AP EAMCET · PHYSICS · Capacitance
A parallel plate capacitor has plates of area \(0.4 \pi \mathrm{~m}^2\) and spacing of 0.5 mm. If a slab of thickness 0.5 mm and dielectric constant 4.5 is introduced in between the plates of the capacitor, then the capacitance of the capacitor is
- A 100 nF
- B 60 pF
- C 100 pF
- D 60 nF
Answer & Solution
Correct Answer
(A) 100 nF
Step-by-step Solution
Detailed explanation
\( C = \frac{k \epsilon_0 A}{d} \) \( C = \frac{4.5 \times 8.854 \times 10^{-12} \, \mathrm{F/m} \times 0.4 \pi \, \mathrm{m}^2}{0.5 \times 10^{-3} \, \mathrm{m}} \) \( C \approx 100 \times 10^{-9} \, \mathrm{F} = 100 \, \mathrm{nF} \)
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