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Unformatted text preview: 51. (a) The potential diﬀerence V across the plates of a capacitor is related to the charge q on the positive
plate by V = q/C , where C is capacitance. Since the charge on a discharging capacitor is given by
q = q0 e−t/τ , this means V = V0 e−t/τ where V0 is the initial potential diﬀerence. We solve for the
time constant τ by dividing by V0 and taking the natural logarithm:
τ =− t
10.0 s
=−
= 2.17 s .
ln (V /V0 )
ln [(1.00 V)/(100 V)] (b) At t = 17.0 s, t/τ = (17.0 s)/(2.17 s) = 7.83, so
V = V0 e−t/τ = (100 V) e−7.83 = 3.96 × 10−2 V . ...
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This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.
 Fall '08
 SPRUNGER
 Physics, Capacitance, Charge

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