Question

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 :

• A. $2\, \mu F$
• B. $32\, \mu F$
• C. $6\, \mu F$
• D. $8 \, \mu F$

Answer 1

Answer: (C)

$ C _{1}=\frac{ A \varepsilon_{0}}{ d / 2}=\frac{2 A \varepsilon_{0}}{ d }= C $
$C _{2}=\frac{ KA \varepsilon_{0}}{ d / 2}=\frac{2 KA \varepsilon_{0}}{ d }=\frac{6 A \varepsilon_{0}}{ d }=3 C$
$C _{1} \,\& \,C _{2}$ 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 }} $
$=\frac{3}{2} \times 4=6 \mu F$