Published by:
CGP EDU Academic Team
Published on: September 12, 2026
An oil drop having charge $2e$ is kept stationary between two parallel horizontal plates 2.0 cm apart when a potential difference of 12000 volts is applied between them. If the density of oil is 900 kg/m 3 , the radius of the drop will be
Text Solution
Verified by ExpertsThe correct answer is:
B
In equilibrium QE = mg
⇒ ⇒ $Q \cdot \frac{v}{d} = mg = \left(\frac{4}{3} \pi r^{3} \rho \right) g$
⇒ ⇒ $2 \times 1.6 \times 10^{-19} \times \frac{12000}{2 \times 10^{-2}} = \frac{4}{3} \pi r^{3} \times 900 \times 10^{-3}$
⇒ ⇒ r = 1.7 × × 10 –6 m
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