Therefore the electric field at the center P of the semicircle is given by EP)=,1 ∫2：=cod0 (-sine'i-cose'j) oJ8=-π/2R2 There are two separate integrals for the x-and y-components.The x-component of the electric field at the center P of the semicircle is given by B(P)=1cos'sine'decos =0 4e。Je-r12 R 8πeoR lgr-元2 We expected this result by the symmetry of the charge distribution about the y-axis. The y-component of the electric field at the center P of the semicircle is given by E,(P)=- 1 co cos e'de'1 8=r2(1+cos20)d0' 4πe。J0=-π12 R e。J0r-r2 2R 8πeR lg=-元216πeo sin 20x12 g'=-π12 w 8EoR Therefore the force on the charged particle at the point P is given by F(P)=qE(P)=-94o_j 8ER 1111 Therefore the electric field at the center P of the semicircle is given by / 2 0 2 2 / 2 0 0 1 1 cos ˆ ˆ ( ) ˆ ( sin cos ) 4 4 wire ds Rd P r R ! " ! " # # ! ! ! ! "$ "$ %= %=& % % = = & % & % ' ' E r i j ! . There are two separate integrals for the x - and y -components. The x -component of the electric field at the center P of the semicircle is given by / 2 2 / 2 0 0 / 2 0 0 / 2 1 cos sin cos ( ) 0 4 8 x d E P R R ! " ! " ! " ! " # ! ! ! # ! "$ "$ %= %= %=& %=& % % % % = & = = ' . We expected this result by the symmetry of the charge distribution about the y-axis. The y -component of the electric field at the center P of the semicircle is given by 2 / 2 / 2 0 0 / 2 / 2 0 0 / 2 / 2 0 0 / 2 / 2 0 0 0 0 1 cos 1 (1 cos 2 ) ( ) 4 4 2 sin 2 8 16 8 y d d E P R R R R R ! " ! " ! " ! " ! " ! " ! " ! " # ! ! # ! ! "$ "$ # # ! ! "$ "$ # $ %= %= %=& %=& %= %= %=& %=& % % + % % = & = & = & % & = & ' ' . Therefore the force on the charged particle at the point P is given by 0 0 ˆ ( ) ( ) 8 q P q P R ! " F = E = # j ! !