Question

You decide to build your own RC circuit out of scrap. To build the capacitor you use two square sheets of aluminum foil 10cm on a side) with cellophane sandwiched between them (e.g., Saran Wrap). Your cellophane has a dielectric constant K=3.5 and a thickness 0.0125mm. You also connect the plates together with 25m of 30 gauge (0.255mm diameter) copper wire (of resistivity rho=1.72×10^−8Ωm). Required:

Find the RC~time constant τ that describes how a charge on the capacitor would decay with time. (You may ignore resistance within the aluminum foil. The vacuum permittivity of free space is ϵ0=8.854×10−12C^2/Nm^2.)

Answer 1

The RC time constant τ for an RC circuit is given by the product of the resistance R and the capacitance C of the circuit: τ = RC.

Let's start by finding R, the resistance of the copper wire. The resistance of a cylindrical conductor is given by R = ρl/A, where ρ is the resistivity, l is the length, and A is the cross-sectional area. The cross-sectional area of a wire with diameter d is given by A = π(d/2)^2.

R = ρl/π(d/2)^2 = 1.72×10^−8 Ωm * 25m / π(0.255mm/2)^2 = 0.878 Ω.

Now let's find the capacitance C of the capacitor. The capacitance of a parallel plate capacitor is given by C = εA/d, where ε is the permittivity of the dielectric, A is the area of one plate, and d is the separation between the plates. In this case, the permittivity of the dielectric is ε = Kε0, where K is the dielectric constant and ε0 is the vacuum permittivity. The area A of a square plate with side length s is given by A = s^2.

C = Kε0 * A/d = 3.5 * 8.854×10^−12 C^2/Nm^2 * (10cm)^2 / 0.0125mm = 2.51 × 10^-9 F.

Finally, we can find the RC time constant τ:

τ = RC = 0.878 Ω * 2.51 × 10^-9 F = 2.20 × 10^-9 s = 2.20 ns.

So the RC time constant for this circuit is 2.20 nanoseconds.