P24_045

# P24_045 - 45 To ﬁnd an expression for the electric ﬁeld...

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

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 45. To ﬁnd an expression for the electric ﬁeld inside the shell in terms of A and the distance from the center of the shell, select A so the ﬁeld does not depend on the distance. We use a Gaussian surface in the form of a sphere with radius rg , concentric with the spherical shell and within it (a < rg < b). Gauss’ law will be used to ﬁnd the magnitude of the electric ﬁeld a distance rg from the shell center. The charge that is both in the shell and within the Gaussian sphere is given by the integral qs = ρ dV over the portion of the shell within the Gaussian surface. Since the charge distribution has spherical symmetry, we may take dV to be the volume of a spherical shell with radius r and inﬁnitesimal thickness dr : dV = 4πr2 dr . Thus, r r r g g g A2 2 qs = 4π ρr2 dr = 4π r dr = 2πA(rg − a2 ) . r dr = 4πA r a a a 2 The total charge inside the Gaussian surface is q + qs = q + 2πA(rg − a2 ). The electric ﬁeld is radial, so 2 the ﬂux through the Gaussian surface is Φ = 4πrg E , where E is the magnitude of the ﬁeld. Gauss’ law yields 2 2 4πε0 Erg = q + 2πA(rg − a2 ) . We solve for E : E= 2πAa2 1 q + 2πA − 2 2 4πε0 rg rg . For the ﬁeld to be uniform, the ﬁrst and last terms in the brackets must cancel. They do if q − 2πAa2 = 0 or A = q/2πa2 . ...
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