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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=d2√2As 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×x√x2+d24⎞⎟ ⎟ ⎟ ⎟⎠=0 =2kq2(x2+d24)32+2kq2(x2+d24)32×(−32)×x On solving we get, x2+d24=3x2 Or, x=d2√2

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