Question

A 500pF capacitor is charged by a 2V battery.(1) how much electrostatic energy is stored by the capacitor?(ii) the capacitor is disconnected from the battery and connected in parallel to another 500pF capacitor. calculate the electrostatic energy stored by the system. (iii)in going from(i) to (ii) where has the remaining energy gone?

Answer 1

Dear student,

$$\begin{gathered} energyincapacitor=\frac{1}{2}c{v}^2=0.5*500pF*4=1000pJ. \\ thechargeremainconserved \\ Q=500pF*2=1000pC \\ chargewilldistributeequally. \\ chargeoneachwillbeQ\text{'}=500pC \\ energyinsystem=2*\frac{Q{\text{'}}^2}{2c}=\frac{Q{\text{'}}^2}{c}=\frac{{\left(500pC\right)}^2}{500pF}=500pJ \\ hencewelost500pJofenergy. \\ energylostinredistributionofthecharges. \\ Regards \end{gathered}$$