A thin rod lies on the \(x\) -axis between \(x=0\) and \(x=L\) and carries total charge \(Q\) distributed uniformly over its length. Show that the electric field strength for \(x>L\) is given by \(E=k Q /[x(x-L)]\)

Short Answer

Expert verified
Therefore, for \(x > L\), the electric field strength is indeed given by \(E = k Q /[x (x - L)]\), where \(k\) is Coulomb's constant.

Step by step solution

01

Define the charge distribution

First, describe the charge on the rod as a continuous distribution. Define the charge density \(\lambda = Q/L\), where \(Q\) is the total charge and \(L\) is the length of the rod.
02

Consider an infinitesimal element

Consider an infinitesimal element dx at a distance x from the origin (0 on the x-axis). The charge on this infinitesimal element can be given as \(dq = \lambda dx = (Q/L) dx\).
03

Find the contribution to the electric field

The contribution dE to the electric field at a point on the x-axis beyond the rod (\(x > L\)) due to this charge is given by Coulomb's Law expressed in differential form, \(dE = k dq / r^2\), where \(r = x - x'\), \(x'\) being the variable of integration indicating position of the infinitesimal charge along the rod.
04

Integrate to find the total field

Integrate this from 0 to \(L\) to get the total electric field, \(E = \int_0^L k (Q/L) dx / (x - x')^2\). The integral, when evaluated yields \(E = k Q /[x (x - L)]\), confirming the given expression.

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