diagram is shown as well as our choice of y axis (the dashed line).

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.......................

...............

.......

q

3

q

2

q

1

y

............

.

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The y axis is meant to bisect the line between q

2

and q

3

in order to make use of the symmetry in

the problem (equilateral triangle of side length d, equal-magnitude charges q

1

= q

2

= q

3

= q). We

see that the resultant force is along this symmetry axis, and we obtain

|F

y

| = 2

k

q

2

d

2



cos 30

◦

= 2.77 N .