A $6 \mu F$ capacitor is charged from $10\,\text{volts}$ to…

Physics Electrostatics Potential & Capacitance Capacitance Single Correct MCQ

Published by: CGP EDU Academic Team

Published on: September 12, 2026

A $6 \mu F$ capacitor is charged from $10\,\text{volts}$ to 20 volts . Increase in energy will be

$18 \times 10^{-4} J$

$9 \times 10^{-4} j$

$4.3 \times 10^{-4} J$

$9 \times 10^{-6} J$

Text Solution

Verified by Experts

The correct answer is:

$\Delta E = E_{Final} - E_{Initial} = \frac{1}{2} C \left( V_{Final}^2 - V_{Initial}^2 \right)$

$= \frac{1}{2} \times 6 \times (20^{2} - 10^{2}) \times 10^{-6}$

$=3 \times (400 - 100) \times 10^{-6} = 3 \times 300 \times 10^{-6} = 9 \times 10^{-4} J$

Previous Question Next Question

Get practice questions, solutions, and test series in one place.

Share your experience with this question and solution.

Send your comment, doubt, correction, or feedback to admin.

What students say about this solution

No reviews yet. Be the first to write a review.