Question 397: Electrostatics & Capacitance

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 = $${{\sqrt 3 } \over 2}L$$.

Accepted Answer

$$	an 	heta  = {{L/2} \over {{{\sqrt 3 } \over 2}L}} \Rightarrow {1 \over {\sqrt 3 }}$$
$$	heta  = 30^\circ $$
$${E_{net}} = {{K\lambda } \over {{{\sqrt 3 } \over 2}L}}(\sin 30^\circ  + \sin 30^\circ ) = {{2KQ} \over {\sqrt 3 {L^2}}}\left( {{1 \over 2} + {1 \over 2}} ight)$$
$${E_{net}} = {1 \over {4\pi {\varepsilon _0}}}{{2Q} \over {\sqrt 3 {L^2}}}$$
$${E_{net}} = {Q \over {2\sqrt 3 \pi {\varepsilon _0}{L^2}}}$$

Answer

$${Q \over {4\pi {\varepsilon _0}{L^2}}}$$

Answer

$${Q \over {3\pi {\varepsilon _0}{L^2}}}$$

Answer

$${Q \over {2\sqrt 3 \pi {\varepsilon _0}{L^2}}}$$

Answer

$${{\sqrt 3 Q} \over {4\pi {\varepsilon _0}{L^2}}}$$

Electrostatics & Capacitance

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 = ${{\sqrt 3 } \over 2}L$.

Choose the correct answer: