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Question

Electric field at $\mathrm{p}=2 \mathrm{E}_{1} \cos 	heta_{1}-2 \mathrm{E}_{2} \cos 	heta_{2}$

$=\frac{2 \mathrm{Kq}}{\left(\mathrm{d}^{2}+\mathr

Vedclass · Accepted Answer

$E \propto \frac{1}{D^4}$

Vedclass · Answer

Electric field at $\mathrm{p}=2 \mathrm{E}_{1} \cos 	heta_{1}-2 \mathrm{E}_{2} \cos 	heta_{2}$

$=\frac{2 \mathrm{Kq}}{\left(\mathrm{d}^{2}+\mathrm{D}^{2}ight)} 	imes \frac{\mathrm{D}}{\left(\mathrm{d}^{2}+\mathrm{D}^{2}ight)^{1 / 2}}-\frac{2 \mathrm{Kq}}{\left[(2 \mathrm{d})^{2}+\mathrm{D}^{2}ight]} 	imes \frac{\mathrm{D}}{\left[(2 \mathrm{d})^{2}+\mathrm{D}^{2}ight]^{1 / 2}}$

$=2 \mathrm{KqD}\left[\left(\mathrm{d}^{2}+\mathrm{D}^{2}ight)^{-3 / 2}-\left(4 \mathrm{d}^{2}+\mathrm{D}^{2}ight)^{-3 / 2}ight]$

$=\frac{2 \mathrm{KqD}}{\mathrm{D}^{3}}\left[\left(1+\frac{\mathrm{d}^{2}}{\mathrm{D}^{2}}ight)^{-3 / 2}-\left(1+\frac{4 \mathrm{d}^{2}}{\mathrm{D}^{2}}ight)^{-3 / 2}ight]$

Applying binomial approximation $\because \mathrm{d}<<\mathrm{D}$

$=\frac{2 \mathrm{KqD}}{\mathrm{D}^{3}}\left[1-\frac{3}{2} \frac{\mathrm{d}^{2}}{\mathrm{D}^{2}}-\left(1-\frac{3 	imes 4 \mathrm{d}^{2}}{2 \mathrm{D}^{2}}ight)ight]$

$=\frac{2 \mathrm{KqD}}{\mathrm{D}^{3}}\left[\frac{12}{2} \frac{\mathrm{d}^{2}}{\mathrm{D}^{2}}-\frac{3}{2} \frac{\mathrm{d}^{2}}{\mathrm{D}^{2}}ight]$

$=\frac{9 \mathrm{kqd}^{2}}{\mathrm{D}^{4}}$

Similar Questions

दो बिन्दु आवेश $Q$ व $ - 3Q$ एक-दूसरे से कुछ दूरी पर रखे हैं। यदि $Q$ स्थिति पर विद्युत क्षेत्र $E$ हो तो स्थिति $ - 3Q$ पर यह होगा