Solution: dE=. do 4兀Ex 1d0 X de= dE cos a 4丌Ex2+a √x2+a24zs(x2+a2)2 do E 4z5(x2+a)24z6(2+a2) Because the symmetry, Ey=0 E=Ei= 4兀E0
( ) 3 2 2 2 0 2 2 2 2 0 1 4 1 4 1 cos x a x a x x a dQ dEx dE + = + + = = ( ) ( ) 3 2 2 2 0 3 2 2 2 0 4 1 4 1 x a Qx x a xdQ Ex + = + = Because the symmetry , Ey=0. ( ) i x a Qx E E i x ˆ 4 1 ˆ 3 2 2 2 0 + = = Solution: 2 2 0 4 1 x a dQ dE + =
+ Discussion: r=0,E=0 2).x>>R,E 4兀E0x 3).x=±R,E=Emx E y R dEX X P dE ⊥
Discussion: x = 0 ，E = 0 dl y z x R r O P q dE⊥ dEx 2 4 0 1 2). x q x R E = ， max 2 2 3). x = R ，E = E E O x R 2 2 − R 2 2
Find the electric field caused by a disk of radius r with a uniform position surface charge density at a point along the axis of the disk a distance x from its center do
x R O Q x Find the electric field caused by a disk of radius R with a uniform position surface charge density , at a point along the axis of the disk a distance x from its center. r drdQ
Solution: d@=odA=o(2rrdr)=2ordr 1(2ordr)x dE 4兀E R 1(2Tordr)x Ox cR x14x)92 2 y2√R2/x)+l
Solution: dQ =dA =(2rdr) = 2 rdr ( ) ( ) 3 2 2 2 0 2 4 1 x r rdr x dEx + = ( ) ( ) ( ) ( ) + = − + = + = 1 1 1 2 2 2 4 1 2 2 0 0 3 2 2 2 0 0 3 2 2 2 0 R x x x r x rdr x r rdr x E R R x