Question

An insulated material of dielectric constant K is inserted into half of the space between the parallel plates. If initial capacitance was C, then its new capacitance will be?

Answer 1

Dear Student

When dielectric of thickness t is inserted between plates then new capacitance will become
$$\begin{gathered} Initialcapaci\tan ceC=\frac{{\epsilon }_{o}A}{d} \\ FinalcapacitanceC\text{'}=\frac{{\epsilon }_{o}A}{d-t\left(1-{\displaystyle \frac{1}{k}}\right)} \\ C\text{'}=\frac{{\epsilon }_{o}A}{d-{\displaystyle \frac{d}{2}}\left(1-{\displaystyle \frac{1}{k}}\right)}=\frac{{\epsilon }_{o}A}{d-{\displaystyle \frac{d}{2}+\frac{d}{2k}}}=\frac{{\epsilon }_{o}A}{\frac{d}{2}+\frac{d}{2k}}=\frac{{\epsilon }_{o}A}{\frac{dk+d}{2k}}=\frac{{\epsilon }_{o}A}{\frac{d(k+1)}{2k}} \\ C\text{'}=\frac{2k{\epsilon }_{o}A}{d(k+1)}=\frac{2k}{(k+1)}C \end{gathered}$$
Regards