JEE Mains · Physics · STD 12 - 2. Electric potential and capacitance
The material filled between the plates of a parallel plate capacitor has resistivity \(200 \Omega \, {m}\). The value of capacitance of the capacitor is \(2\, {pF}\). If a potential difference of \(40 \,{V}\) is applied across the plates of the capacitor, then the value of leakage current flowing out of the capacitor is (given the value of relative permitivity of material is \(50\) )
- A \(9.0\, \mu {A}\)
- B \(9.0\, {mA}\)
- C \(0.9 \,{mA}\)
- D \(0.9 \,\mu {A}\)
Answer & Solution
Correct Answer
(C) \(0.9 \,{mA}\)
Step-by-step Solution
Detailed explanation
\(\rho=200\, \Omega {m}\) \({C}=2 \times 10^{-12}\, {F}\) \({V}=40\, {V}\) \({K}=56\) \(C =\frac{ K \varepsilon_0 A }{ d }\) (\(K\) is the dielectric constant or relative permittivity) and \(R =\frac{\rho d }{ A }\) Now, charge will be discharged through the resistance between…
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