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Callix
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Homework Statement
A capacitor connected to a battery initially holds a charge of +q on its positive plate and -q on its negative plate. The electric field between the plates is initially E. A dielectric material is then inserted that polarizes in such a way as to produce an electric field of -0.30E (where the - sign indicates that this field opposites that of E). Determine the new charge stored on the positive plate.
Homework Equations
Eo-E=Ef
E=kq^2/r^2
The Attempt at a Solution
I did my work on Word using the equation tool just to make the notation format easier to read.
Is my logic correct here?
Any help would be greatly appreciated! :)