Two square parallel conducting surfaces are separated by d=40 time sign 10^-6 m.

Question

Two square parallel conducting surfaces are separated by d=40 time sign 10^-6 m. Given that air breaks down at 3 MV/m, what is the maximum potential difference that can be applied to these plates before arcing will occur? If the capacitance is 50 time sign 10^-6 F, what is the charge on the "positive" plate at the potential difference found in part a? To be safe, the capacitor will be rated for use at one-half the potential difference found in part a. When charged to its rated voltage, what is the energy stored in the capacitor? If the separation between the plates were doubled, but the potential difference were maintained at the rated value, what would be the effect on the capacitance? (circle one) increase decrease stay the same

Explanation / Answer

a)

potential difference dV = E*d = 3*10^6*40*10^-6 = 120 V   <<----------answer

(b)

Q = C*dV = 50*10^-6*120 = 0.006 C   <<----------answer

(c)

energy stored E = 0.5*c*v^2 = 0.5*50*10^-6*(120/2)^2 = 0.09 J <<<----------answer

(d)

capacitance C = eo*A/d

as d increases the C value decreases

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