While measuring the acceleration due to gravity by a simple pendulum , a student makes a positive error of `1%` in the length of the pendulum and a negative error of `3%` in the value of time period . His percentage error in the measurement of `g` by the relation ` g = 4 pi^(2) ( l // T^(2))` will be

Answer

`(##ALN_PHY_C01_E01_334_A01##)` , `E=0`

Explanation

For minimum potential energy of the system, q should be placed in the middle

`U_("system")=(K2qxxq)/(x) xx100+(Kq8q)/((9-x))xx100+((Kxx2qxx8q)/9)xx100`
`(dU)/(dx)=Kq^(2)xx100 [(-2)/x^(2)+8/((9-x)^(2))+0] rArr x=3 cm`
Electric field at the position of
`q=(Kxx2q)/((0.03)^(2))-(Kxx8q)/((0.06)^(2))=0`