Capacitors
A parallel-plate capacitor has plate area (0.02) m² and plate separation (2) mm. What is its capacitance in vacuum?
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About This Question
- Subject
- physics
- Chapter
- electrostatics
- Topic
- capacitors
- Difficulty
- Easy
- Year
- 2025
Solution
Correct Answer:
88.5 pF
The capacitance of a parallel-plate capacitor is given by (C = \varepsilon_0 A/d), where (\varepsilon_0 = 8.85 \times 10^{-12}) F/m is the permittivity of free space, (A) is the plate area, and (d) is the separation between plates. This formula arises from applying Gauss's Law to find the uniform field between the plates and then using (V = Ed). Here, (A = 0.02) m² and (d = 2 \times 10^{-3}) m. Substituting: (C = \frac{8.85 \times 10^{-12} \times 0.02}{2 \times 10^{-3}} = \frac{1.77 imes 10^{-13}}{2 imes 10^{-3}} = 8.85 imes 10^{-11}) F (= 88.5) pF. Option 177 pF is incorrect because it results from halving the separation to (d = 1) mm. Option 44.25 pF is incorrect because it results from doubling the separation to (d = 4) mm. Option 354 pF is incorrect because it uses four times the actual area. This is a standard NCERT-level JEE Main capacitance computation. Plausibility check: ~100 pF for a small plate area and millimetre separation is physically reasonable and consistent with typical textbook examples.
This easy difficulty physics question is from the chapter electrostatics, covering the topic of capacitors. It appeared in the 2025 exam.
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