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Module 2 : Electrostatics

Lecture 8 : Coulomb's Law

where the upper limit of integration $\theta_0$ is given by

$\begin{displaymath}\tan\theta_0 =\frac{(L/2)}{x}\end{displaymath}$

or equivalently,

$\begin{displaymath}\sin\theta_0 = \frac{(L/2)}{(\frac{L^2}{4}+ x^2)^{1/2}}\end{displaymath}$

The field due to the rod, therefore, is

$\begin{displaymath}E = \frac{1}{2\pi\epsilon_0}\frac{Q}{x}\frac{1}{(L^2+4x^2)^{1/2}}\end{displaymath}$

directed along OP.