Answer in Electric Circuits for Tony #159583

(a) The energy stored in the capacitor can be calculated as follows:

“E=dfrac{1}{2}CV^2,”“E=dfrac{1}{2}cdot72cdot10^{-12} Fcdot(12 V)^2=5.18cdot10^{-9} J.”

b) Let’s first calculate the charge stored on the capacitor:

“Q=CV,”“Q=72cdot10^{-12} Fcdot12 V=8.64cdot10^{-10} C.”

Let’s find the new capacitance of the capacitor. The capacitance can be calculated as follows:

“C=dfrac{epsilon_0A}{d}.”

From the conditions of the question we know that the separation between the plates of the capacitor is doubled, therefore, the new capacitance will be:

“C_{new}=dfrac{1}{2}cdotdfrac{epsilon_0A}{d}=dfrac{1}{2}cdot C=dfrac{1}{2}cdot72cdot10^{-12} F=36cdot10^{-12} F.”

Finally, we can find the energy stored in the capacitor:

“E=dfrac{1}{2}dfrac{Q^2}{C}=dfrac{1}{2}cdotdfrac{(8.64cdot10^{-10} C)^2}{36cdot10^{-12} F}=1.04cdot10^{-8} J.”