Three charges of 6μC ,8μC and 10μ C।are placed।at।points।A,।B and C respectively as
shown in the figure Determine force on A dueto B.
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Explanation:
Given four charges placed at corners A, B, C and D of square of side 0.2m.
AD=CD=0.2m
BD=
(o.2)
2
+(0.2)
2
=0.283m
So, force on +8μC due to charge at A
F
1
=
4πε
0
1
(0.2)
2
2×8×10
−12
=9×10
9
×4×10
−10
=3.6N along AD
Force on +8μC due to charge at B
F
2
=
4πε
0
1
(0.2)
2
32×8×10
−12
=9×10
9
×408.16×10
−12
=3673.47×10
−3
=3.673N along BD
Force on +8μC due to charge at C
F
3
=9×10
9
×
(0.2)
2
48×10
−12
=10800×10
−3
=10.8N along AD
Resultant of F
1
and F
2
is
(3.6)
2
+(10.8)
2
F
′
=
12.96+116.64
=
129.6
=11.38N
Now, net force = Resultant of F
′
and F
2
=
(F
′
)
2
+(F
2
)
2
=
(11.38)
2
+(3.673)
2
along F
1
=
129.50+1.491
=
142.99
=11.958N
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