A parallel-plate air capacitor is to store charge of magnitude 290 pC on each pl

Question

A parallel-plate air capacitor is to store charge of magnitude 290 pC on each plate when the potential difference between the plates is 45.0 V. If the area of each plate is 6.80 cm^2, what is the separation between the plates? Express your answer with the appropriate units. If the separation between the two plates is double the value calculated in part (a), what potential difference is required for the capacitor to store charge of magnitude 290 pC on each plate? Express your answer with the appropriate units.

Explanation / Answer

here,

the magnitude of charge , Q = 290 pC = 2.9 * 10^-10 C

potential difference , V = 45 V

a)

let the sepration between the plates be d

C = Q/V = area * e0/d

2.9 * 10^-10 /45 = 6.8 * 10^-4 * 8.85 * 10^-12 /d

d = 9.3 * 10^-4 m

b)

if sepration is doubled

the capacitance is halved

and for constant charge stored

potential is doubled

so, the voltage , V' = 2 * V = 90 V

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