{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

1_10 - PHY 5346 HW Set 1 Solutions Kimel 2 1.10 I will base...

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: PHY 5346 HW Set 1 Solutions , Kimel 2. 1.10 I will base the solution on the application of Green’s theorem, which results in eq. 1.36 from the textbook: 1 (w) 1 1 EMS 8 1 —¢ 10 \$ 3 I I = — d — — — — — — d ¢(x) 471—80 V R :17 + 471' S |: <R):| a Since the volume includes no charge, the ﬁrst term on the rhs vanishes. For the second term 8gb —» A -’ A WZV¢'TL/=—E‘nl Note % E - ﬁ’da’ = / 6/ - EdP’x’ by the divergence theorem S V Using the fact that v4]? = pea/so then the second term of the ﬁrst equation also vanishes, since the volume integrated over contains no charge. % (i) = — #, Where R is the radius of the sphere, and I’m taking the origin at the center of the sphere, 1 Mg) 2 47rR2 f ¢(f’)da’ = mean value of the potential over the sphere. S Since ...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online