A parallel plate capacitor whose capacitance C is 14 pF is charged by a battery to a potential difference V=12 V between its plates. The charging battery is now disconnected and a porcelain plate with k=7 is inserted between the plates, then the plate would oscillate back and forth between the plates with a constant mechanical energy of _______ pJ. (Assume no friction)

see full answer

Start JEE / NEET / Foundation preparation at rupees 99/day !!

21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!

An Intiative by Sri Chaitanya

answer is 864.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

Initial energy stored in the capacitor,
$U_{i} = \frac{1}{2} \times 14 \times 12 \times 12 pJ = 1008 pJ$
After inserting porcelain plate, k=7
$\begin{array}{l} U_{f} = \frac{1}{2} \frac{{CV}^{2}}{k} = 144 pJ \\ △ U = U_{i} - U_{f} = 1008 - 144 \\ = 864 pJ \end{array}$
Constant M.E. with which plate oscillates back and forth

Watch 3-min video & get full concept clarity

hear from our champions

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Books for IIT JEE

Best Books for NEET

🔥 Start Your JEE/NEET Prep at Just ₹1999 / month - Limited Offer! Check Now!

Get Expert Academic Guidance – Connect with a Counselor Today!

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Books for IIT JEE

Best Books for NEET