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Question

Two equal charges are separated by a distance d . A third charge placed on a perpendicular bisector at x distance will experience maximum coulomb force when :

A
x=d2
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B
x=d2
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C
x=d22
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D
x=d23
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Solution

The correct option is C x=d22

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=2kq2r2+d24sinθ

Put d(2Fsinθ)dx=0

We get,

ddx⎜ ⎜ ⎜ ⎜2kq2x2+d24×xx2+d24⎟ ⎟ ⎟ ⎟=0

=2kq2(x2+d24)32+2kq2(x2+d24)32×(32)×x

On solving we get,

x2+d24=3x2

Or, x=d22


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