JEE Mains · Physics · STD 12 - 2. Electric potential and capacitance
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 \(........... \mu F\)
- A \(4\)
- B \(5\)
- C \(2\)
- D \(3\)
Answer & Solution
Correct Answer
(D) \(3\)
Step-by-step Solution
Detailed explanation
\(\text { For } x =\frac{ d }{3}\) \(C _1=\frac{\epsilon_0 A }{\left(\frac{ d / 3}{ k }+\frac{2 d }{3}\right)}=\frac{\epsilon_0 A }{\frac{ d }{12}+\frac{2 d }{3}}\) \(=\frac{\epsilon_0 A }{ d } \times\left(\frac{12}{9}\right)\)…
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