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Seawater has permittivity eps = 80*eps0 and resistivity rho = 0.25 Ohm*m. A parallel-plate capacitor is submerged in this seawater and energised by an alternating supply V(t) = V0*sin(2*pi*f*t) of frequency f = 9*10² Hz. At the instant t = 1/800 s, the conduction current density equals 10^x times the displacement current density. Find x. (Given 1/(4*pi*eps0) = 9*10⁹ N*m²*C⁻²)

1. x = 6
2. x = 5
3. x = 7
4. x = 4

Correct answer: x = 6

Solution

The field inside is E = E0*sin(wt) with w = 2*pi*f. Conduction current density Jc = E/rho = (E0/rho)*sin(wt). Displacement current density Jd = eps*dE/dt = eps*E0*w*cos(wt). Their ratio is Jc/Jd = sin(wt)/(rho*eps*w*cos(wt)) = tan(wt)/(rho*eps*w). At t = 1/800 s with f = 900, wt = 2*pi*900/800 = (9/4)*pi, so tan(wt) = tan(pi/4) = 1.

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