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

To determine the correct statement for this problem, we need to understand how the system behaves when an electric field is applied. Let's break down the details step by step:

1. Initial System: The body of mass \(M\) and charge \(q\) is attached to a spring with spring constant \(k\), and it oscillates about its equilibrium at \(x = 0\) with amplitude \(A\).
2. Electric Force: When the electric field \(E\) is applied along the x-direction, the charge experiences an electric force \(F_e = qE\).
3. New Equilibrium Position: The spring force, which balances the electric force at equilibrium, can be expressed as: \(F_s = kx_{\text{eq}}\). Setting \(F_s = F_e\), we have: \(kx_{\text{eq}} = qE\). Solving for \(x_{\text{eq}}\) gives: \(x_{\text{eq}} = \frac{qE}{k}\).
4. Effect on Total Energy:
   • The potential energy due to the spring and electric field affects the total energy. The spring's contribution remains unchanged for a given amplitude, but the electric field adds an additional potential energy.
   • At the new equilibrium, potential energy due to the electric field is given by the work done to bring the charge to the new equilibrium position: \(U_e = \frac{1}{2} \frac{q^2E^2}{k}\).
   • The total mechanical energy of the system at maximum amplitude and including the electric field is the original energy plus the electric potential energy: \(\frac{1}{2} m \omega^2 A^2 + \frac{1}{2} \frac{q^2 E^2}{k}\).

Considering these analyses, we can conclude:

• Incorrect Option 1: The new equilibrium position is at \(\frac{qE}{k}\), not \(\frac{qE}{2k}\) or \(\frac{2qE}{k}\).
• Correct Option: The total energy of the system includes both the mechanical energy of oscillation and the potential energy due to the electric field. Hence, \(\frac{1}{2} m \omega^2 A^2 + \frac{1}{2} \frac{q^2 E^2}{k}\) is the correct expression for the energy.

Therefore, the correct statement is: The total energy of the system is \(\frac{1}{2} m \omega^2 A^2 + \frac{1}{2} \frac{q^2 E^2}{k}\).