A parallel plate capacitor is charged to a potential \(\mathrm{V}\) using a battery which is later…
A parallel plate capacitor is charged to a potential \(\mathrm{V}\) using a battery which is later disconnected.
A dielectric slab is inserted between the plates such that it completely fills the space between the plates. If \(A\) is the area of the plates and \(d\) is the plate separation, then work done in inserting the dielectric is (K is the dielectric constant)
Before insert dielectric, capacitance, \(\mathrm{C}_{0}=\frac{\mathrm{A} \epsilon_{0}}{\mathrm{~d}}\)
Magnitude of charge on each plate, \(\mathrm{Q}=\mathrm{C}_{0} \mathrm{~V}=\frac{\mathrm{A} \epsilon_{0}}{\mathrm{~d}} \mathrm{~V}\)
Before disconnect the battery, filed between the plates, \(\mathrm{E}_{0}=\frac{\mathrm{V}}{\mathrm{d}}\)
Afe insert dielectric, electric filed, \(\mathrm{E}=\frac{\mathrm{E}_{0}}{\mathrm{~K}}=\frac{\mathrm{V}}{\mathrm{Kd}}\)
Before insert dielectric, work done, \(=\mathrm{W}_{\mathrm{i}}=\frac{1}{2} \mathrm{C}_{0} \mathrm{~V}^{2}\)
After insert work done, \(\mathrm{W}_{\mathrm{f}}=\frac{1}{2}\left(\mathrm{KC}_{0}\right)\left(\frac{\mathrm{V}}{\mathrm{K}}\right)^{2} .\left(\mathrm{C}_{\mathrm{f}}=\frac{\mathrm{A} \epsilon_{0} \mathrm{~K}}{\mathrm{~d}}, \mathrm{~V}_{\mathrm{f}}=\frac{\mathrm{E}}{\mathrm{d}}\right)\)
So net work done, \(\mathrm{W}=\mathrm{W}_{\mathrm{i}}-\mathrm{W}_{\mathrm{f}}=\frac{1}{2} \mathrm{C}_{0} \mathrm{~V}^{2}-\frac{1}{2}\left(\mathrm{KC}_{0}\right)\left(\frac{\mathrm{V}}{\mathrm{K}}\right)^{2}=\frac{1}{2} \mathrm{C}_{0} \mathrm{~V}^{2}\left(1-\frac{1}{\mathrm{~K}}\right)=\)
\(\frac{1}{2} \frac{\mathrm{A} \epsilon_{0} \mathrm{~V}^{2}}{\mathrm{~d}}\left(1-\frac{1}{\mathrm{~K}}\right)\)