Question

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In this question we will consider the electric field of a charged rod of length \(L\) at a point \(P\) located a distance \(b\) from the center of the rod along its perpendicular bisector, as illustrated in Figure \(1 .\)

It can be shown that the magnitude of the electric field at \(P(0, b)\) is given by the following integral

$$ E(b)=\int_{-\frac{L}{2}}^{\frac{L}{2}} \frac{\lambda b}{4 \pi \varepsilon_{0}\left(x^{2}+b^{2}\right)^{3 / 2}} d x $$

where \(\lambda\) is the charge density per unit length on the rod and \(\varepsilon_{0}\) is called the permittivity of free space (it is a universal constant with the value \(8.854 \times 10^{-12} \mathrm{~F} / \mathrm{m}\) (farads per metre)).

Answer 1