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