two parallel plates (each of area 200cm2) of a ca

Question

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

Susmita · Accepted Answer

Induced charge on dielectric is given by the formulaQ'=(k-1)Q/kAQ 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 boundaryEo=kE where Eo is the applied field and E is the resultant field.E=Eo-Ep,where Ep is the polarization field inside the dielectric,that is given in this question.Eo is the applied field,or field between the plates when the dielectric is absent.You can find it out from the given data asI hope now you can solve it.

Electrostatics> two parallel plates (each of area 200cm2)...

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

Bhuvanesh Kumar , 6 Years ago

Susmita

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Electrostatics> two parallel plates (each of area 200cm2)...

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

Bhuvanesh Kumar , 6 Years ago

Susmita

LIVE ONLINE CLASSES

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