# Find the electric field at point P (as shown in figure) on the perpendicular

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 $a=\frac{\sqrt{3}}{2} L$.

1. $\frac{\sqrt{3} Q}{4 \pi \varepsilon_{0} L^{2}}$

2. $\frac{\mathrm{Q}}{3 \pi \varepsilon_{0} \mathrm{~L}^{2}}$

3. $\frac{Q}{2 \sqrt{3} \pi \varepsilon_{0} L^{2}}$

4. $\frac{Q}{4 \pi \varepsilon_{0} L^{2}}$

Correct Option: , 3

Solution:

$\mathrm{E}=\frac{\mathrm{k} \lambda}{\mathrm{a}}\left(\sin \theta_{1}+\sin \theta_{2}\right)$

$\mathrm{E}=\frac{1}{4 \pi \varepsilon_{0}} \times \frac{\mathrm{Q}}{\mathrm{L}} \times \frac{1}{\left(\frac{\sqrt{3} \mathrm{~L}}{2}\right)} \times(2 \sin \theta)$

$\tan \theta=\frac{\mathrm{L} / 2}{\frac{\sqrt{3} \mathrm{~L}}{2}}=\frac{1}{\sqrt{3}}$

$\sin \theta=\frac{1}{2}$

$E=\frac{1}{4 \pi \varepsilon_{0}} \times \frac{2 Q}{\sqrt{3} \mathrm{~L}^{2}} \times\left(2 \times \frac{1}{2}\right)$

$\mathrm{E}=\frac{\mathrm{Q}}{2 \sqrt{3} \pi \varepsilon_{0} \mathrm{~L}^{2}}$