Help me doing this problem: 12) Shown in Figure 4 below is a parallel-plate capa
ID: 2247926 • Letter: H
Question
Help me doing this problem:
12) Shown in Figure 4 below is a parallel-plate capacitor that is half filled with a nonlinear dielectric. The other part of the capacitor is air-filled. The capacitor is charged by being connected to a voltage source. The source is then disconnected, and the capacitor electrodes are short-circuited. In the new electrostatic state, there is a remanent uniform polarization throughout the volume of the dielectric, with the polarization vector being normal to the capacitor plates and its magnitude being P. Determine the electric field intensity vector between the capacitor plates in (a) air and (b) dielectric. Fringing can be neglected. (Note: the dielectric is nonlinear so you cannot use the permittivity of the dielectric.)Explanation / Answer
The electric field strength you will get by dividing the electric displacement by the absolute dielectric permittivity of air an the material
E=D/0r
r=permitivity of air
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Consequently, the electric field in the dielectric is times weaker than the electric field in the air gaps between the plates and the dielectric. Specifically, for the plates of area A, E[in the gaps] = Qplate /0A
E[in the dielectric] = (Qplate + Qsurface )/0A = 1/ × Qplate / 0 A ,
= 1/ × E[in the gaps].
The constant — which depends on a dielectric in question but not on the charges — is called the relative permittivity or the dielectric constant.
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