JEE Mains · Physics · STD 12 - 2. Electric potential and capacitance
A parallel plate capacitor has plate area \(100\, m ^{2}\) and plate separation of \(10\, m\). The space between the plates is filled up to a thickness \(5\, m\) with a material of dielectric constant of \(10 .\) The resultant capacitance of the system is \('x'\) \(pF\). The value of \(\varepsilon_{0}=8.85 \times 10^{-12} F \cdot m ^{-1}\) The value of \('x'\) to the nearest integer is............
- A \(144\)
- B \(161\)
- C \(169\)
- D \(152\)
Answer & Solution
Correct Answer
(B) \(161\)
Step-by-step Solution
Detailed explanation
\(A =100 m ^{2}\) Using \(C =\frac{ k \in_{0} A }{ d }\) \(C _{1}=\frac{10 \epsilon_{0}(100)}{5}\) \(=200 \in_{0}\) \(C _{2}=\frac{\epsilon_{0}(100)}{5}=20 \epsilon_{0}\) \(C _{1} \& C _{2}\) are in series so \(C _{ eqv. }=\frac{ C _{1} C _{2}}{ C _{1}+ C _{2}}\)
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