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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anonymous · Answer

@IrishBoy123

irishboy123 · Answer

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arindameducationusc · Answer

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

anonymous · Answer

@arindameducationusc It is appreciated!! :)

irishboy123 · Answer

http://farside.ph.utexas.edu/teaching/302l/lectures/node45.html

radar · Answer

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.