JEE Mains · Physics · STD 12 - 2. Electric potential and capacitance
A parallel plate capacitor has capacitance C , when there is vacuum within the parallel plates.
A sheet having thickness \(\left(\frac{1}{3}\right)^{ rd }\) of the separation between the plates and relative permittivity K is introduced between the plates. The new capacitance of the system is :
- A \(\frac{3 KC }{2 K+1}\)
- B \(\frac{ CK }{2+ K }\)
- C \(\frac{3 CK ^2}{(2 K+1)^2}\)
- D \(\frac{4 KC }{3 K-1}\)
Answer & Solution
Correct Answer
(A) \(\frac{3 KC }{2 K+1}\)
Step-by-step Solution
Detailed explanation
\(C_1=\frac{3 A \epsilon_0}{2 d}\)\(\quad\)\(C _2=\frac{3 A \in_0 \times K }{ d }\) \(C _1=\frac{3}{2} C\)\(\quad\)\(C _2=3 KC\) \(Ceq =\frac{ C _1 C _2}{ C _1+ C _2}=\frac{\frac{3}{2} C \times 3 KC }{\frac{3}{2} C +3 KC }\)…
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