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 $\frac{2qE}{k}$ from $x=0$

Answer 1

Answer: (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
$kx=qE\Rightarrow x=\frac{qE}{k}$
Thus, new total energy after electric field is applied is
$\frac{1}{2}m{\omega }^2{A}^2+\frac{1}{2}k{x}^2$
$=\frac{1}{2}m{\omega }^2{A}^2+\frac{1}{2}k{\left(\frac{qE}{k}\right)}^2$
Total energy $=\frac{1}{2}m{\omega }^2{A}^2+\frac{1}{2}\frac{q^2{E}^2}{k}$