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STD 12 - 2. Electric potential and capacitance — NEET STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceNEETSTD 12 - 2. Electric potential and capacitance4 Marks
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 $\frac A2$ and thickness $\frac d2$ are inserted in the space between the plates. The capacitance of the capacitor will be given by :
✓

Answer

$C_{eq}=\frac{\frac{A}{2} \varepsilon_{0}}{d}+\frac{A \varepsilon_{0}}{d} \frac{K_{1} K_{2}}{K_{1}+K_{2}}$

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

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