Answer:
t = 1.06 sec
Explanation:
Once disconnected from the battery, the capacitor discharges through the internal resistance of the dielectric, which can be expressed as follows:
R = (1/σ)*d/A, where d is is the separation between plates, and A is the area of one of the plates.
The capacitance C , for a parallel plates capacitor filled with a dielectric of a relative permittivity ε, can be expressed in this way:
C = ε₀*ε*A/d = 8.85*10⁻¹² *12*A/d
The voltage in the capacitor (which is proportional to the residual charge as it discharges through the resistance of the dielectric) follows an exponential decay, as follows:
V = V₀*e(-t/RC)
The product RC (which is called the time constant of the circuit) can be calculated as follows:
R*C = (1/10⁻¹⁰)*d/A*8.85*10⁻¹² *12*A/d
Simplifying common terms, we finally have:
R*C = 8.85*10⁻¹² *12 / (1/10⁻¹⁰) sec = 1.06 sec
If we want to know the time at which the voltage will decay to 3.67 V, we can write the following expression:
V= V₀*e(-t/RC) ⇒ e(-t/RC) = 3.67/10 ⇒ -t/RC = ln(3.67/10)= -1
⇒ t = RC = 1.06 sec.
