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WBJEE 2019 Physics Solved Paper
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© examsnet.com
Question : 20
Total: 40
A negative charge is placed at the midpoint between two fixed equal positive charges, separated by a distance 2d. If the negative charge is given a small displacement x (x << d) perpendicular to the line joining the positive charges, how the force (F)developed on it will approximately depend on x ?
F ∝ x
F ∝
1
x
F ∝
x
2
F ∝
1
x
2
Validate
Solution:
According to the question,
Force experienced by the charge
−
q
due to charge
Q
F
=
−
2
k
Q
q
r
2
cos
θ
.
.
.
(i) [ where,
.
k
=
1
4
π
ε
0
]
Fron diagram,
cos
θ
=
x
r
By substituting the value of
cos
θ
in
E
q
. (i)
F
=
−
2
k
Q
q
r
2
⋅
x
r
or
F
=
−
2
k
Q
x
r
3
or
F
=
−
2
k
Q
x
(
x
2
+
d
2
)
3
/
2
[
∵
r
2
=
x
2
+
d
2
r
=
√
x
2
+
d
2
]
For,
x
<
<
d
,
so
x
2
can be neglected
F
=
−
2
k
Q
x
d
3
So, the force developed by negative charge
(
−
q
)
due to the system of the charges as shown in the figure is.
F
=
−
4
k
Q
q
x
d
3
⇒
F
∝
x
So, the forced developed by negative charge i directly proportional to the distance x.
© examsnet.com
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