Question Number 40786 by Necxx last updated on 27/Jul/18
$${A}\:{parallel}\:{plate}\:{capacitor}\:{of}\:{plate} \\ $$$${spacing},\:\mathrm{1}{mm}\:{is}\:{charged}\:{to}\:{a}\:{potential} \\ $$$${of}\:\mathrm{50}{V}.{Find}\:{the}\:{energy}\:{density}\:{in} \\ $$$${the}\:{capacitor} \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 27/Jul/18
$${energy}=\frac{\mathrm{1}}{\mathrm{2}}{CV}\:^{\mathrm{2}} \: \\ $$$${energy}\:{density}=\frac{\frac{\mathrm{1}}{\mathrm{2}}{CV}\:^{\mathrm{2}} }{{Ad}}\:\:{C}={capscitance} \\ $$$${V}={potential}\:{difference} \\ $$$${A}={area}\:{of}\:{plate}\:{of}\:{cspacitor} \\ $$$${d}={separation}\:{between}\:{plate} \\ $$$${C}=\frac{\epsilon_{\mathrm{0}} {A}}{{d}} \\ $$$${energy}\:{density}=\frac{\mathrm{1}}{\mathrm{2}}×\frac{\frac{\epsilon_{\mathrm{0}} {A}}{{d}}.{V}\:^{\mathrm{2}} }{{Ad}} \\ $$$$\:\:\:=\frac{\mathrm{1}}{\mathrm{2}}.\epsilon_{\mathrm{0}} .\frac{{V}\:^{\mathrm{2}} \:}{{d}^{\mathrm{2}} } \\ $$$$ \\ $$