A thin rod 39.2 cm long is charged uniformly with a positive charge density of 72.0 mC/m. The rod is placed along the y-axis and is centered at the origin. A charge of +46.5 mC is placed 68.4 cm from the midpoint of the rod on the positive x-axis. Calculate the electric field at a point on the x-axis, which is halfway between the point charge and the center of the rod. (Express your answer in terms of the unit vector x. For example, if the electric field is -10.5x N/C, then input -10.5 N/C.)(adsbygoogle = window.adsbygoogle || []).push({});

ok..so i drew a picture and i see that the E field of te line charge will be in the x direction becuase that i perpendicular to the line

i also see that the point charge will be in the x direction as well becuase it is located on the x-axis

so since both charges are postive and the point where i want to know the E field at is between them i am going to subtract their individual E fields at that point to get the net

E=Eline-Epoint

Eline=2K(lamda)/r where r= half the distance betwen the line and point charge

Epoint=kq/r^2

so i have E=(2k(lamda)/r)-(kq/r^2)

does this look right?....if not, please tell me what i am doing wrong

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# Electric field of a line charge and point charge

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