Mutual inductance as a psuedo-capacitor??? 1. I am testing how reduced the mutual inductance btwn two parallel ring-shaped inductors is when a material is placed btwn them. The primary coil is connected to an AC generator in series with an ammeter (on the order of mA), and the secondary in series with an ammeter. [/b] What I'm trying to do is calculate either the permittivity of the material, it's dielectric constant, or some ratio of them, by measuring the reduction in current thru the secondary. [/b] 2. I may have a solution to determining the ratio of epsilon/kappa by the equation epsilonnaught = epsilon / kappa: 3. The primary inductor L1 directs a B-field through the coils of L2, the strength of the field along the axis of the inductor is (munaught / 4 pi) [(2pi*R^2*I)/(x^2 + R^2)^3/2]. But I see elements of a parallell-plate capacitor btwn these coils:[/b] [/b] The primary current is alternating, but at a given time t the current dq1/dt will possess q1 charges involved in inducting the current dq2/dt, and the secondary will possess q2 charge, which must be less than q1. There should be a potential difference btwn this charge difference, and by the electromagnetic eqn E = -grad(V) there is an electric field pointing along the same axis as the gradient of this potential difference. [/b] [/b] Now a parallel-plate capacitor has a charge difference and so it also possesses a potential difference and has a (ideally) uniform electric field through it's seperation distance. So if I CAN treat my mutual inductors as a cap, THEN I can use cap-like equations. By the duality relationship btwn inductance and capacitance, the work stored in the electric field of a cap is equal to the work stored in the magnetic field of an inductor: [/b] (1/2)(L I^2) = (1/2)(C V^2) [/b] Now V = E*d, where d is sepeartion btwn inductors. [/b] Also, E = Q /(A*epsilonnaught) [/b] My prof doesnt really trust my logic on this one. Any help?