## A community for students. Sign up today

Here's the question you clicked on:

## PhoenixFire 2 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?

• This Question is Closed
1. Pompeii00

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.

#### Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy