A rod of length L carries a charge Q uniformly distributed along its length.

Find the electrical field at point P on the axis of the rod, a distance a away from the end of the rod.

###### Solution.

The linear charge density λ is the quantity of charge per unit length, so

λ = Q/L (1)

We can imagine that the rod divided into infinitesimal (extremely small) elements dq.

dq = λdx = Qdx/L (2)

Each element dq makes an infinitesimal contribution to the electric field at the point P.

dE = kdq/x^{2} (3)

We substitute equation (2) into equation (3) and we get

dE = kQdx/(Lx^{2}) (4)

Now we can add up contributions to the electric field from all locations of dq along the rod between a and L+a by taking a definite integral.

E = _{a}∫^{L+a}[kQ/(Lx^{2})]dx

or

E = (kQ/L)*_{a}∫^{L+a}dx/x^{2}

E = (kQ/L)*[-1/x]_{a}^{L+a}

E = (kQ/L)*[-1/(L+a)+1/a]

and finally,

E = kQ/(a(L+a))

It is interesting that the rod can become a point charge

If L → 0 or a>>L then E = kQ/a^{2}