Search any question & find its solution
Question:
Answered & Verified by Expert
A 12 pF capacitor is connected to a 50 V battery, the electrostatic energy stored in the capacitor in nJ is
Options:
Solution:
1092 Upvotes
Verified Answer
The correct answer is:
15
Electrostatic energy stored $U=\frac{1}{2} C V^2$
$=\frac{1}{2} \times 12 \times 10^{-12} \times(50)^2$
$=6 \times 25 \times 10^{-10}$
$=15 \times 10^{-9} \mathrm{~J}$
$=15 \mathrm{~nJ}$
$=\frac{1}{2} \times 12 \times 10^{-12} \times(50)^2$
$=6 \times 25 \times 10^{-10}$
$=15 \times 10^{-9} \mathrm{~J}$
$=15 \mathrm{~nJ}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.