Prof. David R. Jackson ECE Dept. Fall 2014 Notes 11 ECE 2317 Applied Electricity and Magnetism 1

Example Assume Infinite uniform line charge Find the electric field vector 2 x y z S  l =  l0 [C/m] h r 

Example (cont.) 3 h StSt SbSb ScSc  r z Side view

Example (cont.) Hence, We then have 4

5 Note About Cylindrical Coordinates Note: In cylindrical coordinates, the LHS never changes. Hence, the left-hand side of Gauss’s law will always be the same. Assume:

Example x y z l0l0 -h h When Gauss’s Law is not useful E has more than one component ! E is not a function of only  ! 6 Finite uniform line charge Although Gauss’s law is still valid, it is not useful in helping us to solve the problem. We must use Coulomb’s law.

Example  v = 3  2 [C/m 3 ],  < a Infinite cylinder of non-uniform volume charge density x y z S h a  r Find the electric field vector everywhere 7 Note: This problem would be very difficult to solve using Coulomb’s law!

Example (cont.) (a)  < a 8 S  h r z Side view so

Example (cont.) Hence so 9 x y z a r 

Example (cont.) (b)  > a S  h r z 10 so

Example (cont.) 11 x y z a r  Hence, we have

Example y z x  s =  s0 [C/m 2 ] Assume S A r Find the electric field vector everywhere Consider first z > 0 Infinite sheet of uniform surface charge density 12

Example (cont.) Assume S A r Then we have 13 so

Example (cont.) Hence, we have S A r For the charge enclosed we have Hence, from Gauss's law we have so 14 Therefore

Example 15 x x = h x = 0 (a) x > h (c) x < 0 (b) 0 < x < h From superposition:

Example (cont.) (a) x > h (c) x < 0 (b) 0 < x < h Choose: 16 x x = h x = 0

Example (cont.) 0 < x < h Ideal parallel-plate capacitor 17 Metal plates x h

Example Infinite slab of uniform volume charge density Assume (since E x ( x ) is a continuous function) y x d r Find the electric field vector everywhere 18

Example (cont.) (a) x > d / 2 A S x x r d 19 Alternative choice: Another choice of Gaussian surface would be a symmetrical surface, symmetrical about x = 0 (as was done for the sheet of charge).

Example (cont.) Note: If we define QQ  s eff QQ v0v0 (sheet formula) then d A A 20

Example (cont.) (b) 0 < x < d / 2 d y x x = 0 x = x S r 21

Example (cont.) d / 2  v0 d / (2  0 ) x ExEx - d / 2 Summary y x d 22 Note: In the second formula we had to introduce a minus sign, while in the third one we did not.