A small charged ball lies within the hollow of a metallic spherical shell of radius R. For three situations, the net charges on the ball and

Question

A small charged ball lies within the hollow of a metallic spherical shell of radius R. For three situations, the net charges on the ball and shell, respectively, are (1) +4q, 0; (2) -6q, +10q; (3) +16q, -12q. Rank the situations according to the charge on (a) the inner surface of the shell and (b) the outer surface, most positive first.

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Sapo 3 years 2021-08-31T22:52:19+00:00 1 Answers 3 views 0

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  1. King
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    2021-08-31T22:54:09+00:00
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    Answer:

    a) (2) > (1) > (3)

    b) (1) = (2) = (3)

    Explanation:

    a)

    • As the electric field inside a conductor, in electrotatic conditions, must be zero, if we apply the Gauss law to a spherical surface which radius falls witthin the conductor, total electric flux through this surface, must be zero also.
    • Now, as the electric  flux is proportional to the total charge enclosed by the surface, if  the flux is zero, this means that the total charge must be zero.
    • So, for the three cases, as total charge must be zero, there must be a charge on the inner surface of the shell, equal and opposite to the charge of the ball:
    1. q = +4q ⇒ qin = – 4 q (1)
    2. q = -6q ⇒ qin = +6q
    3. q = +16 q ⇒ qin = -16q
    • So, ranking the situations according to the charge on the inner surface of the shell, most positive first, we have:
    • +6q > -4q > -16q
    • ⇒(2) > (1) > (3)

    b)

    • Once we know the charges on the inner surface, as total charge on the shell must be conserved, the charge on the outer surface must meet the following condition:
    • qin + qou = q₀
    • ⇒ qou = qo-qin
    1. q₀ = 0 , qin = -4q ⇒ qou = + 4q
    2. q₀ = +10q, qin = +6q ⇒ qou = +4q
    3. q₀ =-12 q, qin =-16 q ⇒ qou = +4q
    • As it can be seen, the charge on the outer surface is the same for the three cases, so (1) =(2) = (3).

         

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