=========================preview======================
(phys223)[2009](f)midterm~ma_yxf^_10545.pdf
Back to PHYS223 Login to download
======================================================
PHYS223
Intermediate Electricity and Magnetism I
Midterm Examination Solution

1. (a) (6 pts)
.......2zxzyxyAAAAAxyzyzxzxyyxz................................................xyzvxyzz0

(b) (6 pts)
..22223coscos1cos1cos...sin.. 2cossinAAAArrrrrrrAr................................................Err.


(c) (14 pts)
For , 0Rs..
......1111sincossinsin0sAAAAsssss........................E

For , sR...

2222222222211sincos1sin1cos 1sin1sin 0RRsAAsssssARARsssssARARssss.............................................E

Hence there is no volume charge density.
The E field is discontinuous at s = R. Hence there is a surface charge density given by
. ....0abovebelow00.sinsin2sinAAA..................EEs




2. (a) (6 pts)
5224005444554RRARQArrdrArdrQAR................


(b) (14 pts)
Method 1:
By Gausss law, the electric field is , where Qenc is the amount of charge enclosed by a concentric spherical Gaussian surface with radius r. ..enc20.4Qrr...rE


For , Qenc = Q. rR...

For , 0Rr..
55224enc55500554445rrQQrrQArrdrrdrQRRR................

Hence,
..35020.041.4QrRrRrQrRr.................rEr

The potential at the center is
340025500000154444416RRQQrQQRQVddrdrrRRRR........................El

Method 2:
014dV.......r

For observation at the center, r = 0, and hence , therefore r..r

22000022200004435000001sin411 445 4445 16RRRRVrdrddrArrdrrdrrrAARQRrdrRQR.....................................................

3. (a) (2 pts) dlRd...

(b) (5 pts) ........cossincossinDRRRRD..............rrzxyxyzr

(c) ....3/2300220011...cossin44RdRdRRDDR....................Exyzrr

......23/23/200222200cossin044xRRERdDRDR...................

(or by symmetry)
(13 pts) ........23/23/20022220023/2220sincos44 2yRRERdDRDRRDR.......................
....3/23/2022220044zRDREDdDRDR............


4. (a) (6 pts)

q





D

.qR/D



R2/D








R2/D


D qR/D





.q
(b) (i) E = 0 because it is inside a conductor. (2 pts)
(ii) E = 0 because it is inside a conductor. (2 pts)
(iii) ........22222201//.4//qqRDqRDqRDDRRRDRRD....................Ez

(5 pts)

(c) (3 pts)
................2222220222222011//.24//1//1. 8//qqRDqRDqQRDDRRRDRRDQqRDRDRDDRRRDRRD.................................................F0zz

(d) (10 pts)
The E field just above the plate is
The charge density is hence ............21/21/2222422224200323/23/2222420033/23/222224011//..2244//11/.. 22/1. 2qDqRDRDrDrRDrDrRDqDqRDrDrRDqDRrDDrR.....................................Ezzzzz

. ....303/23/22222412qDRErDDrR..................


(e) (6 pts)
The force is the same as that due to the three image charges:
............22222022222201//.42//1//. 42//qqRDqRDqDDRDDRDqRDRDDDRDDRD..................................Fzz