Case: Read the passage and answer the question.
An empty 20 pF capacitor is charged to a potential difference of 40 V. The charging battery is then disconnected, and a piece of Teflon with a dielectric constant of 2.1 is inserted to completely fill the space between the capacitor plates.
What is the energy stored in the capacitor with the dielectric and without the dielectric, respectively?
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 7.6 nJ, 16 nJ
Without the dielectric, the energy stored in a capacitor is \( U_0=\frac{1}{2}CV^{2} \). The battery is disconnected before the Teflon is inserted, so the charge Q on the plates stays constant. The dielectric raises the capacitance from \( C \) to \( KC \). Since \( U=Q^{2}/2C \), the energy falls by the factor \( K \).
The data are:
\( C=20\,\mathrm{pF}=20\times10^{-12}\,\mathrm{F} \)
\( V=40\,\mathrm{V} \)
\( K=2.1 \)
The energy without the dielectric is:
\( \begin{aligned} U_0 &= \frac{1}{2}CV^{2} \\ &= \frac{1}{2}\times20\times10^{-12}\,\mathrm{F}\times(40\,\mathrm{V})^{2} \\ &= 1.6\times10^{-8}\,\mathrm{J} \\ &= 16\,\mathrm{nJ} \end{aligned} \)
With the dielectric, the charge is the same and the capacitance is \( KC \), so the energy is:
\( \begin{aligned} U &= \frac{Q^{2}}{2KC} \\ &= \frac{U_0}{K} \\ &= \frac{16\,\mathrm{nJ}}{2.1} \\ &\approx 7.6\,\mathrm{nJ} \end{aligned} \)
Therefore, the energy stored is 7.6 nJ with the dielectric and 16 nJ without it.