MCQ
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 $\frac A2$ and thickness $\frac d2$ are inserted in the space between the plates. The capacitance of the capacitor will be given by :
  • $\frac{\varepsilon_{0} {A}}{{d}}\left(\frac{1}{2}+\frac{{K}_{1} {K}_{2}}{{K}_{1}+{K}_{2}}\right)$
  • 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

Correct option: A.
$\frac{\varepsilon_{0} {A}}{{d}}\left(\frac{1}{2}+\frac{{K}_{1} {K}_{2}}{{K}_{1}+{K}_{2}}\right)$
a
$C_{eq}=\frac{\frac{A}{2} \varepsilon_{0}}{d}+\frac{A \varepsilon_{0}}{d} \frac{K_{1} K_{2}}{K_{1}+K_{2}}$

$=\frac{A \varepsilon_{0}}{d}\left(\frac{1}{2}+\frac{K_{1} K_{2}}{K_{1}+K_{2}}\right)$

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