- Home
- Standard 12
- Physics
चार बिन्दु आवेशों $- q ,+ q ,+ q$ और $- q$ को $y$-अक्ष पर, क्रमश: $y =-2 d , y =- d , y =+ d$ तथा $y =+2 d$ पर रखा गया है। $x$-अक्ष पर उपस्थित एक बिन्दु $x = D$, जहाँ $D \gg d$ है, पर विधुत क्षेत्र के परिमाण $E$ का व्यवहार होगा?
$E \propto \frac{1}{D}$
$E \propto \frac{1}{D^3}$
$E \propto \frac{1}{D^2}$
$E \propto \frac{1}{D^4}$
Solution
Electric field at $\mathrm{p}=2 \mathrm{E}_{1} \cos \theta_{1}-2 \mathrm{E}_{2} \cos \theta_{2}$
$=\frac{2 \mathrm{Kq}}{\left(\mathrm{d}^{2}+\mathrm{D}^{2}\right)} \times \frac{\mathrm{D}}{\left(\mathrm{d}^{2}+\mathrm{D}^{2}\right)^{1 / 2}}-\frac{2 \mathrm{Kq}}{\left[(2 \mathrm{d})^{2}+\mathrm{D}^{2}\right]} \times \frac{\mathrm{D}}{\left[(2 \mathrm{d})^{2}+\mathrm{D}^{2}\right]^{1 / 2}}$
$=2 \mathrm{KqD}\left[\left(\mathrm{d}^{2}+\mathrm{D}^{2}\right)^{-3 / 2}-\left(4 \mathrm{d}^{2}+\mathrm{D}^{2}\right)^{-3 / 2}\right]$
$=\frac{2 \mathrm{KqD}}{\mathrm{D}^{3}}\left[\left(1+\frac{\mathrm{d}^{2}}{\mathrm{D}^{2}}\right)^{-3 / 2}-\left(1+\frac{4 \mathrm{d}^{2}}{\mathrm{D}^{2}}\right)^{-3 / 2}\right]$
Applying binomial approximation $\because \mathrm{d}<<\mathrm{D}$
$=\frac{2 \mathrm{KqD}}{\mathrm{D}^{3}}\left[1-\frac{3}{2} \frac{\mathrm{d}^{2}}{\mathrm{D}^{2}}-\left(1-\frac{3 \times 4 \mathrm{d}^{2}}{2 \mathrm{D}^{2}}\right)\right]$
$=\frac{2 \mathrm{KqD}}{\mathrm{D}^{3}}\left[\frac{12}{2} \frac{\mathrm{d}^{2}}{\mathrm{D}^{2}}-\frac{3}{2} \frac{\mathrm{d}^{2}}{\mathrm{D}^{2}}\right]$
$=\frac{9 \mathrm{kqd}^{2}}{\mathrm{D}^{4}}$
