Question

A capacitor of capacitance 1 $ \mu F $ is filled with two dielectrics of dielectric constants 4 and 6. What is the new capacitance?

• A. 10 $ \mu F $
• B. 5 $ \mu F $
• C. 4 $ \mu F $
• D. 7 $ \mu F $

Answer 1

Answer: (B)

Initially, the capacitance of capacitor
$ C=\frac{{{\varepsilon }_{0}}A}{d} $
$ \therefore $ $ \frac{{{\varepsilon }_{0}}A}{d}=1\mu F $ ... (i)
When it is filled with dielectrics of dielectric constants $ {{K}_{1}} $ and $ {{K}_{2}} $ as shown, then there are two capacitors connected in parallel. So, $ C=\frac{{{K}_{1}}{{\varepsilon }_{0}}(A/2)}{d}+\frac{{{K}_{2}}{{\varepsilon }_{0}}(A/2)}{d} $ (as area becomes half)
$ C=\frac{4{{\varepsilon }_{0}}A}{2d}+\frac{6{{\varepsilon }_{0}}A}{2d} $
$=\frac{2{{\varepsilon }_{0}}A}{d}+\frac{3{{\varepsilon }_{0}}A}{d} $
Using Eq. (i),
we obtain $ C=2\times 1+3\times 1=5\mu F $