A capacitor connected to a battery initially holds a charge +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 the field opposes that of E). Determine the new charge stored on the positive plate

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A capacitor connected to a battery initially holds a charge +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 the field opposes that of E). Determine the new charge stored on the positive plate

Physics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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I am very bad in Electrostatics.. But I will try.

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@arindameducationusc It is appreciated!! :)
http://farside.ph.utexas.edu/teaching/302l/lectures/node45.html
Does the battery with E voltage remain connected (problem doesn't say otherwise) then the value of E would remain the same, and the Capacitance increases due to added dielectric, then the charge will increase. Study the link provided by @IrishBoy123 to determine the percentage increase of the charge.

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