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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$
$2$
$32$
$6$
$8$
Solution
$C _{\text {original }}=\frac{ A \varepsilon_{0}}{ d }$
$C _{1}=\frac{ A \varepsilon_{0}}{ d / 2}=\frac{2 A \varepsilon_{0}}{ d }= C$
$C _{2}=\frac{ KA \varepsilon_{0}}{ d / 2}=\frac{2 KA \varepsilon_{0}}{ d }=\frac{6 A \varepsilon_{0}}{ d }=3 C$
$C_{1}$ and $C_{2}$ are in series
$C _{\text {new }}=\frac{ C _{1} C _{2}}{ C _{1}+ C _{2}}=\frac{ C \times 3 C }{ C +3 C }=\frac{3 C }{4}$
$=\frac{3}{4} \times \frac{2 A \varepsilon_{0}}{d}=\frac{3}{2} \times \frac{A \varepsilon_{0}}{d}$
$C _{\text {new }}=\frac{3}{2} C _{\text {original }}$
$=\frac{3}{2} \times 4=6\, \mu\, F$
