Module 2 : Electrostatics
Lecture 8 : Coulomb's Law
 

By symmetry, corresponding to the element $dy$ at A, there is an element $dy$ at B at a depth $y$ which produces a field of the same magnitude at P directed along $\vec{BP}$. When we add fields due to such symmetric pairs, the component parallel to the rod cancel and the net non-zero component is along $\vec{OP}$. The net field at P is, therefore,

\begin{eqnarray*} E &=& \int dE\cos\theta\\ &=& \frac{2}{4\pi\epsilon_0}\frac{Q}{L} \int_0^{L/2}\frac{\cos\theta} {(x^2+y^2)}dy \end{eqnarray*}

where, we have used $\cos\theta = x/(x^2+y^2)^{1/2}$. To evaluate the integral, note that $ y = x\tan\theta$, so that $x^2+y^2 = x^2\sec^2\theta$and $dy = x\sec^2\theta$. Here, $x$ is constant as the position P is fixed. Thus

\begin{displaymath} E = \frac{1}{2\pi\epsilon_0}\frac{Q}{L}\int_0^{\theta_0}\fra... ...= \frac{1}{2\pi\epsilon_0} \left[\sin\theta\right]_0^{\theta_0}\end{displaymath}

 

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