Question

A parellel plate capacitor is charged by a battery. After some time the battery is disconnected and a dielectric slab of dielectric constant K is inserted between the plates. How would (i) the capacitance, (ii) the electric field between the plates and (iii) the energy stored in the capacitor, be affected? Justify your answer.

Answer 1

$$\begin{gathered} Derastudent \\ capaciatnce,C=\frac{A{\in }_0}{d} \\ ifdielectricisinserted,C=\frac{Ak{\in }_0}{d} \\ capaci\tan ceincresesbyktimes \\ Electricfield=E=\frac{\sigma }{{\in }_0} \\ ifdielectrcisinserted,E=\frac{\sigma }{k{\in }_0} \\ electricfielddecreasesbyktimes \\ energystoredincapacitor=\frac{1}{2}{\in }_0{E}^2 \\ ifdielectrcisinserted,\frac{1}{2}k{\in }_0{\left(\frac{\sigma }{k{\in }_0}\right)}^2 \\ energystoreddecreasesbyktimes \\ Regards \end{gathered}$$