Question 498: Electrostatics & Capacitance

Question

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
?

Accepted Answer

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.
$$	herefore\,\,\,\,$$  kx
eq
= qE
$$ \Rightarrow $$$$\,\,\,$$  x
eq
= $${{qE} \over K}$$
$$	herefore\,\,\,\,$$ Total energy of the system
= $${1 \over 2}m{\omega ^2}{A^2} + {1 \over 2}K\,x_{eq}^2$$
= $${1 \over 2}$$ m$${\omega ^2}$$A
2
+ $${1 \over 2}.{{{q^2}{E^2}} \over K}$$

Answer

The new equilibrium position is at a distance $${{qE} \over {2k}}$$ from $$x=0.$$

Answer

The total energy of the system is $${1 \over 2}m{\omega ^2}{A^2} + {1 \over 2}{{{q^2}{E^2}} \over k}.$$

Answer

The total energy of the system is $${1 \over 2}m{\omega ^2}{A^2} - {1 \over 2}{{{q^2}{E^2}} \over k}.$$

Answer

The new equilibrium position is at a distance $${{2qE} \over k}$$ from $$x=0.$$

Electrostatics & Capacitance

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 ?

Choose the correct answer: