A parallel plate capacitor with plate area \(A\) and plate separation \(d =2 \,m\) has a capacitance of \(4 \,\mu F\). The new capacitance of the system if half of the space between them is filled with a dielectric material of dielectric constant \(K =3\) (as shown in figure) will be .........\( \mu \,F\)

\(C _{1}=\frac{ A \varepsilon_{0}}{ d / 2}=\frac{2 A \varepsilon_{0}}{ d }= C\)

\(C _{2}=\frac{ KA \varepsilon_{0}}{ d / 2}=\frac{2 KA \varepsilon_{0}}{ d }=\frac{6 A \varepsilon_{0}}{ d }=3 C\)

\(C_{1}\) and \(C_{2}\) are in series

\(C _{\text {new }}=\frac{ C _{1} C _{2}}{ C _{1}+ C _{2}}=\frac{ C \times 3 C }{ C +3 C }=\frac{3 C }{4}\)

\(=\frac{3}{4} \times \frac{2 A \varepsilon_{0}}{d}=\frac{3}{2} \times \frac{A \varepsilon_{0}}{d}\)

\(C _{\text {new }}=\frac{3}{2} C _{\text {original }}\)

\(=\frac{3}{2} \times 4=6\, \mu\, F\)