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  • 3 years ago

How do I use Gauss' Law to show that the electric field of any point inside (r<R ) the hollow sphere is 0?

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  1. anonymous
    • 3 years ago
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    The simple answer is because charges are present on the outer surface of the hollow sphere. The mathematical proof of this is as follows: Gauss' Law states: \[\int\limits_{}^{}E*dA=\frac{ q^{enc} }{ \epsilon _{0} }\]In most cases, Gauss' Law is useful because the electric field is constant, and so the integral simplifies immensely: \[\int\limits_{}^{}E*dA=E \int\limits_{}^{}dA=EA=\frac{ q^{enc} }{ \epsilon _{0} }\]Now the one thing we know is that the total charge enclosed is 0! There are no charges within the hollow sphere. Therefore: \[EA=\frac{ q^{enc} }{ \epsilon _{0} }=\frac{ 0 }{ \epsilon _{0} }=0\]So now, we have the expression: \[EA = 0\]And since there is obviously a positive value of area (non-zero), E has to be 0. Thus, we have proven conceptually and mathematically how Gauss' Law describes the interior of a hollow sphere as having E=0.

  2. PhoenixFire
    • 3 years ago
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    @Pompeii00 Thank you! I understand now why it's 0.

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