Question

Find the electric field at a point on the perpendicular bisector of unifromly charged rod, The length of the rod is L. the charge on it is Q and the distance of P from the centre of the rod is a,

• A. 0
• B. $\frac{Q}{4\pi {\epsilon }_{0}a\sqrt{L^2+4a^2}}$
• C. $\frac{Q}{2\pi {\epsilon }_{0}a\sqrt{L^2+4a^2}}$
• D. $\frac{Q}{2\pi {\epsilon }_{0}a\sqrt{L^2+a^2}}$

Answer 1

The correct option is C $\frac{Q}{2\pi {\epsilon }_{0}a\sqrt{L^2+4a^2}}$

Let $\lambda$ is the charge density

say $\lambda =\frac{Q}{L}$

So,

${dE}_P=dE\cos \theta$

${dE}_P=\frac{Kdq}{(y^2+x^2)}\cos \theta$

$E_P={\int }_{-\frac{L}{2}}^{\frac{L}{2}}\frac{KQdx}{L(y^2+x^2)}.\frac{y}{\sqrt{y^2+x^2}}=\frac{K\lambda y}{L}{\int }_{-\frac{L}{2}}^{\frac{L}{2}}\frac{dx}{{(y^2+x^2)}^{3/2}}$

$=K\lambda y{\left[\frac{x}{y^2\sqrt{y^2+x^2}}\right]}_{-\frac{L}{2}}^{\frac{L}{2}}$

$E_P=\frac{2KQ}{a\sqrt{4a^2+L^2}}$