- #1

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i don't know what i wrong..but here is what i am doing

linecharge(change in lenght)= change in Q

E=Kdq/r^2

E=k*linecharge(dL)/r^2

i drew a free body diagram and the x components cancle out so i only have to worry about y

Ey=k*linecharge(dL)(costhea)/r^2 *costheta=height in y direction/r

Ey=k*linecharge(dL)(h/r)/r^2

so now i am going to integrate only half of the rod becuase of symetry

so i have the integral from 0 to L/2 of k*linecharge(dL)(h/r)/r^2

i simplify and i have Ey=k*linecharge(h) integral from 0 to L/2 of 1/r^3

i simplify again to get =k*linecharge(L/2)/h(h^2+(L/2)^2)^1/2

i believe this is the correct formula...but when i put the numbers in i get the wrong answer

please help