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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

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  1. anonymous
    • one year ago
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    @IrishBoy123

  2. IrishBoy123
    • one year ago
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    .

  3. arindameducationusc
    • one year ago
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    I am very bad in Electrostatics.. But I will try.

  4. anonymous
    • one year ago
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    @arindameducationusc It is appreciated!! :)

  5. IrishBoy123
    • one year ago
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    http://farside.ph.utexas.edu/teaching/302l/lectures/node45.html

  6. radar
    • one year ago
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    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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