Lect_5_note

Lect_5_note - E reaches zero. So, E(r) = 0 r<a...

Info iconThis preview shows pages 1–17. Sign up to view the full content.

View Full Document Right Arrow Icon
Chapter 24 -2 Gauss’s Law
Background image of page 1

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

View Full DocumentRight Arrow Icon
Gauss’s Law ± q in is the net charge inside the surface ± E represents the electric field at any point on the surface EA in E o q d ε Φ= = ±
Background image of page 2
charged spherical shell Q r
Background image of page 3

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

View Full DocumentRight Arrow Icon
Field inside and outside a uniforml charged spherical shell Q Consider a spherical surface centred on the charged shell. Due to spherical symmetry, at every point on the sphere, E field is perpendicular to the sphere surface and has equal value. 2 0 4 1 | | r Q E πε = out Φ in Φ EA in E o q d ε Φ= = ± E 4 π r 2 =Q/ ε 0
Background image of page 4
charged spherical shell Q 0 out ε Q = Φ 2 0 4 1 | | r Q E πε = 0 = E Outside out Φ in Φ Inside 0 in = Φ charge within surface = 0
Background image of page 5

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

View Full DocumentRight Arrow Icon
Applying Gauss’s Law ± To use Gauss’s law, you want to choose a gaussian surface over which the surface integral can be simplified and the electric field determined ± Take advantage of symmetry ± Remember, the gaussian surface is a surface you choose, it does not have to coincide with a real surface ± Movie 5_1
Background image of page 6
insulating sphere Movie 5_2
Background image of page 7

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

View Full DocumentRight Arrow Icon
A charged conducting sphere In a conductor, net charges will move under a non-zero field until
Background image of page 8
Background image of page 9

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

View Full DocumentRight Arrow Icon
Background image of page 10
Background image of page 11

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

View Full DocumentRight Arrow Icon
Background image of page 12
Background image of page 13

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

View Full DocumentRight Arrow Icon
Background image of page 14
Background image of page 15

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

View Full DocumentRight Arrow Icon
Background image of page 16
Background image of page 17
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: E reaches zero. So, E(r) = 0 r<a Q(r) = 0 at r<a-δ s Charge is accumulated at the outer surface ! 2 4 1 | | r Q E πε = r >a A conductor hollow sphere E(r) = ? 2D Problem in 3D Space x y z r Apply Coulomb’s Law =k e λ ∫ r /(x 2 +r ) 3/2 dx = 2k λ / r E + ∞-∞ θ x Apply Gauss’s Law Solution Movie 5_3 Gauss‘s law in 1D q E= (q/ ε ) / 2 1D Problem in 3D Space x y z Apply Coulomb’s Law r =k e σ ∫ dx ∫ dy cos φ cos θ /(x 2 +y +r ) = k 2 π σ = σ / 2ε E + ∞-∞ x + ∞-∞ y Apply Gauss’s Law Φ E =E p A cylinder +2E n disc Constant! Parallel Plates ± The electric field is uniform in the central region, but not at the ends of the plates ± If the separation between the plates is small compared with the length of the plates, the effect of the non-uniform field can be ignored...
View Full Document

Page1 / 17

Lect_5_note - E reaches zero. So, E(r) = 0 r<a...

This preview shows document pages 1 - 17. Sign up to view the full document.

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