The dielectric medium gets polarized by say two randomly shaped charged bodies placed some distance apart.How does the resultant fields from dielectric diminish the coulomb force .There are a bunch of questions which follow up which I think won't be too much for this question.
Will there be a difference if the slab is placed at different locations along the line joining..if not , why ?
given that thickness of slab is constant only the surface is being enlarged , will it affect the force?
I am unable to understand the mechanism of this force reduction by dielectric, I have just found " field from dielectric reduces force " or some solved questions which convert the dielectric into vacuum of equivalent length to solve problems , no satisfying explanation yet. Please explain me the mechanism , gauss law and relevant things are known to me in case.

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    $\begingroup$ The prime reason is dielectric polarization. $\endgroup$ – user8718165 May 4 '19 at 9:35

Consider the simple case, A capacitor with dielectric placed between it. The negative plate of the capacitor polarize the dielectric end near to it and make it positive. Similarly the positive plate polarize the dielectric end near to it, negative. If the field due to the capacitor plates is in positive direction, then the field due to dielectric will be in the negative direction. Adding these two fields we will get a electric field less than the field due to the capacitor plate alone. Thus force $(F=qE)$ get reduced when dielectric is inserted. Approach this method to answer your questions.

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  • $\begingroup$ I understand this line of reasoning , I tried to ask wether the field will change if I inserted a dielectric medium which is of less thickness than thecseparation between charges , will the field change here , in the region where still vacuum is present $\endgroup$ – Aditya Prakash May 4 '19 at 20:02
  • $\begingroup$ I'm sorry. I didn't get you, what do you mean by thickness? If you meant increasing surface area, the electric field due to a parallel plate depends on the surface charge density. If area is increased the charge may also increase due to polarization. If the surface charge density, charge per unit area remains same, while you vary the thickness, then the electric field inside dielectric remains same. $\endgroup$ – walber97 May 5 '19 at 2:31
  • $\begingroup$ I'll give a picture $\endgroup$ – Aditya Prakash May 5 '19 at 16:42
  • $\begingroup$ there are two point charges r distance apart , and a dielectric slab of a cuboidal shape of thickness d along the line joining the charges.The charges are point charges. d is less than r $\endgroup$ – Aditya Prakash May 5 '19 at 16:55
  • $\begingroup$ now if I change dimensions of cuboid, say increase area of the faces which face the charges ,or move the slab to different positions along the line joining charges...how will it affect force $\endgroup$ – Aditya Prakash May 5 '19 at 17:00

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