Why is induced electric field due to bound charges in dielectric always less than the external electric field?
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4 Answers
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When a dielectric is placed in an external electric field $\vec E_{\rm external}$ a change is induced on the surface of the dielectric as shown in the diagram below in red.
This may be due to the formation of temporary electric dipoles in the dielectric or the rotation of permanent electric dipoles.
The number if induced charges cannot exceed the number of external charges and is usually less.
When the electric fields are shown you will see that the induced charges on the dielectric produce an induced electric field within the dielectric $\vec E_{\rm induced}$.
By superposition the electric field within the dielectric $\vec E_{\rm external} + \vec E_{\rm induced}$ will be less than $\vec E_{\rm external}$.
The very best that can be done is to reduce the field in the dielectric to zero.
This occurs when the number of external charges is equal the number of induced charges ie $|\vec E_{\rm external}|=|\vec E_{\rm induced}|$.
You might no longer call the material a dielectric but rather call it a conductor.
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The permittivity of free space determines the electric field lines passing through any space of the body. For dielectrics the permittivity of free space is always more than that of vacuum as per the definition of permittivity of free space which is the measure of resistance to electric flux.
As you know field is inversely proportional to permittivity of free space, therefore the above explanation explains the phenomenon.
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Why would it? If the dielectric produced an opposing field greater than that of the external field. The resulting external field would appear in the other direction, causing the dielectric to flip. It's like asking why drag is always less than thrust.
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Think of a dielectric inside a parallel plate capacitor with one positive and one negative plate. The negative charges inside the dielectric will be attracted towards the positive plate of the capacitor while the positive charges will be attracted towards the negatively charged plate of the capacitor, however these charges can travel only so far inside the dielectric, but if you see now the electric field created by the charges setup inside the dielectric will oppose the external electric field set up by the capacitor plates. Hence, this opposition reduces the electric field inside the dielectric.
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$\begingroup$ ok, but i think we need to make a distinction as to whether he is asking about the size of the induced field, or the size of the resulting field $\endgroup$– user86425Commented Apr 29, 2017 at 15:47
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$\begingroup$ in any case, the answer should be obvious, especially if the equation is derived $\endgroup$– user86425Commented Apr 29, 2017 at 15:48
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$\begingroup$ resulting field=external field +induced field, induced field being in the opposite direction, while it retains the properties of a vector, hence vector addition implies subtraction in this case, but some times it leads to some other effects. $\endgroup$– Ghosal_CCommented Apr 29, 2017 at 15:50