6
Solution In the given equilateral triangle ABC of sides of length l, if
we draw a perpendicular AD to the side BC,
AD = AC cos 30º = ( 3 /2 ) l  and the distance AO of the centroid O
from A is (2/3) AD = (1/ 3 ) l By symmatry AO = BO = CO Rationalised 2023-24
Electric Charges
and Fields
13
 EXAMPLE 1 5
Thus,
Force F1 on Q due to charge q at A  = 
43
0
2
πε
lQq
 along AO
Force F2 on Q due to charge q at B  = 
43
0
2
πε
lQq
 along BO
Force F3 on Q due to charge q at C  = 
43
0
2
πε
lQq
 along CO
The resultant of forces F2 and F3 is 
43
0
2
πε
lQq
 along OA, by the
parallelogram law