Two particles A and B having charges 8 x 10^-6C and -2 x10^-6C, respectively, are held
fixed with a separation of 20 cm. Where should a third charged particle be placed so that
it does not experience a net electric force?
Answers
Answered by
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Explanation:
As the net electric force C should be equal to zero, the force due to A and B must be opposite in direction. Hence, the particle should be placed on the line AB. As A and B have changes of opposite signs, C cannot be between A and B.
Also A has larger magnitude of charge than B. Hence, C should be placed closer to B that A. The situation is shown in figure. Suppose BC=x and the charge on C is Q
F
CA
=
4π∈
0
1
(0.2+x)
2
(8.0×10
−6
)Q
i
^
and
F
CB
=
4π∈
0
−1
x
2
(2.0×10
−6
)Q
i
^
F
C
=
F
CA
+
F
CB
=
4π∈
0
1
[
(0.2×10
−6
)
(8.0×10
−6
)Q
−
x
2
(2.0×10
−6
)Q
]
i
^
But ∣
F
C
∣=0
Hence
4π∈
0
1
[
(0.2×x)
2
(8.0×10
−6
)Q
−
x
2
(2.0×10
−6
)Q
]=0
Which gives x=0.2m
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