- #1
GravityGirl
- 29
- 0
A rod of length 80 cm has a uniform linear charge density of 5 mC/m. Determine the Electric Field at a point P located at a perpendicular distance 57 cm along a line of symmetry of the rod
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
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