The electric field in the dielectric is wasted?

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Homework Help Overview

The discussion revolves around the behavior of electric fields in dielectrics, specifically focusing on a scenario where an electric field decreases when passing through glass. Participants are exploring concepts related to permittivity and the effects of polarization on electric fields.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants are questioning whether the electric field in the dielectric is "wasted" and why there is a decrease in the electric field strength. There are inquiries about the potential for permittivity to act as an electric field booster and how this might work in practice.

Discussion Status

The discussion is active, with multiple participants raising questions and exploring different interpretations of how electric fields interact with dielectrics. Some have provided insights into the role of polarization surface charges and their effects on the electric field, while others are probing the implications of changing permittivity.

Contextual Notes

Participants are considering the implications of boundary effects between different materials and the nature of electric dipoles in dielectrics. There is an ongoing exploration of the assumptions regarding electric field behavior in various contexts, particularly in relation to capacitors and dielectric materials.

baby_1
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Hello
as you see the problem and solution (see the attachment)the input electric filed is 2K v/m but in the glass the electric filed decrease approximately to 705.9 v/m.i want to know
1- The electric field in the dielectric is wasted?
2-why electric field decrease?

Thanks
 

Attachments

Physics news on Phys.org
and another question
permittivity can be changed as electric field booster.it means we can set the permittivity of the second environment that electric filed increase(like an electric filed amplifier)?if yes how it does?
 
baby_1 said:
Hello
as you see the problem and solution (see the attachment)the input electric filed is 2K v/m but in the glass the electric filed decrease approximately to 705.9 v/m.i want to know
1- The electric field in the dielectric is wasted?
2-why electric field decrease?

Thanks

Because at the boundary between the air and the glass, polarization surface charges are set up. These charges set up an opposing E field to the external E field so the sum of the two E fields is less than the external E field alone.
 
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Thanks dear rude man
what about the next question
"and another question
permittivity can be changed as electric field booster.it means we can set the permittivity of the second environment that electric filed increase(like an electric filed amplifier)?if yes how it does? "

is it possible the permittivity can increase electric filed?is possible that polarization surface charges setup an agree electric filed that increase E filed in the second environment?
 
using a dielectric (in a capacitor gap) means that you can store κ times as much charge on the plates
(which would otherwise fill the gap with tremendously intense E-field) with the same voltage supply.
Because the charge in each potential carries Energy : (½QaVa + ½QbVb) , the stored Energy is also increased by factor κ .

At the molecular level, electric dipoles arise with their (+) end in the E-field's direction,
(so their internal E will always tend to cancel the external E-field).
Molecules that have permanent electric dipole try to rotate so as to reduce their PE in the external E-field ... this rotation always will reduce the local Electric field (to the extent that it occurs; solids severely limit rotation)

We can not force (torque?) molecular dipoles to orient themselves along the external E-field, the way magnets do naturally.
 
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baby_1 said:
and another question
permittivity can be changed as electric field booster.it means we can set the permittivity of the second environment that electric filed increase(like an electric filed amplifier)?if yes how it does?

True. Assume you have a slab of dielectric between two parallel plates. The plates are charged to some level Q. Then it takes positive work to remove the slab from between the plates. The new E field where the dielectric was is now larger than the E field when the dielectric was in place. This assumes Q did not change.
 
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