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## Homework Statement

The electric field between two circular plates of a capicator is changing at a rate of 1.5 x 10^6 V/m per Second. If the displacement current this instant is ID = 0.80 x 10^-8 A find the dimensions of the plates

## Homework Equations

## The Attempt at a Solution

The capacitance of a parallel plate capacitor is C = A*er*eo/t where A is the area of the plates, e is the relative permittivity of the dielectric, eo is the permeability of free space and t is the separation of the plates.

Q= V*C, where Q is the charge of the capacitor, V the voltage,

so we can write Q = A*E*er*eo, or Q/E = A*er*eo which implies (dQ/dt)/(dE/dt) = A*er*eo

I = dQ/dt, leading to I/(dE/dt) = A*er*eo = 0.80*10^-8/(1.5*10^6) = 5.33*10^-15

A = 5.33*10^-15/(er*eo)

If A is known the diameters of the circular plates could be determined, but A depends on er, and no information is given concerning this....I dont think I did this right.