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प्रश्न
Two charges – q each are separated by distance 2d. A third charge + q is kept at mid point O. Find potential energy of + q as a function of small distance x from O due to – q charges. Sketch P.E. v/s x and convince yourself that the charge at O is in an unstable equilibrium.
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उत्तर
U = `1/(4piε_0) {(-q^2)/((d - x)) + (-q^2)/((d - x))}`
U = `(-q^2)/(4piε_0) (2d)/((d^2 - x^2))`
`(dU)/(dx) = (-q^2 . 2d)/(4piε_0) . (2x)/((d^2 - x^2)^2`
U0 = `(2q^2)/(4piε_0d) (dU)/(dx)` = 0 at x = 0
x = 0 is an equilibrium point.
`(d^2U)/(dx^2) = ((-2dq^2)/(4piε_0)) [2/((d^2 - x^2)^2) - (8x^2)/((d^2 - x^2)^3)]`
= `((-2dq^2)/(4piε_0)) 1/((d^2 - x^2)^3) [2(d^2 - x^2)^2 - 8x^2]`
At x = 0
`(d^2U)/(dx^2) = ((-2dq^2)/(4piε_0)) (1/d^6) (2d^2)`, which is < 0.
Hence, unstable equilibrium.
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