Question

A parallel plate capacitor whose capacitance $C$ is $14pF$ is charged by a battery to a potential difference $V=12V$ 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)

Answer 1

Step 1: Given Data

Capacitance, $C=14pF$

Potential difference, $V=12V$

Dielectric constant, $k=7$

Step 2: Find initial energy stored in the capacitor

Initial energy stored in the capacitor,

$$\begin{gathered} U_i=(½)c{v}^2 \\ {U}_i=(½)\times 14\times {\left(12\right)}^{2}pJ \\ {U}_i=1008pJ \end{gathered}$$

Step 3: Find the final energy stored in the capacitor

The final energy stored in the capacitor is

$$\begin{gathered} U_f=Q^2/2kC \\ {U}_f=\frac{{\left(14\times 12\right)}^2}{2\times 7\times 14} \\ {U}_f=144pJ \end{gathered}$$

Step 4: Find oscillating energy

oscillating energy is,

$$\begin{gathered} U_o=U_i–U_f \\ {U}_o=1008-144 \\ {U}_o=864pJ \end{gathered}$$

Hence, the plate would oscillate back and forth between the plates, with constant mechanical energy of $864pJ.$