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

Answer 1

Answer: (A)

When electric field is applied, at equilibrium

$F=kx=qE$

$x=\frac{qE}{k}$, x- extension of spring

Total Energy of system = Kinetic energy +Potential energy

$T.E.=\frac{1}{2}m{v}^2+\frac{1}{2}k{x}^2$

$=\frac{1}{2}m{\omega }^2{A}^2+\frac{1}{2}\frac{q^2{E}^2}{k}$