Published by:
CGP EDU Academic Team
Published on: September 12, 2026
A particle A has charge $+q$ and a particle B has charge + 4q with each of them having the same mass m. When allowed to fall from rest through the same electric potential difference, the ratio of their speed $\frac{V_A}{V_B}$ will become
Text Solution
Verified by ExpertsThe correct answer is:
B
Using $v = \sqrt{\frac{2QV}{m}}$ ⇒ ⇒ $v \propto \sqrt{Q}$ ⇒ ⇒ $\frac{v_A}{v_B} = \sqrt{\frac{Q_A}{Q_B}} = \sqrt{\frac{q}{4q}} = \frac{1}{2}$
Prepare Smarter with CGP Edu
Get practice questions, solutions, and test series in one place.
Write a Review
Share your experience with this question and solution.
Commentary
Send your comment, doubt, correction, or feedback to admin.
Similar Questions
Explore conceptually related problems
A charge q is placed at the center of the line joining two equal charges Q. The system of the three…
Inside a hollow charged spherical conductor, the potential
Two small spheres each carrying a charge q are placed r meter apart. If one of the spheres is taken…
The electric charge in uniform motion produces
Two charged spheres of radii 10 cm and 15 cm are connected by a thin wire. No current will flow, if…
The electric field inside a spherical shell of uniform surface charge density is