Question

A parallel - plate capacitor with plate area A has separation $d$ between the plates. Two dielectric slabs of dielectric constant $K_{1}$ and $K_{2}$ of same area $A / 2$ and thickness $d / 2$ are inserted in the space between the plates. The capacitance of the capacitor will be given by:

• A. $\frac{\varepsilon_{0} A }{ d }\left(\frac{1}{2}+\frac{ K _{1} \,K _{2}}{ K _{1}+ K _{2}}\right)$
• B. $\frac{\varepsilon_{0} A }{ d }\left(\frac{1}{2}+\frac{ K _{1}\, K _{2}}{2\left( K _{1}+ K _{2}\right)}\right)$
• C. $\frac{\varepsilon_{0} A }{ d }\left(\frac{1}{2}+\frac{ K _{1}+ K _{2}}{ K _{1} \,K _{2}}\right)$
• D. $\frac{\varepsilon_{0} A }{ d }\left(\frac{1}{2}+\frac{2\left( K _{1}+ K _{2}\right)}{ K _{1}\, K _{2}}\right)$

Answer 1

Answer: (A)

$C _{\text{eq }}=\frac{\frac{ A }{2} \varepsilon_{0}}{ d }+\frac{ A \varepsilon_{0}}{ d } \frac{ K _{1}\, K _{2}}{ K _{1}+ K _{2}}$
$=\frac{ A \varepsilon_{0}}{ d }\left(\frac{1}{2}+\frac{ K _{1} \,K _{2}}{ K _{1}+ K _{2}}\right)$