two parallel plates (each of area 200cm2) of a capacitor are given equal and opposite charge of magnitude 1.75microC.The space between the plates is filled with dielectric.If the electric field in the dielectric is 1.5*10^6 V/m what will be the magnitude of chare induced on each dielelectric plate

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Susmita · Answer

Induced charge on dielectric is given by the formula Q'=(k-1)Q/kA Q is charges on capacitor plate,A area,k dielectric constant.so basically you have to find out k. From the continuity of normal component of electric field across the boundary E o =kE where E o is the applied field and E is the resultant field. E=E o -E p ,where E p is the polarization field inside the dielectric,that is given in this question. E o is the applied field,or field between the plates when the dielectric is absent.You can find it out from the given data as I hope now you can solve it.

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two parallel plates (each of area 200cm2) of a capacitor are given equal and opposite charge of magnitude 1.75microC.The space between the plates is filled with dielectric.If the electric field in the dielectric is 1.5*10^6 V/m what will be the magnitude of chare induced on each dielelectric plate

two parallel plates (each of area 200cm2) of a capacitor are given equal and opposite charge of magnitude 1.75microC.The space between the plates is filled with dielectric.If the electric field in the dielectric is 1.5*10^6 V/m what will be the magnitude of chare induced on each dielelectric plate

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