Two charge particles of charges +4 *C and +9 C
are fixed at A and B separated by a distance 20 cm.
Where should be a third point charge 1 C placed
such that net electrostatic force on third charge
becomes zero?
Answers
Answer:
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
Answer:
Two charge particles of charges +4 *C and +9 C
are fixed at A and B separated by a distance 20 cm.
Where should be a third point charge 1 C placed
such that net electrostatic force on third charge
becomes zero?