P24_037 - 37. The field is radially outward and takes on...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 37. The field is radially outward and takes on equal magnitude-values over the surface of any sphere centered at the atom’s center. We take the Gaussian surface to be such a sphere (of radius r). If E is the magnitude of the field, then the total flux through the Gaussian sphere is Φ = 4πr2 E . The charge enclosed by the Gaussian surface is the positive charge at the center of the atom plus that portion of the negative charge within the surface. Since the negative charge is uniformly distributed throughout the large sphere of radius R, we can compute the charge inside the Gaussian sphere using a ratio of volumes. That is, the negative charge inside is −Zer3 /R3 . Thus, the total charge enclosed is Ze − Zer3 /R3 for r ≤ R. Gauss’ law now leads to r3 r Ze 1 4πε0 r2 E = Ze 1 − 3 −3. =⇒ E = 2 R 4πε0 r R ...
View Full Document

This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.

Ask a homework question - tutors are online