Question

A body of mass $M$ and charge $q$ is connected to a 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 ?

• A. The new equilibrium position is at a distance $\frac{qE}{2k}$ from $x = 0 $
• B. The total energy of the system is $\frac{1}{2} m \omega^2 A^2 + \frac{1}{2} \frac{q^2 E^2}{k}$
• C. The total energy of the system is $\frac{1}{2} m \omega^2 A^2 - \frac{1}{2} \frac{q^2 E^2}{k}$
• D. The new equilibrium position is at a distance $\frac{2 q E}{k}$ from $x = 0 $

Answer 1

The Correct Option is B

Total energy of the system = kinetic energy + potential energy
Given that amplitude of oscillation is $A$.
Therefore, energy at extreme point is $\frac{1}{2} k A^{2}= \frac{1}{2} m \omega^{2} A^{2}$.
When electric field is applied, new mean position is
$k x=q E \Rightarrow x=\frac{q E}{k}$
Thus, new total energy after electric field is applied is
$\frac{1}{2} m \omega^{2} A^{2}+\frac{1}{2} k x^{2}$
$=\frac{1}{2} m \omega^{2} A^{2}+\frac{1}{2} k\left(\frac{q E}{k}\right)^{2}$
Total energy $=\frac{1}{2} m \omega^{2} A^{2}+\frac{1}{2} \frac{q^{2} E^{2}}{k}$