Question

Two-point charges, each Q, are fixed at separation 2d. A charged particle having charge q and mass m is placed between them. (a) Now this charged particle is slightly displaced along the line joining the charges, show that it will execute simple harmonic motion and find the time period of oscillation. (b) If charge q is negative and it is displaced slightly perpendicular to the line joining the charges, repeat the part (a).

Answer 1

a) T = 2\pi \sqrt{\frac{m d^3 \pi\epsilon_0}{Qq}}, b) T = 2\pi \sqrt{\frac{2m d^3 \pi\epsilon_0}{Q|q|}}

Answer 2

The problem involves analyzing the motion of a charged particle under the influence of two fixed charges. For simple harmonic motion (SHM) to occur, the net force on the displaced particle must be a restoring force, i.e., proportional to the displacement and directed towards the equilibrium position (F = -kx).

Let the two fixed charges Q be located at (-d, 0) and (d, 0). The separation between them is 2d.

Part (a): Charged particle q (mass m) displaced along the line joining the charges.

1. Equilibrium and Stability: For the particle q to execute SHM when displaced along the line joining the charges, the equilibrium position at (0, 0) must be stable. This happens if Q and q have the same sign (e.g., both positive). If q is positive and displaced to (x, 0) (where x << d), the repulsive force from the Q at (d, 0) will increase, and the repulsive force from the Q at (-d, 0) will decrease. This creates a net force pushing q back towards (0, 0).

2. Forces: Let q be displaced by a small distance x along the x-axis, so its position is (x, 0).

   • The force F_1 on q due to Q at (-d, 0) is F_1 = \frac{kQq}{(d+x)^2} (directed to the right, away from -d).
   • The force F_2 on q due to Q at (d, 0) is F_2 = \frac{kQq}{(d-x)^2} (directed to the left, away from d).

   Where k = \frac{1}{4\pi\epsilon_0}.

3. Net Force: The net force F_net on q is F_net = F_1 - F_2.

   F_net = kQq \left[ \frac{1}{(d+x)^2} - \frac{1}{(d-x)^2} \right]

   F_net = kQq \left[ \frac{(d-x)^2 - (d+x)^2}{(d+x)^2 (d-x)^2} \right]

   F_net = kQq \left[ \frac{(d^2 - 2dx + x^2) - (d^2 + 2dx + x^2)}{(d^2 - x^2)^2} \right]

   F_net = kQq \left[ \frac{-4dx}{(d^2 - x^2)^2} \right]

4. Small Displacement Approximation (x << d): For small x, (d^2 - x^2)^2 \approx (d^2)^2 = d^4.

   So, F_net \approx kQq \left( \frac{-4dx}{d^4} \right) = - \left( \frac{4kQq}{d^3} \right) x.

5. SHM and Time Period: The net force is of the form F_net = -K_{eff} x, where K_{eff} = \frac{4kQq}{d^3}. This indicates simple harmonic motion.

   The angular frequency \omega = \sqrt{\frac{K_{eff}}{m}} = \sqrt{\frac{4kQq}{md^3}}.

   The time period T = \frac{2\pi}{\omega} = 2\pi \sqrt{\frac{md^3}{4kQq}}.

   Substituting k = \frac{1}{4\pi\epsilon_0}:

   T = 2\pi \sqrt{\frac{md^3}{4 \left(\frac{1}{4\pi\epsilon_0}\right) Qq}} = 2\pi \sqrt{\frac{m d^3 \pi\epsilon_0}{Qq}}.

Part (b): If charge q is negative and it is displaced slightly perpendicular to the line joining the charges.

1. Equilibrium and Stability: Since q is negative and Q is positive, the forces are attractive. When q is displaced perpendicular to the line joining the charges, say to (0, y) (where y << d), the attractive forces from Q at (-d, 0) and (d, 0) will have components that pull q back towards (0, 0). This ensures stable equilibrium.

2. Forces: Let q be displaced by a small distance y along the y-axis, so its position is (0, y).

   • The distance from each Q to q is r = \sqrt{d^2 + y^2}.
   • The magnitude of the force from each Q on q is F = \frac{kQ|q|}{r^2} = \frac{kQ|q|}{d^2 + y^2}.
   • Let \theta be the angle between the line connecting Q and q and the x-axis. Then \sin\theta = \frac{y}{r} = \frac{y}{\sqrt{d^2 + y^2}}.
   • Due to symmetry, the x-components of the forces cancel out.
   • The y-components add up. Since q is negative, the forces are attractive, pulling q towards each Q. Thus, the y-component of each force will be directed downwards (towards the x-axis).
   • Net force in y-direction: F_y = -2 F \sin\theta.
   • F_y = -2 \left( \frac{kQ|q|}{d^2 + y^2} \right) \left( \frac{y}{\sqrt{d^2 + y^2}} \right)
   • F_y = - \frac{2kQ|q|y}{(d^2 + y^2)^{3/2}}.

3. Small Displacement Approximation (y << d): For small y, (d^2 + y^2)^{3/2} \approx (d^2)^{3/2} = d^3.

   So, F_y \approx - \frac{2kQ|q|y}{d^3}.

4. SHM and Time Period: The net force is of the form F_y = -K_{eff} y, where K_{eff} = \frac{2kQ|q|}{d^3}. This indicates simple harmonic motion.

   The angular frequency \omega = \sqrt{\frac{K_{eff}}{m}} = \sqrt{\frac{2kQ|q|}{md^3}}.

   The time period T = \frac{2\pi}{\omega} = 2\pi \sqrt{\frac{md^3}{2kQ|q|}}.

   Substituting k = \frac{1}{4\pi\epsilon_0}:

   T = 2\pi \sqrt{\frac{md^3}{2 \left(\frac{1}{4\pi\epsilon_0}\right) Q|q|}} = 2\pi \sqrt{\frac{2m d^3 \pi\epsilon_0}{Q|q|}}.

The motion will be simple harmonic in both cases under the specified conditions.

The final answer is $\boxed{a) T = 2\pi \sqrt{\frac{m d^3 \pi\epsilon_0}{Qq}}, \quad b) T = 2\pi \sqrt{\frac{2m d^3 \pi\epsilon_0}{Q|q|}}}$

Explanation of the solution:

(a) For longitudinal displacement, assuming Q and q have the same sign for stability, the net force on q displaced by x is F_x = kQq [1/(d+x)^2 - 1/(d-x)^2]. For x << d, this simplifies to F_x = -(4kQq/d^3)x. This is a restoring force F=-Kx. The time period is T = 2\pi \sqrt{m/K} = 2\pi \sqrt{md^3/(4kQq)} = 2\pi \sqrt{m d^3 \pi\epsilon_0 / Qq}.

(b) For perpendicular displacement, with q negative, the attractive forces from Q on q (displaced by y) have y-components adding up to F_y = -2kQ|q|y / (d^2+y^2)^{3/2}. For y << d, this simplifies to F_y = -(2kQ|q|/d^3)y. This is a restoring force F=-K'y. The time period is T = 2\pi \sqrt{m/K'} = 2\pi \sqrt{md^3/(2kQ|q|)} = 2\pi \sqrt{2m d^3 \pi\epsilon_0 / (Q|q|)}.

Answer:

(a) The time period of oscillation is T = 2\pi \sqrt{\frac{m d^3 \pi\epsilon_0}{Qq}}. (b) The time period of oscillation is `T = 2\pi \sqrt{\frac{2m d^3 \pi\epsilon_0}{Q|q|}}$.