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JEE Mains · Physics · STD 12 - Electric potential and capacitance
  • JEE Mains 2023
  • Hard
  • Numerical

A parallel plate capacitor with plate area \(A\) and plate separation \(d\) is filled with a dielectric material of dielectric constant \(K =4\). The thickness of the dielectric material is \(x\), where \(x < d\). Let \(C_1\) and \(C_2\) be the capacitance of the system for \(x =\frac{1}{3} d\) and \(x =\frac{2 d }{3}\), respectively. If \(C _1=2 \mu F\) the value of \(C _2\) is \(........... \mu F\)

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Answer & Solution

Correct Answer

(D) \(3\)

Step-by-step Solution

Detailed explanation

\(\text { For } x =\frac{ d }{3}\)

\(C _1=\frac{\epsilon_0 A }{\left(\frac{ d / 3}{ k }+\frac{2 d }{3}\right)}=\frac{\epsilon_0 A }{\frac{ d }{12}+\frac{2 d }{3}}\)

\(=\frac{\epsilon_0 A }{ d } \times\left(\frac{12}{9}\right)\)

\(C _1=\frac{4}{3} \frac{\epsilon_0 A }{ d }=2 \mu F\)

\(\text { for } x =\frac{2 d }{3}\)

\(C _2=\frac{\epsilon_0 A }{\left(\frac{2 d / 3}{ k }+\frac{ d }{3}\right)}=\frac{\epsilon_0 A }{ d } \times 2\)

\(\Rightarrow \frac{6}{4} \times 2=3 \mu F\)