Coulomb's Law

Question

Rod with length L has net charge -Q distributed along its length. The rod is placed with one end at the origin and the other at point (0,L) on the y-axis. Find the components of the electric field at the point (L,0) on the x-axis. 

 

It would help if the answer could include a diagram of sorts. Thank you.

 

Answer 1

It is hard to draw a diagram in this forum, but I will set up the math for you.

 

First, realize that E_x must be negative, since the negative charge along the y axis will attract a positive test charge at (L,0) to the -x direction.  Similarly, E_y must be positive, since a positive test charge at (L,0) will be attracted toward the +y direction.  I suggest finding the magnitudes of E_x and E_y, and then adding the correct signs to both as the last step.

 

Consider an infinitesimal source charge dq on the rod.  The linear charge density of the rod is

 

λ = -Q/L

 

Then

 

dq = λdy

 

The strength of the infinitesimal field at the field point (L,0) is

 

dE = kdq / (L² + y²) = kλdy / (L² + y²)

 

Let θ = angle between x axis and the line from the source point (0,y) to the field point (L,0).  Then

 

cosθ = L / (L² + y²)^1/2, sinθ = y / (L² + y²)^1/2

 

dE_x = dE cosθ = kλLdy / (L² + y²)^3/2

 

E_x = ∫dE_x = kλL ∫₀^L [dy / (L² + y²)^3/2]

 

Evaluate the integral (or find it in tables), and substitute λ = -Q/L.  Remember E_x must be negative, so if the integral comes out positive, you need to put in the negative sign yourself to properly align the direction of E in this problem.

 

Similarly,

 

dE_y = dE sinθ = kλydy / (L² + y²)^3/2

 

E_y = ∫dE_y = kλ ∫₀^L [ydy / (L² + y²)^3/2]

 

Recall E_y must be positive, so if evaluation of the integral comes out negative, you need to take the absolute value to properly align E in this problem.

 

That covers the physics of the problem.  The rest is to just do the math.  You should be able to finish from here.  If you need to see a diagram, send an email to my WyzAnt page, and I will reply with a diagram.