COMEDK · Physics · 18. Capacitance
A parallel plated capacitor has area \(2 \mathrm{~m}^{2}\) separated by 3 dielectric slabs. Their relative permittivity is \(2,3,6\) and thickness is \(0.4 \mathrm{~mm}\), \(0.6 \mathrm{~mm}, 1.2 \mathrm{~mm}\), respectively. The capacitance is
- A \(5 \times 10^{-8} \mathrm{~F}\)
- B \(11 \times 10^{-8} \mathrm{~F}\)
- C \(2.95 \times 10^{-8} \mathrm{~F}\)
- D \(10 \times 10^{-8} \mathrm{~F}\)
Answer & Solution
Correct Answer
(C) \(2.95 \times 10^{-8} \mathrm{~F}\)
Step-by-step Solution
Detailed explanation
The given situation is shown below, \(d_{1}=0.4 \mathrm{~mm}=4 \times 10^{-4} \mathrm{~m}\) \(d_{2}=0.6 \mathrm{~mm}=6 \times 10^{-4} \mathrm{~m}\) \(d_{3}=1.2 \mathrm{~mm}=1.2 \times 10^{-3} \mathrm{~m}\) The given arrangement will be equal to three capacitors connected in…
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