Topic: Electronics Engineering
A parallel plate capacitor of 100 pF having an air dielectric is charged to 10 kilovolts. It is then electrically isolated. The plates are pulled away from each other until the distance is ten times more than before. Estimate the energy needed to pull the plates.
- A. 0.05 Joules
- B. 50 Joules
- C. 500 Joules
- D. -50 Joules
Answer: A. 0.05 Joules
Explanation: $E_{1}=\frac{1}{2}C_{1}V_{1}^{2}=\frac{1}{2}\times100\times10^{-12}\times10^{8} = \frac{1}{2}\times10^{-2}$
$C_{2}=\frac{C_{1}}{10}=100\times10^{-13}$
$E_{2}=\frac{1}{2}C_{2}V_{2}^{2}=\frac{1}{2}\times100\times10^{-13}\times10^{8} = \frac{1}{2}\times10^{-3}$
$\Delta E=E_{1}-E_{2}$
$= \frac{1}{2}\times\frac{1}{100}-\frac{1}{2}\times\frac{1}{1000}$
$= \frac{1}{2\times100}[1-\frac{1}{10}]$
$= \frac{1}{2\times100}\times\frac{9}{10}=4.5\times10^{-3}$