Question

The direction of e at point p due to uniformly charged finite rod will be

Answer 1

A. At angle 30° from x-axis

B. 45° from x-axis

C. 60° from x-axis

D. one of these

The direction of e at point p is C. 60° from x-axis.

Explanation:

The angle suspended by the finite line charge at P=60°  

So the resultant electric field due to line charge will be at $\frac{60}{20}$⇒30°  

Since we can assume the charge concentrated at the centre of finite line charge.

$E_{x}=\frac{k \lambda}{d}(\sin 60+\sin 0)$

$=\frac{14}{d} \sin 60=\frac{k \lambda}{d} \sqrt{3} / 2$

$E_{y}=\frac{f_{r}}{d}\left[\cos {0^{\circ} -\cos 60^{\circ})=\frac{k \lambda}{d}(1-1 / 2)$

$=\frac{K \lambda}{2 d}$

So the direction of e at point p=60°.

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