Example 1.6 Consider three charges each equal to q at the
vertices of an equilateral triangle of side 1. What is the force on a
charge (with the same sign as q) placed at the centrold of the
triangle, as shown in Fig. 1.99
9,9
D
с
9.9
FIGURE 1.9
Answers
Answer:
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Consider three charges q
1
,q
2
,q
3
each equal to q at the vertices of an equilateral triangle of side l. What is the force on a charge Q( with the same sign as q) placed at the centroid of the triangle, as shown in Fig.?
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Hard
Solution
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Hint : Total force acting on a given charge is equal to the vector sum of forces exerted on it by all the other charges.
Step :1 calculate magnitude and direction of force by charge q
1
,q
2
,q
3
.
force due to charge on another charge is given by coloumb's law
F=
4πϵ
0
r
2
q
1
q
2
along the direction of line joining q
1
andq
2
now in the given problem
∵ ABC is an equilateral triangle and O is it's centroid
∴OA = OB = OC
from figure
BD =
2
BC
=
2
l
=DC
∠ OCD =
2
∠BCA
= 30
o
in triangle OCD
∵
OC
DC
= cos∠OCD = cos30
o
=
2
3
∴ OC =
3
2
DC
OC=
3
l
=OB=OA
force on Q due toq
1
F
1
=
4πϵ
0
(OA)
2
q
1
Q
in the direction of AO
F
1
=
4πϵ
o
l
2
3Qq
similarly F
2
=
4πϵ
o
l
2
3Qq
;along BO
F
3
=
4πϵ
o
l
2
3Qq
;along CO
magnitude of F
1
=F
2
=F
3
=F
Step :2 Find the direction and magnitude resultant force.
The resultant Force of F
2
and F
3
can be found by parallelogram law
Angle between F
2
and F
3
is 120
o
(from fig)
magnitude of F
1
=F
2
=F
3
=F
F' =
F
2
2
+F
3
2
+2F
1
F
2
cos120
o
=
F
2
+F
2
+2F.F.(
2
−1
)
= F
direction of the resultant will be along OA (from fig)
now the resultant of F' and F
1
∵ they are opposite in direction and equal in magnitude
∴ the resultant of the forces= F
1
-F' = F-F = 0
Hence the the net force on a charge Q=0N