estat2 - Physics 54 Electrostatics 2 The sooner you fall...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
Physics 54 Electrostatics 2 The sooner you fall behind the more time you have to catch up. — Anonymous Calculating E from Gauss’s law There are a few charge distributions having sufFcient symmetry that one can use Gauss's law to calculate the E-Feld. The trick is to choose the closed “Gaussian surface” through which the ±ux is calculated in such a way as to exploit the symmetry, making E d A simple enough that E can be taken outside the integral. The total ±ux then becomes a simple product of E and a surface area. Spherical symmetry . In these cases there is a point, called the “center of symmetry” about which the physical situation is the same in all directions. These are very important cases, among other reasons because isolated atoms have this symmetry. A charge distribution is spherically symmetric if its volume charge density ρ depends only on the distance from the center of symmetry but not on the direction. The E-Feld produced by this charge re±ects the symmetry: its direction is radial relative to the center of symmetry — outward/inward for positive/negative net charge — and its magnitude E depends only on the distance from that center. To Fnd the Feld at distance r one evaluates the ±ux through a Gaussian surface which is a sphere of radius r about the center of symmetry. Then, since d A is radially outward, E d A = E r dA , where E r is the outward radial component of the Feld; E r is positive if the Feld lines go outward through the surface. By the symmetry E r also has the same value at all points on the spherical surface, so the ±ux will be E d A = E r dA = E r 4 π r 2 . On the right side of Gauss's law we have 4 π k times the net amount of charge inside the sphere of radius r . Call this amount of charge Q ( r ). (This may or may not be the total charge of the distribution.) Then from Gauss’s law E r 4 r 2 = 4 k Q ( r ) , or E r ( r ) = k Q ( r ) r 2 . If Q ( r ) is negative, this gives a negative answer, showing that E is directed inward . PHY 54 1 Electrostatics 2
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Note that this formula for E r has exactly the same form as that for a point charge of amount Q ( r ) placed at the center of the distribution. This shows an important and very useful fact about spherically symmetric distributions: The E-feld at distance r From the center oF a spherically symmetric distribution is the same as that oF a point charge located at the center, with charge Q ( r ) equal to the total charge contained in a sphere oF radius r about the center. If the Feld point is outside the entire charge distribution , then Q ( r ) will be simply the total charge of the distribution. This means that the E-Feld outside any spherical charge distribution is that of the total charge, treated as a point charge at the center. These conclusions allow for easy calculation of the E-Feld for any spherically symmetric
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 10/19/2011 for the course PHYSICS 54L taught by Professor Thomas during the Summer '09 term at Duke.

Page1 / 10

estat2 - Physics 54 Electrostatics 2 The sooner you fall...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online