A rod carries a charge uniformly distributed along its length

A rod of length L carries a charge Q uniformly distributed along its length.ROD
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/x2    (3)

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

dE = kQdx/(Lx2)    (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 = aL+a[kQ/(Lx2)]dx

or

E = (kQ/L)*aL+adx/x2

E = (kQ/L)*[-1/x]aL+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/a2

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