JEE Mains · Physics · STD 12 - 2. Electric potential and capacitance
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 :

- 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 & Solution
Correct Answer
(A) \(\frac{\varepsilon_{0} {A}}{{d}}\left(\frac{1}{2}+\frac{{K}_{1} {K}_{2}}{{K}_{1}+{K}_{2}}\right)\)
Step-by-step Solution
Detailed explanation
\(C_{eq}=\frac{\frac{A}{2} \varepsilon_{0}}{d}+\frac{A \varepsilon_{0}}{d} \frac{K_{1} K_{2}}{K_{1}+K_{2}}\) \(=\frac{A \varepsilon_{0}}{d}\left(\frac{1}{2}+\frac{K_{1} K_{2}}{K_{1}+K_{2}}\right)\)
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