electric field \(x-\) axis at a point \(=p\)
angle \(=\theta\) with \(x-\) axis
\(\tan a=\frac{1}{2} \tan \theta\)
\((\theta+\alpha)=\) the value of the position vector makes an angle \(\theta\)
\(\theta=60^{\circ}+\alpha\)
now resolving \(E\) into its components
\(E \cos \alpha=\frac{2 p \cos 60^{0}}{4 \pi \epsilon_{0} r^{3}} \quad \longrightarrow(1)\)
\(E \sin \alpha=\frac{p \sin 60^{0}}{4 \pi \epsilon_{0} r^{3}} \quad \longrightarrow(2)\)
Dividing \(2\) by \(1\)
\(\tan \alpha=\frac{1}{2} \tan \theta\)
\(\tan \alpha=\tan (\theta+\alpha)\)
\(\alpha=\theta+\alpha\)
[$g =9.8 \,m / s ^{2}$ આપેલા ]
$R ( R > > L )$ અંતરે વિદ્યુતક્ષેત્રની તીવ્રતાનું મૂલ્ય $.....$ પ્રમાણે બદલાશે.