A parallel plates capacitor is filled with a dielectric of relative permittivity ε = 12 and a conductivity σ = 10^-10 S/m. The capacitor is fully charged at a voltage V=10V and then disconnected from the battery. Find the time after wich the voltage in the capacitor decays to 3.67 V?

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.