(1) If a capacitor has a capacitance oc 10.2 pF and we wish to lower the potenti
ID: 1320240 • Letter: #
Question
(1) If a capacitor has a capacitance oc 10.2 pF and we wish to lower the potential difference across the plates by 60.0 V, what magnitude of charge will we have to remove from each plate?
(2) A parallel plate capacitor has a capacitance of 2.0 pF and plate seperation of 1.0 mm. (A) How mush potential difference can be placed across the capacitor before dielectric breakdown of air occours(Emax = 3 x 10^6 V/m)? (B) what is the magnitude of the greatest charge the capacitor can store before breakdown?
please show the work so i can learn it.
Explanation / Answer
1)
we know,
Q = C*V
so, the charge that shold be removed,
dQ = C*delta V
= 10.2*10^-12*60
= 6.12*10^-10 C or 0.612 nC or 612 pC
2)
a)
V(max) = Emax*d
= 3*10^6*1*10^-3
= 3*10^3 volts
b) Q(max) =C*V(max)
= 2*10^-12*3*10^3
= 6*10^-9 C or 6 nC
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