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Asked: 2022-04-13T07:02:46+00:00 In:

A conductor with an inner cavity, like that shown in Fig.22.23c from the textbook, carries a…

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A conductor with an inner cavity, like that shown in Fig22.23c from the textbook, carries a total charge of +5.40 nC. The charge within the cavity, insulated from the conductor is -7.80 nC. How much charge is on
Part A
How much charge is on the inner surface of the conductor?
qinner =            nC
Part B
How much charge is on the outer surface of the conductor?
qouter =           nC

♦ Relevant knowledge

An uncharged or charged conductor can cause the electric field within the conductor to be zero.A conductor charged with charge rearranges its charge, while uncharged conductors split the negative and positive charges so such that both the both the internal and external electric field becomes zero.
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2 Answers

  1. Best Answer
    Expert

    Charge residing on the inside of the conductor is -7.80nC. This will induce an equal and opposite charge of +7.80nC uniformly over the inner surface of the conductor. This can be shown to be true if you create a Gaussian Surface of radius bigger than the inner radius of the conductor but smaller than the outer radius. Then by
    Gauss’ Law the charge enclosed should be zero since the Electric Field inside a conductor must be zero.

    So the charge on the inner surface of the conductor is
    qinner = +7.80 nC

    Now this +7.80 nC induces an equal and opposite charge of -7.80 nC
    uniformly over the outer surface of the conductor.

    But we know that there is already a charge of +5.40 nC residing
    on the outer surface of the conductor therefore the net charge on
    the outer surface now will be:

    qouter = -( 7.80 + 5.40) nC = -2.40 nC

    • 16

  2. Charge residing on the inside of the conductor is -7.80nC. This
    will induce an equal and opposite charge of +7.80nC uniformly over
    the inner surface of the conductor. This can be shown to be true if
    you create a Gaussian Surface of radius bigger than the inner
    radius of the conductor but smaller than the outer radius. Then by
    Gauss’ Law the charge enclosed should be zero since the Electric
    Field inside a conductor must be zero.

    So the charge on the inner surface of the conductor is
    qinner = +7.80 nC

    Now this +7.80nC induces an equal and opposite charge of -7.80nC
    uniformly over the outer surface of the conductor.

    But we know that there is already a charge of +5.40nC residing
    on the outer surface of the conductor therefore the net charge on
    the outer surface now will be: qouter = -( 7.80 + 5.40)
    nC = -2.40nC

    • 19

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