Concepts Of Physics MCQ Edition [Volume 2]PhysicsAlternating Current
A parallel-plate air-capacitor has an area of 20 cm ^2 and a plate separation of 0.10 mm . The dielectric strength of air is 3.0 10^6 V/m . Determine the maximum rms voltage of an AC source that can be safely connected to this capacitor.
Options
- A150 V
- B300 V
- C212 V
- D424 V
Correct answer
C. 212 V
Step-by-step solution
The maximum electric field (dielectric strength) is given by E_ max = 3.0 10^6 V/m . The maximum peak voltage V_ max that can be applied across the plates without breakdown is V_ max = E_ max d Substituting the given value d = 0.10 mm = 10⁻⁴ m V_ max = (3.0 10^6) 10⁻⁴ = 300 V For an AC source, the peak voltage is related to the rms voltage by V_ max = 2 V_ rms . V_ rms = V_ max 2 = 300 2 = 150 2 V V_ rms 150 1.414 = 212.1 V