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Question
The dielectric strength of air is 3.0 × 106 V/m. A parallel-plate air-capacitor has area 20 cm2 and plate separation 0.10 mm. Find the maximum rms voltage of an AC source that can be safely connected to this capacitor.
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Solution
Given:
Area of parallel-plate air-capacitor, A = 20 cm2
Separation between the plates, d = 0.1 mm
Dielectric strength of air, E= 3 × 106 V/m
E = `V/d`,
where V = potential difference across the capacitor
`therefore V = Ed`
`= 3xx10^6xx0.1xx10^-3`
`= 3xx10^2=300 V`
Thus, peak value of voltage is 300 V.
Maximum rms value of voltage `(V_{rms})` is given by,
`V_{rms} = V_0 /sqrt2`
= `300/sqrt2 = 212 V`
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