Question

Two charges (-q, m) and (+q, m) are connected by a massless rod of length $\ell$ and released in a region of uniform electric field as shown :

Find the maximum speed of the charges :

• A. $\sqrt{\frac{qEl}{2m}}$
• B. $\sqrt{\frac{3q\ell E}{2m}}$
• C. $\sqrt{\frac{3qE\ell}{m}}$
• D. $\sqrt{\frac{qE\ell}{3m}}$

Answer 1

Answer: (A)

$\sqrt{\frac{qEl}{2m}}$

Answer 2

The system's initial potential energy is $U_i = -pE \cos\theta_i$. The final potential energy at maximum speed is $U_f = -pE \cos\theta_f$, where $\theta_f = 0^\circ$ (dipole aligned with the field). Assuming the initial angle between the dipole moment and the electric field is $\theta_i = 60^\circ$, the change in potential energy is $\Delta U = U_f - U_i = -q\ell E - (-q\ell E \cos 60^\circ) = -q\ell E + q\ell E/2 = -q\ell E/2$.

By conservation of energy, the maximum kinetic energy gained is $K_{max} = -\Delta U = q\ell E/2$.

The total kinetic energy of the two charges is $K_{max} = 2 \times \frac{1}{2} m v_{max}^2 = m v_{max}^2$.

Equating the two expressions for kinetic energy: $m v_{max}^2 = q\ell E/2$.

Solving for $v_{max}$: $v_{max} = \sqrt{\frac{q\ell E}{2m}}$.