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^{5}V/m. when the area is filled with a dialectric substance, the field is reduced to 1.2*10

^{5}V/m. what is the density of the induced charge??

this is what i have done so far, could someone tell me if it is correct, and if not where i am going wrong.

E=1.2*10

^{5}

E

_{0}=2*10

^{5}

K=E

_{0}/E=5/3

∫Kε

_{0}EdA=Q

_{free}

==>Q

_{free}=ε

_{0}KEA

∫ε

_{0}EdA=Q

_{free}+Q

_{induced}

==>Q

_{induced}=ε

_{0}KEA-ε

_{0}EA

Q

_{ind}/A=ε

_{0}E(1-K)

=ε

_{0}(1.2-2)*10

^{5}

=7.083*10

^{-7}C/m

^{2}