Question

If the total charge on a rod of length 0.4 m is 2.6 nc, what is the magnitude of the electric field at a location 3 cm from the midpoint of the rod?

Answer 1

Let the rod be on the x-axes with endpoints -L/2 and L/2 and uniform charge density lambda (2.6nC/0.4m = 7.25 nC/m). 

The point then lies on the y-axes at d = 0.03 m. 

from symmetry, the field at that point will be ascending along the y-axes. 


A charge element at position x on the rod has distance sqrt(x^2 + d^2) to the point. 

Also, from the geometry, the component in the y-direction is d/sqrt(x^2+d^2) times the field strength. 


All in all, the infinitesimal field strength from the charge between x and x+dx is: 


dE = k lambda dx * 1/(x^2+d^2) * d/sqrt(x^2+d^2) 


Therefore, upon integration, 


E = k lambda d INTEGRAL{dx / (x^2 + d^2)^(3/2) } where x goes from -L/2 to L/2. 


This gives:


E = k lambda L / (d sqrt((L/2)^2 + d^2) ) 


But lambda L = Q, the total charge on the rod, so 


E = k Q / ( d * sqrt((L/2)^2 + d^2) )