AbC is a equilateral triangle of side 0.1m charge of +3nc and-3nc are placed at the berries A and B respectively calculate the resultant electric intensity at a point C
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Answer:
AC=BD=
(0.2)
2
+(0.2)
2
=
2
×0.2=0.28m
∴AO=BO=CO=
2
0.28m
=0.14m
V i.e, Potential at O;V
o
=
4πε
0
1
[
AO
q
1
+
BO
q
2
+
CO
q
3
]
V i.e, Potential at D;V
D
=
4πε
0
1
[
AD
q
1
+
BD
q
2
+
CD
q
3
]
Work done =q[V
o
−V
D
]
=
4πε
0
2×10
−9
[
0.14
2×10
−9
+
0.14
4×10
−9
+
0.14
8×10
−9
−
0.2
2×10
−9
−
0.28
4×10
−9
−
0.2
8×10
−9
]
On solving, we get
W=2×10
−9
×2×10
9
[
0.14
14×10
−9
−
0.2
10×10
−9
−
0.28
4×10
−9
]
=4×[10
−7
−0.5×10
−7
−0.14×10
−7
]
=5.40×10
−7
Joules
Explanation:
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