COMEDK · Physics · 18. Capacitance
A parallel plate capacitor having a dielectric constant 5 and dielectric strength \(10^6 \mathrm{~V} \mathrm{~m}^{-1}\) is to be designed with voltage rating of \(2 \mathrm{~kV}\). The field should never exceed \(10 \%\) of its dielectric strength. To have the capacitance of \(60 \mathrm{~pF}\) the minimum area of the plates should be
- A \(27.1 \times 10^{-4} \mathrm{~m}^2\)
- B \(2.71 \times 10^{-4} \mathrm{~m}^2\)
- C \(27.1 \times 10^{-2} \mathrm{~m}^2\)
- D \(2.7 \times 10^{-2} \mathrm{~m}^2\)
Answer & Solution
Correct Answer
(D) \(2.7 \times 10^{-2} \mathrm{~m}^2\)
Step-by-step Solution
Detailed explanation
Working electric field: \(E = 10\%\) of \(10^6 = 10^5\) V m\(^{-1}\) Plate separation: \(d = \dfrac{V}{E} = \dfrac{2000}{10^5} = 2 \times 10^{-2}\) m…
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