Question

Find the electric field at point P (as shown in figure) on the perpendicular bisector of a uniformly charged thin wire of length L carrying a charge Q. The distance of the point P from the centre of the rod is \(a = \frac{\sqrt{3}}{2}L\).

• A. \(\frac{Q}{4\pi\varepsilon_0 L^2}\)
• B. \(\frac{Q}{3\pi\varepsilon_0 L^2}\)
• C. \(\frac{Q}{2\sqrt{3}\,\pi\varepsilon_0 L^2}\)
• D. \(\frac{\sqrt{3}\,Q}{4\pi\varepsilon_0 L^2}\)

Hint:

Integrate the field of line elements along the rod; on the perpendicular bisector \(E = \frac{kQ}{a\sqrt{a^2+L^2/4}}\).

Answer 1

The Correct Option is C

Let \(\lambda = Q/L\). Take an element \(dx\) at distance x from the centre; its field component along the bisector is \(dE = \frac{k\lambda\,dx}{a^2+x^2}\cdot\frac{a}{\sqrt{a^2+x^2}}\) (components along the rod cancel by symmetry).
\(E = 2k\lambda a\int_0^{L/2}\frac{dx}{(a^2+x^2)^{3/2}} = 2k\lambda a\cdot\frac{1}{a^2}\cdot\frac{L/2}{\sqrt{a^2+L^2/4}} = \frac{kQ}{a\sqrt{a^2+L^2/4}}\).
With \(a = \frac{\sqrt{3}}{2}L\): \(\sqrt{a^2+L^2/4} = \sqrt{\frac{3L^2}{4}+\frac{L^2}{4}} = L\).
\(E = \frac{kQ}{\frac{\sqrt{3}}{2}L\cdot L} = \frac{2}{\sqrt{3}}\cdot\frac{Q}{4\pi\varepsilon_0 L^2} = \frac{Q}{2\sqrt{3}\,\pi\varepsilon_0 L^2}\).
Answer: (3) \(\frac{Q}{2\sqrt{3}\,\pi\varepsilon_0 L^2}\)