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anonymous
 one year ago
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
anonymous
 one year ago
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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arindameducationusc
 one year ago
Best ResponseYou've already chosen the best response.0I am very bad in Electrostatics.. But I will try.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@arindameducationusc It is appreciated!! :)

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1http://farside.ph.utexas.edu/teaching/302l/lectures/node45.html

radar
 one year ago
Best ResponseYou've already chosen the best response.1Does 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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