Question

the plates of a parallel plate system are charged to 100V. A 4mm thickness dielectric slab is inserted between the plates. then to maintain the same potential difference, the distance between the system plates is increased by 2mm.find the dielectric constant.

Answer 1

$$\begin{gathered} \mathrm{According}\mathrm{to}\mathrm{charge}\mathrm{conservation},\mathrm{the}\mathrm{charge}\mathrm{before}\mathrm{the}\mathrm{separation}\mathrm{is}\mathrm{increased}\mathrm{and}\mathrm{afterwards}\mathrm{must}\mathrm{be}\mathrm{the}\mathrm{same}. \\ \mathrm{Moreover},\mathrm{for}\mathrm{the}\mathrm{potential}\mathrm{difference}\mathrm{is}\mathrm{to}\mathrm{remain}\mathrm{the}\mathrm{same},\mathrm{the}\mathrm{capacitances}\mathrm{in}\mathrm{the}\mathrm{two}\mathrm{cases}\mathrm{will}\mathrm{be}\mathrm{equal}. \\ \frac{{\in }_0\mathrm{A}}{\mathrm{d}}=\frac{{\in }_0\mathrm{A}}{\mathrm{d}-\mathrm{t}(1-{\displaystyle \frac{1}{\mathrm{k}}})+\mathrm{x}} \\ \mathrm{or}\mathrm{x}=\mathrm{t}(1-\frac{1}{\mathrm{k}}) \\ \mathrm{Putting}\mathrm{the}\mathrm{values},2=4(1-\frac{1}{\mathrm{k}}) \\ \mathrm{or}(1-\frac{1}{\mathrm{k}})=\frac{1}{2} \\ \mathrm{or}\frac{1}{\mathrm{k}}=\frac{1}{2} \\ \mathrm{or}\mathrm{k}=2 \end{gathered}$$