分布电荷的电场（PPT讲稿）Electric field of distributed charges

Electric field of distributed charges

Electric field of distributed charges

Electric field of a uniformly charged thin rod Before we start think about the pattern of the electric field What should it look like? What symmetry does it have?

Electric field of a uniformly charged thin rod Before we start, think about the pattern of the electric field. What should it look like? What symmetry does it have? + + + + + + + +

y obs. location)-(source (x.0)-(0,y0 x,y △O Magnitude: △E Unit vector(direction x,y, x+

y x Q r  y E   , ,0 ,0,0 0, ,0 obs. location source x y x y r = − = − = −  2 2 r = x + (−y)  Magnitude: 2 2 , ,0 ˆ x y x y r r r + − =  =  Unit vector (direction):

△ y △ E 4. △ O y △ O y △ Q D23)32(x,-y:0 △ O △ E △ E 4. 2 y △ △ E Q y 4. y △ E

y x  Q r y E  ( ) ( ) , ,0 4 1 , ,0 4 1 ˆ 4 1 3 2 2 2 0 2 2 2 2 0 2 0 x y x yQ x y x y x yQr rQ E − + = +− + =   =         ( ) ( ) ( ) 04 1 4 1 3 2 2 2 0 3 2 2 2 0  = +  −  = +  = zyx E x y Q y E x y Qx E    

y Next step Add up all the fields contributed by each piece (superposition principle △O E,=∑AE,=0 △E △E E3=∑AE △ △Q 1△Qx 兀C 3/2

y x Q r  y E   Next step: Add up all the fields contributed by each piece (superposition principle) Q r  y E   Ey =Ey = 0 ( )   +  = =  3 2 2 2 4 0 1 x y Qx Ex Ex  

y Charge of each piece: Q △=Ay △O △E linear charge density △E △ △Q X △ 4丌EL(x2

y x Q r  y E   Charge of each piece: Q r  y E   y L Q Q        = ( ) y x y x L Q Ex  + = 3 2 2 2 4 0 1   “linear charge density

y Do the integration magic. △O △y→> very small. △E △E △ △Q +L/2 4丌EL L/2x+y

y x Q r  y E   Do the integration magic: Q r  y E   y → very small... ( )  + − + = / 2 / 2 3 2 2 2 0 1 4 1 L L x dy x y x L Q E  

y Result of the integration E E 4z52|x√x2+(L2)2

y x Result of the integration: E          + = 2 2 4 0 ( 2) 1 x x L Q E  

y We can change x to r, since the field is rotationally symmetrIc. E E 4n5|r√2+(L2)2

y r We can change x to r, since the field is rotationally symmetric: E          + = 2 2 4 0 ( 2) 1 r r L Q E  

y What if we go very far away from the rod? r>>l E E ACSO r√r2+(L/2) E 4兀 Just like a point charge. makes sense

y r What if we go very far away from the rod? E        = 2 4 0 1 r Q E   r  L Just like a point charge… makes sense!         + = 2 2 4 0 ( 2) 1 r r L Q E  