Midway between two equal and similar charges in third equal and similar charge is placed then ticket charge
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0
Explanation:
Correct option is
C
x=
2
2
d
As cos components cancel each other sin components of force add up. We can find the maxima of the equation to find the place of maximum value of force.
2Fsinθ=2
r
2
+
4
d
2
kq
2
sinθ
Put
dx
d(2Fsinθ)
=0
We get,
dx
d
⎝
⎜
⎜
⎜
⎛
x
2
+
4
d
2
2kq
2
×
x
2
+
4
d
2
x
⎠
⎟
⎟
⎟
⎞
=0
=
(x
2
+
4
d
2
)
2
3
2kq
2
+
(x
2
+
4
d
2
)
2
3
2kq
2
×(
2
−3
)×x
On solving we get,
x
2
+
4
d
2
=3x
2
Or, x=
2
2
d
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