## PhoenixFire 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?

1. anonymous

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

@Pompeii00 Thank you! I understand now why it's 0.