JEE Challenger
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PhyMedAdv 2015 • Paper 1

Motion of Charges Between Infinite Line Charges After Displacement

The diagrams below show two scenarios where two static, infinitely long line charges possessing uniform positive charge density λ\lambda are positioned parallel to one another. Within their net electric field, point charges +q+q and −q-q are situated in equilibrium between the lines. These point charges can only move along the xx-axis. If a small perturbation from their equilibrium state is imparted to each charge, select the true statement(s):

Question Diagram 1

Options

A

Both point charges undergo simple harmonic motion.

B

Both point charges keep moving along the direction of their initial displacement.

C

The +q+q charge undergoes simple harmonic motion, whereas the −q-q charge keeps moving along its displacement direction.

Correct
D

The −q-q charge undergoes simple harmonic motion, whereas the +q+q charge keeps moving along its displacement direction.

Topics & Concepts

Electric Charges and FieldsElectric FieldSHM

Step-by-Step Solution

To determine the nature of motion of the charges after a small displacement along the xx-axis, let us analyze the net electric field between the two parallel, infinitely long line charges.

1. Setup and Electric Field Calculation

Let the two infinitely long line charges, each having a uniform positive charge density λ\lambda, be located at x=−dx = -d and x=+dx = +d. The midpoint between them is chosen as the origin x=0x = 0, which is the equilibrium position for both point charges.

For any point xx between the line charges (−d<x<d-d < x < d), the electric field due to each line charge at distance rr is given by Gauss's Law as E=λ2πε0rE = \frac{\lambda}{2\pi \varepsilon_0 r}.

  • The electric field due to the left line charge at x=−dx = -d is directed along the +x+x-direction: E1=λ2πε0(d+x)i^E_1 = \frac{\lambda}{2\pi \varepsilon_0 (d + x)} \hat{i}

  • The electric field due to the right line charge at x=+dx = +d is directed along the −x-x-direction: E2=λ2πε0(d−x)i^E_2 = \frac{\lambda}{2\pi \varepsilon_0 (d - x)} \hat{i}

The net electric field E⃗net\vec{E}_{\text{net}} at position xx is: E⃗net=E1−E2=λ2πε0[1d+x−1d−x]i^\vec{E}_{\text{net}} = E_1 - E_2 = \frac{\lambda}{2\pi \varepsilon_0} \left[ \frac{1}{d + x} - \frac{1}{d - x} \right] \hat{i}

E⃗net=λ2πε0[(d−x)−(d+x)d2−x2]i^=−λxπε0(d2−x2)i^\vec{E}_{\text{net}} = \frac{\lambda}{2\pi \varepsilon_0} \left[ \frac{(d - x) - (d + x)}{d^2 - x^2} \right] \hat{i} = -\frac{\lambda x}{\pi \varepsilon_0 (d^2 - x^2)} \hat{i}

For a very small displacement x≪dx \ll d, we can approximate d2−x2≈d2d^2 - x^2 \approx d^2: E⃗net≈−(λπε0d2)xi^\vec{E}_{\text{net}} \approx -\left( \frac{\lambda}{\pi \varepsilon_0 d^2} \right) x \hat{i}

2. Analysis of the Motion of +q+q

The electrostatic force acting on the charge +q+q when displaced by a small distance xx from the origin is: F⃗+q=(+q)E⃗net=−(qλπε0d2)xi^\vec{F}_{+q} = (+q) \vec{E}_{\text{net}} = -\left( \frac{q\lambda}{\pi \varepsilon_0 d^2} \right) x \hat{i}

  • Notice that the restoring force is directed opposite to the displacement (F⃗∝−x\vec{F} \propto -x).
  • Hence, the equilibrium is stable, and the charge +q+q undergoes Simple Harmonic Motion (SHM) with an angular frequency ω=qλπε0md2\omega = \sqrt{\frac{q\lambda}{\pi \varepsilon_0 m d^2}}.

3. Analysis of the Motion of −q-q

The electrostatic force acting on the charge −q-q when displaced by a small distance xx from the origin is: F⃗−q=(−q)E⃗net=+(qλπε0d2)xi^\vec{F}_{-q} = (-q) \vec{E}_{\text{net}} = +\left( \frac{q\lambda}{\pi \varepsilon_0 d^2} \right) x \hat{i}

  • Notice that the force is in the same direction as the displacement (F⃗∝+x\vec{F} \propto +x).
  • Hence, the equilibrium is unstable, and the force accelerates the −q-q charge further away from the equilibrium position.
  • Consequently, the charge −q-q continues moving along the direction of its initial displacement.

Conclusion

The +q+q charge undergoes simple harmonic motion, whereas the −q-q charge keeps moving along its displacement direction.

Correct Option: C

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