JEE MAIN - Physics (2018 - 15th April Morning Slot - No. 2)
A body of mass $$M$$ and charge $$q$$ is connected to spring of spring constant $$k.$$ It is oscillating along $$x$$-direction about its equilibrium position, taken to be at $$x=0,$$ with an amplitude $$A$$. An electric field $$E$$ is applied along the $$x$$-direction. Which of the following statements is correct ?
The new equilibrium position is at a distance $${{qE} \over {2k}}$$ from $$x=0.$$
The total energy of the system is $${1 \over 2}m{\omega ^2}{A^2} + {1 \over 2}{{{q^2}{E^2}} \over k}.$$
The total energy of the system is $${1 \over 2}m{\omega ^2}{A^2} - {1 \over 2}{{{q^2}{E^2}} \over k}.$$
The new equilibrium position is at a distance $${{2qE} \over k}$$ from $$x=0.$$
Explanation
On the body of charge q a electric fied E is applied, because of this equilibrium position of body will shift to a point where resulttant force is zero.
$$\therefore\,\,\,\,$$ kxeq = qE
$$ \Rightarrow $$$$\,\,\,$$ xeq = $${{qE} \over K}$$
$$\therefore\,\,\,\,$$ Total energy of the system
= $${1 \over 2}m{\omega ^2}{A^2} + {1 \over 2}K\,x_{eq}^2$$
= $${1 \over 2}$$ m$${\omega ^2}$$A2 + $${1 \over 2}.{{{q^2}{E^2}} \over K}$$
$$\therefore\,\,\,\,$$ kxeq = qE
$$ \Rightarrow $$$$\,\,\,$$ xeq = $${{qE} \over K}$$
$$\therefore\,\,\,\,$$ Total energy of the system
= $${1 \over 2}m{\omega ^2}{A^2} + {1 \over 2}K\,x_{eq}^2$$
= $${1 \over 2}$$ m$${\omega ^2}$$A2 + $${1 \over 2}.{{{q^2}{E^2}} \over K}$$
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