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NEET · Physics · STD 12 - 2. Electric potential and capacitance
A parallel-plate capacitor of area \(A,\) plate separation \(d\) and capacitance \(C\) is filled with four dielectric materials having dielectric constants \(K_1,K_2,K_3\) and \(K_4\) as shown in the figure. If a single dielectric material is to be used to have the same capacitance \(C\) in this capacitor, then its dielectric constant \(K\) is given by

- A \(\frac{2}{K} = \frac{3}{{{K_1} + {K_2} + {K_3}}} + \frac{1}{{{K_4}}}\;\;\;\;\)
- B \(\;\frac{1}{K} = \frac{1}{{{K_1}}} + \frac{1}{{{K_2}}} + \frac{1}{{{K_3}}} + \frac{3}{{2{K_4}}}\)
- C \(K=K_1+K_2+K_3+3K_4\)
- D \(K=\) \(\frac{2}{3}\left[ {{K_1} + {K_2} + {K_3}} \right] + 2{K_4}\)
Answer & Solution
Correct Answer
(A) \(\frac{2}{K} = \frac{3}{{{K_1} + {K_2} + {K_3}}} + \frac{1}{{{K_4}}}\;\;\;\;\)
Step-by-step Solution
Detailed explanation
Here, \(C_{1}=\frac{2 \varepsilon_{0} k_{1} A}{3 d}, C_{2}=\frac{2 \varepsilon_{0} k_{2} A}{3 d}\) \(C_{3}=\frac{2 \varepsilon_{0} k_{3} A}{3 d}, C_{4}=\frac{2 \varepsilon_{0} k_{4} A}{d}\)
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