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1. Electric Charges and Fields
normal
The direction $(\theta ) $ of $\vec E$ at point $P$ due to uniformly charged finite rod will be

A
at angle $30°$ from $x-$axis
B
$45°$ from $x$ $-$ axis
C
$60°$ from $x$-axis
D
none of these
Solution

The angle suspended by the finite line charge at $P=60^{0}$
So the resultant electric field due to line charge will be at $60 / 2 \Rightarrow 30^{\circ}$ since we can
assume the charge concentrated at the centre of finite line charge.
Standard 12
Physics