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Q))Two equal chargs are ixed at x=a and x=-a on d axis.anodr pt charge Q is placed at d orign.The change in d electric potential energy of Q,when it is displaced by a small distance x along d x-axis, is approximately proportional to....
(a)x
(b)x2
(c)x3
(d)1/x
kindly give me the answer f dis ques....m tired of posting ques here....widou geettin ne answer!!
Initial Energy:
K Qq/a
Final Energy
K qq*2a/(a*a-x*x)
Change in energy: Final energy-initial energy
= KQq [2*x*x/(a*a-x*x)a]
The x square term in the denominator can be neglected...So the change in energy is proportional to X square.
The corredc answer is (b)
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